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|---|---|---|---|---|---|---|---|---|---|---|
Kd4je4zs0Nem9nzKX70N6
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
Let S be the set of all real values of k for which the systemof linear equations
<br/>x + y + z = 2
<br/>2x + y $$-$$ z = 3
<br/>3x + 2y + kz = 4
<br/>has a unique solution. Then S is :
|
[{"identifier": "A", "content": "an empty set "}, {"identifier": "B", "content": "equal to {0}"}, {"identifier": "C", "content": "equal to <b>R</b>"}, {"identifier": "D", "content": "equal to <b>R</b> $$-$$ {0}"}]
|
["D"]
| null |
As system of linear equations have unique solutions so, determinant of coefficient $$ \ne $$ 0<br><br>
$$ \therefore $$ $$\left| {\matrix{
1 & 1 & 1 \cr
2 & 1 & { - 1} \cr
3 & 2 & k \cr
} } \right|$$ $$ \ne $$ 0<br><br>
$$ \Rightarrow $$ k + 2 - (2k + 3) + 1 $$ \ne $$ 0<br><br>
$$ \Rightarrow $$ k $$ \ne $$ 0<br><br>
$$ \therefore $$ k $$ \in $$ R - {0}
|
mcq
|
jee-main-2018-online-15th-april-morning-slot
|
cxX8lOt6W4Ubbzk1YTKQz
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
If the system of linear equations
<br/>x + ay + z = 3
<br/>x + 2y + 2z = 6
<br/>x + 5y + 3z = b
<br/>has no solution, then :
|
[{"identifier": "A", "content": "a = $$-$$ 1, b = 9"}, {"identifier": "B", "content": "a = $$-$$ 1, b $$ \\ne $$ 9"}, {"identifier": "C", "content": "a $$ \\ne $$ $$-$$ 1, b = 9"}, {"identifier": "D", "content": "a = 1, b $$ \\ne $$ 9"}]
|
["B"]
| null |
As the given system of equations has no solution then
<br><br> $$\Delta $$ = 0 and at least one of $$\Delta $$<sub>1</sub>, $$\Delta $$<sub>2</sub> and $$\Delta $$<sub>2</sub> should not be zero.
<br><br>$$ \therefore $$ $$\Delta $$ = $$\left| {\matrix{
1 & a & 1 \cr
1 & 2 & 2 \cr
1 & 5 & 3 \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow $$ - $$a$$ - 1 = 0
<br><br>$$ \Rightarrow $$ a = - 1
<br><br>$$\Delta $$<sub>2</sub> = $$\left| {\matrix{
1 & 3 & 1 \cr
1 & 6 & 2 \cr
1 & b & 3 \cr
} } \right| \ne 0$$
<br><br>$$ \Rightarrow $$ b $$ \ne $$ 0
|
mcq
|
jee-main-2018-online-15th-april-evening-slot
|
MtVekwWI8fkPYsMPN0LRs
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
The number of values of k for which the system of linear equations,
<br/>(k + 2)x + 10y = k
<br/>kx + (k +3)y = k -1
<br/>has <b>no solution,</b> is :
|
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "infinitely many"}]
|
["A"]
| null |
System of linear equation have no solution,
<br><br>$$\therefore\,\,\,$$ determinant of coefficient = 0
<br><br>$$\left| {\matrix{
{k + 2} & {10} \cr
k & {k + 3} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow $$ $$\,\,\,\,$$ (k + 2) (k + 3) $$-$$ 10 K = 0
<br><br>$$ \Rightarrow $$ $$\,\,\,\,$$ k<sup>2</sup> $$-$$ 5k + 6 = 0
<br><br>$$\therefore\,\,\,\,$$ k = 2, 3
<br><br>When, k = 2 then equations become,
<br><br>4x + 10y = 2
<br><br>and 2x + 5y = 1
<br><br>It has in finite number of solutions.
<br><br>When k = 3, equations becomes
<br><br>5x + 10y = 3
<br><br>3x + 6y = 2
<br><br>Those equation has no solutions.
<br><br>$$\therefore\,\,\,\,$$ When k = 3, then system of equations have no solutions.
|
mcq
|
jee-main-2018-online-16th-april-morning-slot
|
1wm32XZzhcGEpKn931wYK
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
An ordered pair ($$\alpha $$, $$\beta $$) for which the system of linear equations
<br/>(1 + $$\alpha $$) x + $$\beta $$y + z = 2
<br/>$$\alpha $$x + (1 + $$\beta $$)y + z = 3
<br/>$$\alpha $$x + $$\beta $$y + 2z = 2
<br/>has a unique solution, is :
|
[{"identifier": "A", "content": "(\u20133, 1) "}, {"identifier": "B", "content": "(1, \u20133) "}, {"identifier": "C", "content": "(\u20134, 2) "}, {"identifier": "D", "content": "(2, 4) "}]
|
["D"]
| null |
For unique solution
<br><br>$$\Delta $$ $$ \ne $$ 0 $$ \Rightarrow $$ $$\left| {\matrix{
{1 + \alpha } & \beta & 1 \cr
\alpha & {1 + \beta } & 1 \cr
\alpha & \beta & 2 \cr
} } \right| \ne 0$$
<br><br>$$\left| {\matrix{
1 & { - 1} & 0 \cr
0 & 1 & { - 1} \cr
\alpha & \beta & 2 \cr
} } \right| \ne 0 \Rightarrow \alpha + \beta \ne - 2$$
|
mcq
|
jee-main-2019-online-12th-january-morning-slot
|
JZwkUYQ4zq6QEqebXY3rsa0w2w9jx236oa3
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
Let $$\lambda $$ be a real number for which the system of linear equations x + y + z = 6, 4x + $$\lambda $$y β $$\lambda $$z = $$\lambda $$ β 2,
3x + 2y β 4z = β 5 has infinitely many solutions. Then $$\lambda $$ is a root of the quadratic equation:
|
[{"identifier": "A", "content": "$$\\lambda $$<sup>2</sup> + $$\\lambda $$ - 6 = 0"}, {"identifier": "B", "content": "$$\\lambda $$<sup>2</sup> - $$\\lambda $$ - 6 = 0"}, {"identifier": "C", "content": "$$\\lambda $$<sup>2</sup> - 3$$\\lambda $$ - 4 = 0"}, {"identifier": "D", "content": "$$\\lambda $$<sup>2</sup> + 3$$\\lambda $$ - 4 = 0"}]
|
["B"]
| null |
$$\Delta = 0$$<br><br>
$$\left| {\matrix{
1 & 1 & 1 \cr
4 & \lambda & { - \lambda } \cr
3 & 2 & { - 4} \cr
} } \right| = 0$$<br><br>
On solving we get $$\lambda $$ = 3
|
mcq
|
jee-main-2019-online-10th-april-evening-slot
|
yYEy0R19J8dgQnc40R3rsa0w2w9jwy0m9v1
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
If the system of linear equations
<br/>x + y + z = 5
<br/>x + 2y + 2z = 6
<br/>x + 3y + $$\lambda $$z = $$\mu $$, ($$\lambda $$, $$\mu $$ $$ \in $$ R), has infinitely many solutions, then the value of $$\lambda $$ + $$\mu $$ is :
|
[{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "12"}]
|
["A"]
| null |
x + y + z = 5<br><br>
x + 2y + 2z = 6<br><br>
x + 3y + $$\lambda $$z = $$\mu $$ have infinite solution<br><br>
$$\Delta $$ = 0, $$\Delta $$x = $$\Delta $$y = $$\Delta $$z = 0<br><br>
$$\Delta = \left| {\matrix{
1 & 1 & 1 \cr
1 & 2 & 2 \cr
1 & 3 & \lambda \cr
} } \right| = 0$$<br><br>
$$ \Rightarrow 1(2\lambda - 6) - 1(\lambda - 2) + 1(3 - 2) = 0$$<br><br>
$$ \Rightarrow 2\lambda - 6 - \lambda + 2 + 1 = 0$$<br><br>
$$ \Rightarrow \lambda = 3$$<br><br>
Now, $$\Delta x = \left| {\matrix{
5 & 1 & 1 \cr
6 & 2 & 2 \cr
\mu & 3 & 3 \cr
} } \right| = 0$$,<br><br> $$\Delta $$y = $$\left| {\matrix{
1 & 5 & 1 \cr
1 & 6 & 2 \cr
1 & \mu & 3 \cr
} } \right| = 0$$<br><br>
$$ \Rightarrow \left| {\matrix{
1 & 5 & 1 \cr
0 & 1 & 1 \cr
0 & {\mu - 5} & 2 \cr
} } \right| = 0$$<br><br>
$$ \Rightarrow \mu = 7$$<br><br>
$$\Delta z = \left| {\matrix{
1 & 1 & 5 \cr
1 & 2 & 6 \cr
1 & 3 & \mu \cr
} } \right| = \left| {\matrix{
1 & 1 & 5 \cr
0 & { - 1} & { - 1} \cr
0 & 2 & {\mu - 5} \cr
} } \right|$$<br><br>
$$ \Rightarrow 1(5 - \mu + 2) = 0$$<br><br>
$$ \Rightarrow \mu = 7$$<br><br>
So, $$\lambda + \mu = 10$$
|
mcq
|
jee-main-2019-online-10th-april-morning-slot
|
HMsP8Zu83OwHy7SXri18hoxe66ijvwu7iyt
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
If the system of equations 2x + 3y β z = 0, x + ky
β 2z = 0 and 2x β y + z = 0 has a non-trival solution
(x, y, z), then $${x \over y} + {y \over z} + {z \over x} + k$$
is equal to :-
|
[{"identifier": "A", "content": "-4"}, {"identifier": "B", "content": "$${3 \\over 4}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$$-{1 \\over 4}$$"}]
|
["C"]
| null |
Given 2x + 3y β z = 0,
<br><br>x + ky β 2z = 0
<br><br>2x β y + z = 0
<br><br>For non trivial solution
<br><br>$$\Delta = 0 \Rightarrow \left| {\matrix{
2 & 3 & { - 1} \cr
1 & k & { - 2} \cr
2 & { - 1} & 1 \cr
} } \right| = 0$$<br><br>
$$ \Rightarrow k = {9 \over 2}$$<br><br>
$$ \therefore $$ Equations are 2x + 3y β z = 0 ...(i)
<br><br>
2x β y + z = 0 ...(ii)<br><br>
2x + 9y β 4z = 0 ...(iii)<br><br>
By (i) β (ii) we get,
<br><br>4y - 2z = 0
<br><br>$$ \Rightarrow $$ 2y = z .......(iv)
<br><br>$$ \Rightarrow $$ $${y \over z} = {1 \over 2}$$
<br><br>From equation (i) and (iv)
<br><br>2x + 3y - 2y = 0
<br><br>$$ \Rightarrow $$ 2x + y = 0
<br><br>$$ \Rightarrow $$ $${x \over y} = - {1 \over 2}$$
<br><br>$$ \Rightarrow $$ $${x \over y} \times {y \over z} = - {1 \over 2} \times {1 \over 2} = - {1 \over 4}$$
<br><br>$$ \Rightarrow $$ $${z \over x} = - 4$$
<br><br>$$ \therefore $$ $${x \over y} + {y \over z} + {z \over x} + k = {{ - 1} \over 2} + {1 \over 2} - 4 + {9 \over 2}$$ = $${1 \over 2}$$<br><br>
|
mcq
|
jee-main-2019-online-9th-april-evening-slot
|
V1GwSMBgU8TtzC3pJZMkz
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
The greatest value of c $$ \in $$ R for which the system
of linear equations<br/>
x β cy β cz = 0<br/>
cx β y + cz = 0<br/>
cx + cy β z = 0<br/>
has a non-trivial solution, is :
|
[{"identifier": "A", "content": "-1"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "1/2"}, {"identifier": "D", "content": "2"}]
|
["C"]
| null |
If the system of equations has non-trivial
solutions, then
<br><br>D = 0
<br><br>$$\left| {\matrix{
1 & { - c} & { - c} \cr
c & { - 1} & c \cr
c & c & { - 1} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow $$ (1 - c<sup>2</sup>) + c(-c - c<sup>2</sup>) - c(c<sup>2</sup> + c) = 0
<br><br>$$ \Rightarrow $$ (1 + c)(1 - c - 2c<sup>2</sup>) = 0
<br><br>$$ \Rightarrow $$ (1 + c)<sup>2</sup> (1 - 2c) = 0
<br><br>$$ \Rightarrow $$ c = -1 or $${1 \over 2}$$
<br><br>$$ \therefore $$ Greatest value of c is $${1 \over 2}$$.
|
mcq
|
jee-main-2019-online-8th-april-morning-slot
|
z0Ep8YB1pu7YM0a4MG1hU
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
The set of all values of $$\lambda $$ for which the system of linear equations
<br/>x β 2y β 2z = $$\lambda $$x
<br/>x + 2y + z = $$\lambda $$y
<br/>β x β y = $$\lambda $$z
<br/>has a non-trivial solutions :
|
[{"identifier": "A", "content": "is an empty set"}, {"identifier": "B", "content": "contains more than two elements"}, {"identifier": "C", "content": "is a singleton"}, {"identifier": "D", "content": "contains exactly two elements\n"}]
|
["C"]
| null |
$$\left| {\matrix{
{\lambda - 1} & 2 & 2 \cr
1 & {2 - \lambda } & 1 \cr
1 & 1 & 1 \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow {\left( {\lambda - 1} \right)^3} = 0 \Rightarrow \lambda = 1$$
|
mcq
|
jee-main-2019-online-12th-january-evening-slot
|
SQdlZGDAX2v9SsT2dj0Rf
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
If the system of linear equations
<br/>2x + 2y + 3z = a
<br/>3x β y + 5z = b
<br/>x β 3y + 2z = c
<br/>where a, b, c are non zero real numbers, has more one solution, then :
|
[{"identifier": "A", "content": "b \u2013 c \u2013 a = 0"}, {"identifier": "B", "content": "a + b + c = 0"}, {"identifier": "C", "content": "b \u2013 c + a = 0"}, {"identifier": "D", "content": "b + c \u2013 a = 0"}]
|
["A"]
| null |
P<sub>1</sub> : 2x + 2y + 3z = a
<br><br>P<sub>2</sub> : 3x $$-$$ y + 5z = b
<br><br>P<sub>3</sub> : x $$-$$ 3y + 2z = c
<br><br>We find
<br><br>P<sub>1</sub> + P<sub>3</sub> = P<sub>2</sub> $$ \Rightarrow $$ a + c = b
|
mcq
|
jee-main-2019-online-11th-january-morning-slot
|
MZzFH39aOy3rUWdoZ77Jw
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
The number of values of $$\theta $$ $$ \in $$ (0, $$\pi $$) for which the system of linear equations
<br/><br/>x + 3y + 7z = 0
<br/><br/>$$-$$ x + 4y + 7z = 0
<br/><br/>(sin3$$\theta $$)x + (cos2$$\theta $$)y + 2z = 0.
<br/><br/>has a non-trival solution, is -
|
[{"identifier": "A", "content": "two"}, {"identifier": "B", "content": "one"}, {"identifier": "C", "content": "four"}, {"identifier": "D", "content": "three"}]
|
["A"]
| null |
$$\left| {\matrix{
1 & 3 & 7 \cr
{ - 1} & 4 & 7 \cr
{\sin 3\theta } & {\cos 2\theta } & 2 \cr
} } \right| = 0$$
<br><br>(8 $$-$$ 7 cos 2$$\theta $$) $$-$$ 3($$-$$2 $$-$$ 7 sin 3$$\theta $$)
<br><br> +7 ($$-$$ cos 2$$\theta $$ $$-$$ 4 sin 3$$\theta $$) = 0
<br><br>14 $$-$$ 7 cos 2$$\theta $$ + 21 sin 3$$\theta $$ $$-$$ 7 cos 2$$\theta $$
<br><br> $$-$$ 28 sin 3$$\theta $$ = 0
<br><br>14 $$-$$ 7 sin 3$$\theta $$ $$-$$ 14 cos 2$$\theta $$ = 0
<br><br>14 $$-$$ 7 (3 sin $$\theta $$ $$-$$ 4 sin<sup>3</sup>$$\theta $$ ) $$-$$ 14 (1 $$-$$ 2 sin<sup>2</sup> $$\theta $$) = 0
<br><br>$$-$$ 21 sin $$\theta $$ + 28 sin<sup>3</sup> $$\theta $$ + 28 sin<sup>2</sup> $$\theta $$ = 0
<br><br>7 sin $$\theta $$ [$$-$$ 3 + 4 sin<sup>2</sup> $$\theta $$ + 4 sin $$\theta $$] = 0 sin$$\theta $$,
<br><br>4 sin<sup>2</sup> $$\theta $$ + 6 sin $$\theta $$ $$-$$ 2 sin $$\theta $$ $$-$$ 3 = 0
<br><br>2 sin $$\theta $$(2 sin $$\theta $$ + 3) $$-$$ 1 (2 sin $$\theta $$ + 3) = 0
<br><br>sin $$\theta $$ = $${{ - 3} \over 2}$$; sin$$\theta $$ = $${1 \over 2}$$
<br><br>Hence, 2 solutions in (0, $$\pi $$)
|
mcq
|
jee-main-2019-online-10th-january-evening-slot
|
mpfwNPbYtgolDevb4dZRy
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
If the system of equations
<br/><br/>x + y + z = 5
<br/><br/>x + 2y + 3z = 9
<br/><br/>x + 3y + az = $$\beta $$
<br/><br/>has infinitely many solutions, then $$\beta $$ $$-$$ $$\alpha $$ equals -
|
[{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "21"}, {"identifier": "C", "content": "18"}, {"identifier": "D", "content": "5"}]
|
["A"]
| null |
$$D = \left| {\matrix{
1 & 1 & 1 \cr
1 & 2 & 3 \cr
1 & 3 & \alpha \cr
} } \right| = \left| {\matrix{
1 & 1 & 1 \cr
0 & 1 & 2 \cr
0 & 2 & {\alpha - 1} \cr
} } \right|$$
<br><br> $$ = \left( {\alpha - 1} \right) - 4 = \left( {\alpha - 5} \right)$$
<br><br>for infinite solutions $$D = 0 \Rightarrow \alpha = 5$$
<br><br>$${D_x} = 0 \Rightarrow \left| {\matrix{
5 & 1 & 1 \cr
9 & 2 & 3 \cr
\beta & 3 & 5 \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow \left| {\matrix{
0 & 0 & 1 \cr
{ - 1} & { - 1} & 3 \cr
{\beta - 15} & { - 2} & 5 \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow 2 + \beta - 15 = 0 \Rightarrow \beta - 13 = 0$$
<br><br>on $$\beta = 13$$ we get $${D_y} = {D_z} = 0$$
<br><br>$$\alpha = 5,\beta = 13$$
|
mcq
|
jee-main-2019-online-10th-january-morning-slot
|
HeOmh2hQUK6GpkFIn8YAj
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
If the system of linear equations
<br/>x $$-$$ 4y + 7z = g
<br/>Β Β Β Β Β Β Β 3y $$-$$ 5z = h
<br/>$$-$$2x + 5y $$-$$ 9z = k
<br/>is consistent, then :
|
[{"identifier": "A", "content": "g + 2h + k = 0"}, {"identifier": "B", "content": "g + h + 2k = 0"}, {"identifier": "C", "content": "2g + h + k = 0"}, {"identifier": "D", "content": "g + h + k = 0"}]
|
["C"]
| null |
x $$-$$ $$4y + 7z = g$$
<br> $$3y$$ $$-$$ $$5z = h$$
<br>$$-$$$$2x + 5y$$ $$-$$ $$9z = k$$
<br><br>$$D = \left| {\matrix{
1 & { - 4} & 7 \cr
0 & 3 & { - 5} \cr
{ - 2} & 5 & { - 9} \cr
} } \right|$$
<br><br>$$D = 1\left( { - 27 + 25} \right) - 2\left( {20 - 21} \right)$$
<br><br>$$D = - 2 + 2 = 0$$
<br><br>If system is consistent then $${D_1} = {D_2} = {D_3} = 0$$
<br><br>$$\left| {\matrix{
1 & { - 4} & g \cr
0 & 3 & h \cr
{ - 2} & 5 & k \cr
} } \right| = 0$$
<br><br>$$1\left( {3k - 5h} \right) - 2\left( { - 4h - 3g} \right) = 0$$
<br><br>$$3k - 5h + 8h + 6g = 0$$
<br><br>$$6g + 3h + 3k = 0$$
<br><br>$$2g + h + k = 0$$
|
mcq
|
jee-main-2019-online-9th-january-evening-slot
|
MpyEVVbwxRUBTKEHiR8Ao
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
The system of linear equations
<br/> x + y + z = 2
<br/>2x + 3y + 2z = 5
<br/>2x + 3y + (a<sup>2</sup> β 1) z = a + 1 then
|
[{"identifier": "A", "content": "has infinitely many solutions for a = 4 "}, {"identifier": "B", "content": "has a unique solution for |a| = $$\\sqrt3$$"}, {"identifier": "C", "content": "is inconsistent when |a| = $$\\sqrt3$$"}, {"identifier": "D", "content": "is inconsistent when a = 4"}]
|
["C"]
| null |
$$D = \left| {\matrix{
1 & 1 & 1 \cr
2 & 3 & 2 \cr
2 & 3 & {{\alpha ^2} - 1} \cr
} } \right|$$
<br><br>D = 3$$a$$<sup>2</sup> $$-$$ 3 $$-$$ 6 $$-$$ 2$$a$$<sup>2</sup> + 2 + 4 + 2$$a$$<sup>2</sup> $$-$$ 2 $$-$$ 4
<br><br>D = ($$a$$<sup>2</sup> $$-$$ 3)
<br><br>When D $$ \ne $$ 0 then system of equiation has unique solution.
<br><br>$$ \therefore $$ 3($$a$$<sup>2</sup> $$-$$ 3) $$ \ne $$ 0
<br><br>$$ \Rightarrow $$ $$\left| a \right|$$ $$ \ne \sqrt 3 $$
<br><br>When $$3({a^2} - 3) = 0$$
<br><br>$$ \Rightarrow $$ $$\left| a \right| = \sqrt 3 $$ then D = 0
<br><br>If D = 0 then two cases possible
<br><br>(1) System of equation has infinite many solution.
<br><br>(2) System of equation has no solution and inconsistent.
<br><br>Here D<sub>1</sub> = $$\left| {\matrix{
2 & 1 & 1 \cr
5 & 3 & 2 \cr
{a + 1} & 3 & {{a^2} - 1} \cr
} } \right| = {a^2} - a + 1$$
<br><br>D<sub>2</sub> = $$\left| {\matrix{
1 & 2 & 1 \cr
2 & 5 & 2 \cr
2 & {a + 1} & {{a^2} - 1} \cr
} } \right| = {a^2} - 3$$
<br><br>D<sub>3</sub> = $$\left| {\matrix{
1 & 1 & 2 \cr
2 & 3 & 5 \cr
2 & 3 & {a + 1} \cr
} } \right| = a - 4$$
<br><br>System of equation will have infinite solution if D<sub>1</sub> = D<sub>2</sub> = D<sub>3</sub> = 0.
<br><br>And system of equation will have no solution if at last one of D<sub>1</sub>, D<sub>2</sub>, D<sub>2</sub> is non zero.
<br><br>At $$\left| a \right| = \sqrt 3 $$ we get D = 0
<br><br>But D<sub>1</sub> = 3 $$ \pm $$ $$\sqrt 3 + 1$$ $$ \ne $$ 0
<br><br>and D<sub>3</sub> = $$ \pm $$ $$\sqrt 3 - 4$$ $$ \ne $$ 0
<br><br>So, system of equations has no solution at $$\left| a \right| = \sqrt 3 $$ then system is in consistent
|
mcq
|
jee-main-2019-online-9th-january-morning-slot
|
smoZSqPnEuRLubvtFV7k9k2k5k6n942
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
The following system of linear equations<br/>
7x + 6y β 2z = 0<br/>
3x + 4y + 2z = 0<br/>
x β 2y β 6z = 0, has
|
[{"identifier": "A", "content": "no solution"}, {"identifier": "B", "content": "infinitely many solutions, (x, y, z) satisfying\ny = 2z"}, {"identifier": "C", "content": "infinitely many solutions, (x, y, z) satisfying\nx = 2z"}, {"identifier": "D", "content": "only the trivial solution"}]
|
["C"]
| null |
Given
<br>7x + 6y β 2z = 0 .......(1)<br>
3x + 4y + 2z = 0 ......(2)<br>
x β 2y β 6z = 0 .......(3)
<br><br>$$\Delta $$ = $$\left| {\matrix{
7 & 6 & { - 2} \cr
3 & 4 & 2 \cr
1 & { - 2} & { - 6} \cr
} } \right|$$
<br><br>= 7(β24 + 4) β 6(β18 β 2) β 2(β6 β 4) = 0
<br><br>$$ \therefore $$ $$\Delta $$ = 0
<br><br>The system of equation has infinite non-trival solution.
<br><br>Also adding equation (1) and 3$$ \times $$(3), we get
<br><br>10x = 20z
<br><br>$$ \Rightarrow $$ x = 2z
|
mcq
|
jee-main-2020-online-9th-january-evening-slot
|
Ms4TVcCbnqYMke09ANjgy2xukg4mzvbh
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
The sum of distinct values of $$\lambda $$ for which the
system of equations<br/><br/>$$\left( {\lambda - 1} \right)x + \left( {3\lambda + 1} \right)y + 2\lambda z = 0$$<br/>$$\left( {\lambda - 1} \right)x + \left( {4\lambda - 2} \right)y + \left( {\lambda + 3} \right)z = 0$$<br/>$$2x + \left( {3\lambda + 1} \right)y + 3\left( {\lambda - 1} \right)z = 0$$<br/><br/>
has non-zero solutions, is ________ .
|
[]
| null |
3
|
$$\left| {\matrix{
{\lambda - 1} & {3\lambda + 1} & {2\lambda } \cr
{\lambda - 1} & {4\lambda - 2} & {\lambda + 3} \cr
2 & {3\lambda + 1} & {3\left( {\lambda - 1} \right)} \cr
} } \right|$$ = 0
<br><br>R<sub>2</sub> $$ \to $$ R<sub>2</sub>
β R<sub>1</sub>
<br>R<sub>3</sub> $$ \to $$ R<sub>3</sub>
β R<sub>1</sub>
<br><br>$$\left| {\matrix{
{\lambda - 1} & {3\lambda + 1} & {2\lambda } \cr
0 & {\lambda - 3} & { - \lambda + 3} \cr
{3 - \lambda } & 0 & {\lambda - 3} \cr
} } \right| = 0$$
<br><br>C<sub>1</sub> $$ \to $$ C<sub>1</sub>
+ C<sub>3</sub>
<br><br>$$\left| {\matrix{
{3\lambda - 1} & {3\lambda + 1} & {2\lambda } \cr
{ - \lambda + 3} & {\lambda - 3} & { - \lambda + 3} \cr
0 & 0 & {\lambda - 3} \cr
} } \right| = 0$$
<br>$$ \Rightarrow $$ ($$\lambda $$ - 3) [(3$$\lambda $$ - 1) ($$\lambda $$ - 3) β (3 β $$\lambda $$) (3$$\lambda $$ + 1)] = 0
<br>$$ \Rightarrow $$ ($$\lambda $$ β 3) [3$$\lambda $$<sup>2</sup>
β 10$$\lambda $$ + 3 β(8$$\lambda $$ β3$$\lambda $$<sup>2</sup>
+ 3)] = 0
<br>$$ \Rightarrow $$ ($$\lambda $$ β 3) (6$$\lambda $$<sup>2</sup>
β 18$$\lambda $$) = 0
<br>$$ \Rightarrow $$ (6$$\lambda $$) ($$\lambda $$ β 3)<sup>2</sup> = 0
<br>$$ \Rightarrow $$ $$\lambda $$ = 0, 3
<br>$$ \therefore $$ sum of values of $$\lambda $$ = 0 + 3 = 3
|
integer
|
jee-main-2020-online-6th-september-evening-slot
|
zkhEzaGYQ1BJMqDNbnjgy2xukfuvc6zl
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
The values of $$\lambda $$ and $$\mu $$ for which the system of linear equations
<br/>x + y + z = 2
<br/>x + 2y + 3z = 5
<br/>x + 3y + $$\lambda $$z = $$\mu $$
<br/>has infinitely many solutions are, respectively:
|
[{"identifier": "A", "content": "6 and 8"}, {"identifier": "B", "content": "5 and 8"}, {"identifier": "C", "content": "5 and 7"}, {"identifier": "D", "content": "4 and 9"}]
|
["B"]
| null |
For infinite many solutions
<br><br>D = D<sub>1</sub> = D<sub>2</sub> = D<sub>3</sub> = 0
<br><br>Now D = $$\left| {\matrix{
1 & 1 & 1 \cr
1 & 2 & 3 \cr
1 & 3 & \lambda \cr
} } \right|$$ = 0
<br><br>$$ \Rightarrow $$ 1. (2$$\lambda $$ β 9) β1.($$\lambda $$ β 3) + 1.(3 β 2) = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$ = 5
<br><br>Now D<sub>1</sub> = $$\left| {\matrix{
2 & 1 & 1 \cr
5 & 2 & 3 \cr
\mu & 3 & 5 \cr
} } \right|$$ = 0
<br><br>$$ \Rightarrow $$ 2(10 β 9) β1(25 β 3$$\mu $$) + 1(15 β 2$$\mu $$) = 0
<br><br>$$ \Rightarrow $$ $$\mu $$ = 8
|
mcq
|
jee-main-2020-online-6th-september-morning-slot
|
BSQX3YXOf0Mdbfeyjgjgy2xukfqda872
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
If the system of linear equations
<br/>x + y + 3z = 0
<br/>x + 3y + k<sup>2</sup>z = 0
<br/>3x + y + 3z = 0
<br/>has a non-zero solution (x, y, z) for some k $$ \in $$ R,
then x + $$\left( {{y \over z}} \right)$$ is equal to :
|
[{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "-9"}, {"identifier": "D", "content": "-3"}]
|
["D"]
| null |
x + y + 3z = 0 .....(i)
<br>x + 3y + k<sup>2</sup>z = 0 .........(ii)
<br>3x + y + 3z = 0 ......(iii)
<br><br>$$\left| {\matrix{
1 & 1 & 3 \cr
1 & 3 & {{k^2}} \cr
3 & 1 & 3 \cr
} } \right|$$ = 0
<br><br>$$ \Rightarrow $$ 9 + 3 + 3k<sup>2</sup>
β 27 β k<sup>2</sup>
β 3 = 0
<br><br>$$ \Rightarrow $$ k<sup>2</sup> = 9
<br><br>Perform (i) β (iii),
<br><br>β2x = 0 $$ \Rightarrow $$ x = 0
<br><br>Now from (i), y + 3z = 0
<br><br>$$ \Rightarrow $$ $${y \over z} = - 3$$
<br><br>$$ \therefore $$ x + $$\left( {{y \over z}} \right)$$ = -3
|
mcq
|
jee-main-2020-online-5th-september-evening-slot
|
ILoYUJvJqnt3fXMETijgy2xukfg6tviv
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
Let $$\lambda \in $$ R . The system of linear equations<br/>
2x<sub>1</sub>
- 4x<sub>2</sub> + $$\lambda $$x<sub>3</sub> = 1<br/>
x<sub>1</sub> - 6x<sub>2</sub> + x<sub>3</sub> = 2<br/>
$$\lambda $$x<sub>1</sub> - 10x<sub>2</sub> + 4x<sub>3</sub> = 3<br/>
is inconsistent for:
|
[{"identifier": "A", "content": "exactly one positive value of $$\\lambda $$"}, {"identifier": "B", "content": "exactly one negative value of $$\\lambda $$"}, {"identifier": "C", "content": "exactly two values of $$\\lambda $$\n"}, {"identifier": "D", "content": "every value of $$\\lambda $$"}]
|
["B"]
| null |
D = $$\left| {\matrix{
2 & { - 4} & \lambda \cr
1 & { - 6} & 1 \cr
\lambda & { - 10} & 4 \cr
} } \right|$$ = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$ = 3, $$ - {2 \over 3}$$
<br><br>D<sub>1</sub> = $$\left| {\matrix{
1 & { - 4} & \lambda \cr
2 & { - 6} & 1 \cr
3 & { - 10} & 4 \cr
} } \right|$$
<br><br>= 14 + 4(5) + $$\lambda $$(β2)
<br><br>= β2$$\lambda $$ + 6
<br><br>D<sub>2</sub> = $$\left| {\matrix{
2 & 1 & \lambda \cr
1 & 2 & 1 \cr
\lambda & 3 & 4 \cr
} } \right|$$
<br><br>= β2($$\lambda $$ β 3)($$\lambda $$ + 1)
<br><br>D<sub>3</sub> = $$\left| {\matrix{
2 & { - 4} & 1 \cr
1 & { - 6} & 2 \cr
\lambda & { - 10} & 3 \cr
} } \right|$$
<br><br>= β 2$$\lambda $$ + 6
<br><br>When , $$\lambda $$ = 3 then
<br><br>D = D<sub>1</sub> = D<sub>2</sub> = D<sub>3</sub> = 0
<br><br>$$ \Rightarrow $$ Infinite many solution
<br><br>When $$\lambda $$ = $$ - {2 \over 3}$$ then D<sub>1</sub>, D<sub>2</sub>, D<sub>3</sub> none of them is zero so equations are inconsistant.
<br><br>$$ \therefore $$ $$\lambda $$ = $$ - {2 \over 3}$$
|
mcq
|
jee-main-2020-online-5th-september-morning-slot
|
9Mw6Lbj0gcskJSXlCAjgy2xukfaju0ka
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
If the system of equations<br/>
x+y+z=2<br/>
2x+4yβz=6<br/>
3x+2y+$$\lambda $$z=$$\mu $$<br/>
has infinitely many solutions, then
|
[{"identifier": "A", "content": "2$$\\lambda $$ - $$\\mu $$ = 5"}, {"identifier": "B", "content": "$$\\lambda $$ - 2$$\\mu $$ = -5"}, {"identifier": "C", "content": "2$$\\lambda $$ + $$\\mu $$ = 14"}, {"identifier": "D", "content": "$$\\lambda $$ + 2$$\\mu $$ = 14"}]
|
["C"]
| null |
$$D = 0\,\left| {\matrix{
1 & 1 & 1 \cr
2 & 4 & { - 1} \cr
3 & 2 & \lambda \cr
} } \right| = 0$$<br><br>$$ \Rightarrow $$ $$(4\lambda + 2) - 1(2\lambda + 3) + 1(4 - 12) = 0$$<br><br>$$ \Rightarrow $$ $$4\lambda + 2$$ $$ - 2\lambda - 3$$$$ - $$8$$ = 0$$<br><br>$$ \Rightarrow $$ $$2\lambda = 9 \Rightarrow \lambda = {9 \over 2}$$<br><br>$${D_x} = \left| {\matrix{
2 & 1 & 1 \cr
6 & 4 & { - 1} \cr
\mu & 2 & { - 9/2} \cr
} } \right| = 0$$<br><br>$$ \Rightarrow \mu = 5$$<br><br>By checking all the options we find (C) is correct <br><br>$$2\lambda + \mu = 14$$
|
mcq
|
jee-main-2020-online-4th-september-evening-slot
|
r4m8c8M3NQOXkfjMEDjgy2xukfagtbir
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
Suppose the vectors x<sub>1</sub>, x<sub>2</sub> and x<sub>3</sub> are the <br/>solutions of the system of linear equations,<br/> Ax = b when the vector b on the right side is equal to b<sub>1</sub>, b<sub>2</sub> and b<sub>3</sub> respectively. if<br/><br/>
$${x_1} = \left[ {\matrix{
1 \cr
1 \cr
1 \cr
} } \right]$$, $${x_2} = \left[ {\matrix{
0 \cr
2 \cr
1 \cr
} } \right]$$, $${x_3} = \left[ {\matrix{
0 \cr
0 \cr
1 \cr
} } \right]$$<br/><br/>
$${b_1} = \left[ {\matrix{
1 \cr
0 \cr
0 \cr
} } \right]$$, $${b_2} = \left[ {\matrix{
0 \cr
2 \cr
0 \cr
} } \right]$$ and $${b_3} = \left[ {\matrix{
0 \cr
0 \cr
2 \cr
} } \right]$$, <br/>then the determinant of A is equal to :
|
[{"identifier": "A", "content": "$${3 \\over 2}$$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}]
|
["C"]
| null |
Let A = $$\left[ {\matrix{
{{a_1}} & {{a_2}} & {{a_3}} \cr
{{a_4}} & {{a_5}} & {{a_6}} \cr
{{a_7}} & {{a_8}} & {{a_9}} \cr
} } \right]$$
<br><br>For Ax<sub>1</sub> = b<sub>1</sub> :
<br><br>$$ \Rightarrow $$ $$\left[ {\matrix{
{{a_1}} & {{a_2}} & {{a_3}} \cr
{{a_4}} & {{a_5}} & {{a_6}} \cr
{{a_7}} & {{a_8}} & {{a_9}} \cr
} } \right]\left[ {\matrix{
1 \cr
1 \cr
1 \cr
} } \right] = \left[ {\matrix{
1 \cr
0 \cr
0 \cr
} } \right]$$
<br><br>$$ \therefore $$ $${a_1} + {a_2} + {a_3} = 1$$ ....(1)
<br><br>$${a_4} + {a_5} + {a_6} = 0$$ ......(2)
<br><br>$${a_7} + {a_8} + {a_9} = 0$$ .....(3)
<br><br>For Ax<sub>2</sub> = b<sub>2</sub> :
<br><br>$$\left[ {\matrix{
{{a_1}} & {{a_2}} & {{a_3}} \cr
{{a_4}} & {{a_5}} & {{a_6}} \cr
{{a_7}} & {{a_8}} & {{a_9}} \cr
} } \right]\left[ {\matrix{
0 \cr
2 \cr
1 \cr
} } \right] = \left[ {\matrix{
0 \cr
2 \cr
0 \cr
} } \right]$$
<br><br>$$ \therefore $$ $$2{a_2} + {a_3} = 0$$ .....(4)
<br><br>$$2{a_5} + {a_6} = 2$$ ....(5)
<br><br>$$2{a_8} + {a_9} = 0$$ ....(6)
<br><br>For Ax<sub>3</sub> = b<sub>3</sub> :
<br><br>$$\left[ {\matrix{
{{a_1}} & {{a_2}} & {{a_3}} \cr
{{a_4}} & {{a_5}} & {{a_6}} \cr
{{a_7}} & {{a_8}} & {{a_9}} \cr
} } \right]\left[ {\matrix{
0 \cr
0 \cr
1 \cr
} } \right] = \left[ {\matrix{
0 \cr
0 \cr
2 \cr
} } \right]$$
<br><br>$$ \therefore $$ $${{a_3} = 0}$$
<br><br>$${{a_6} = 0}$$
<br><br>$${{a_9} = 2}$$
<br><br>Putting value of $${{a_3}}$$ in equation (4), we get
<br><br>$${{a_2}}$$ = 0
<br><br>Putting value of $${{a_6}}$$ in equation (5), we get
<br><br>$${{a_5}}$$ = 1
<br><br>Putting value of $${{a_9}}$$ in equation (6), we get
<br><br>$${{a_8}}$$ = -1
<br><br>Putting value of $${{a_2}}$$ and $${{a_3}}$$ in equation (1), we get
<br><br>$${{a_1}}$$ = 1
<br><br>Putting value of $${{a_5}}$$ and $${{a_6}}$$ in equation (6), we get
<br><br>$${{a_4}}$$ = -1
<br><br>Putting value of $${{a_5}}$$ and $${{a_6}}$$ in equation (6), we get
<br><br>$${{a_4}}$$ = -1
<br><br>Putting value of $${{a_8}}$$ and $${{a_9}}$$ in equation (6), we get
<br><br>$${{a_7}}$$ = -1
<br><br>$$ \therefore $$ A = $$\left[ {\matrix{
1 & 0 & 0 \cr
{ - 1} & 1 & 0 \cr
{ - 1} & { - 1} & 2 \cr
} } \right]$$
<br><br>So, |A| = 2(1) = 2
|
mcq
|
jee-main-2020-online-4th-september-evening-slot
|
dWw9wkzgDfk71NagXfjgy2xukf8zx58k
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
If the system of equations<br/>
x - 2y + 3z = 9<br/>
2x + y + z = b<br/>
x - 7y + az = 24, <br/>has infinitely many solutions, then a - b is equal to.........
|
[]
| null |
5
|
D = 0<br><br>$$\left| {\matrix{
1 & { - 2} & 3 \cr
2 & 1 & 1 \cr
1 & { - 7} & a \cr
} } \right| = 0$$<br><br>$$1(a + 7) + 2(2a - 1) + 3( - 14 - 1) = 0$$<br><br>$$a + 7 + 4a - 2 - 45 = 0$$<br><br>$$5a = 40$$<br><br>$$a = 8$$<br><br>$${D_1} = \left| {\matrix{
9 & { - 2} & 3 \cr
b & 1 & 1 \cr
{24} & { - 7} & 8 \cr
} } \right| = 0$$<br><br>$$ \Rightarrow 9(8 + 7) + 2(8b - 24) + 3( - 7b - 24) = 0$$<br><br>$$ \Rightarrow 135 + 16b - 48 - 21b - 72 = 0$$<br><br>$$ \Rightarrow $$ $$15 = 5b$$
<br><br>$$ \Rightarrow b = 3$$<br><br>$$a - b = 5$$
|
integer
|
jee-main-2020-online-4th-september-morning-slot
|
sSAYfOA3tsGReJtLHNjgy2xukezelwq7
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
Let A = {X = (x, y, z)<sup>T</sup>: PX = 0 and
<br/><br/>x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup> = 1} where
<br/><br/>$$P = \left[ {\matrix{
1 & 2 & 1 \cr
{ - 2} & 3 & { - 4} \cr
1 & 9 & { - 1} \cr
} } \right]$$,
<br/><br/>then the set A :
|
[{"identifier": "A", "content": "is an empty set.\n"}, {"identifier": "B", "content": "contains more than two elements."}, {"identifier": "C", "content": "contains exactly two elements."}, {"identifier": "D", "content": "is a singleton."}]
|
["C"]
| null |
Let $$X = \left[ {\matrix{
x \cr
y \cr
z \cr
} } \right]$$<br><br>
PX = O<br><br>
$$\left[ {\matrix{
1 & 2 & 1 \cr
{ - 2} & 3 & { - 4} \cr
1 & 9 & { - 1} \cr
} } \right]\left[ {\matrix{
x \cr
y \cr
z \cr
} } \right] = \left[ {\matrix{
0 \cr
0 \cr
0 \cr
} } \right]$$<br><br>
x + 2y + z = 0........(1)<br><br>
-2x + 3y β 4z = 0....(2)<br><br>
x + 9y - z = 0..........(3)<br><br>
from (1) & (3)<br>
$$ \Rightarrow $$ 2x+11y =0<br><br>
from (1) & (2)<br>
$$ \Rightarrow $$ 2x + 11y = 0<br><br>
from (2) & (3)<br>
β6x β33y = 0<br>
$$ \Rightarrow $$ 2x +11y = 0<br><br>
putting value of x in (1), we get<br>
β7y + 2z = 0<br><br>
Now $${\left( {{{11y} \over 2}} \right)^2} + {y^2} + {\left( {{{7y} \over 2}} \right)^2} = 1$$<br><br>
y<sup>2</sup>(121 + 1 + 49) = 4<br><br>
y<sup>2</sup>(171) = 4<br><br>
$$y = \pm {2 \over {\sqrt {171} }}$$<br><br>
$$ \Rightarrow x = \pm {7 \over {\sqrt {171} }}$$<br><br>
$$ \Rightarrow z = \pm {{11} \over {\sqrt {171} }}$$<br><br>
$$ \therefore $$ So, there are 2 solution set of (x, y, z)
|
mcq
|
jee-main-2020-online-2nd-september-evening-slot
|
xYS1FKI11Dse7mNmPOjgy2xukewnbkm1
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
Let S be the set of all $$\lambda $$ $$ \in $$ R for which the system
of linear equations
<br/><br/>2x β y + 2z = 2
<br/>x β 2y +
$$\lambda $$z = β4
<br/>x +
$$\lambda $$y + z = 4
<br/><br/>has no solution. Then the set S :
|
[{"identifier": "A", "content": "contains more than two elements."}, {"identifier": "B", "content": "contains exactly two elements."}, {"identifier": "C", "content": "is a singleton."}, {"identifier": "D", "content": "is an empty set."}]
|
["B"]
| null |
For no solution :
<br><br>$$\Delta $$ = 0 and $$\Delta $$<sub>1</sub>/$$\Delta $$<sub>2</sub>/$$\Delta $$<sub>3</sub> $$ \ne $$ 0
<br><br>$$\Delta $$ = $$\left| {\matrix{
2 & { - 1} & 2 \cr
1 & { - 2} & \lambda \cr
1 & \lambda & 1 \cr
} } \right|$$ = 0
<br><br>$$ \Rightarrow $$ 2(β2 β $$\lambda $$<sup>2</sup>) + 1 (1 β $$\lambda $$) + 2($$\lambda $$ + 2) = 0
<br><br>$$ \Rightarrow $$ β2$$\lambda $$<sup>2</sup>
+ $$\lambda $$ + 1 = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$ = 1, $$ - {1 \over 2}$$
<br><br>When <b>$$\lambda $$ = 1</b>
<br><br>2x β y + 2z = 2 ...(1)
<br><br>x β 2y + z = β4 ...(2)
<br><br>x + y + z = 4 ...(3)
<br><br>Adding (2) and (3), we get
<br><br>2x β y + 2z = 0 (contradiction) hence no solution.
<br><br>$$ \therefore $$ $$\lambda $$ = 1 belongs to set S.
<br><br>When $$\lambda $$ = $$ - {1 \over 2}$$
<br><br>2x β y + 2z = 2 ...(1)
<br><br>x β 2y $$ - {1 \over 2}$$z
= β4 ...(2)
<br><br>x $$ - {1 \over 2}$$y
+ z = 4 ...(3)
<br><br>(1) and (3) contradict each other, hence no
solution.
<br><br>$$ \therefore $$ $$\lambda $$ = $$ - {1 \over 2}$$ belongs to set S.
|
mcq
|
jee-main-2020-online-2nd-september-morning-slot
|
NGdiQX4MhPGe9ikRQC7k9k2k5itcig0
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
If for some $$\alpha $$ and $$\beta $$ in R, the intersection of the
following three places<br/>
x + 4y β 2z = 1<br/>
x + 7y β 5z = b<br/>
x + 5y + $$\alpha $$z = 5<br/>
is a line in R<sup>3</sup>, then $$\alpha $$ + $$\beta $$ is equal to :
|
[{"identifier": "A", "content": "-10"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "2"}]
|
["C"]
| null |
For planes to intersect on a line there should be infinite solution of the
given system of equations.
<br><br>For infinite solutions
<br><br>$$\Delta $$ = $$\left| {\matrix{
1 & 4 & { - 2} \cr
1 & 7 & { - 5} \cr
1 & 5 & \alpha \cr
} } \right|$$ = 0
<br><br>$$ \Rightarrow $$ 1(7$$\alpha $$ + 25) β 4($$\alpha $$ + 5) β 2(5 β 7) = 0
<br><br>$$ \Rightarrow $$ 7$$\alpha $$ + 25 β 4$$\alpha $$ β 20 + 4 = 0
<br><br>$$ \Rightarrow $$ 3$$\alpha $$ + 9 = 0
<br><br>$$ \Rightarrow $$ $$\alpha $$ = -3
<br><br>Also $$\Delta $$<sub>z</sub> = 0
<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
1 & 4 & 1 \cr
1 & 7 & \beta \cr
1 & 5 & 5 \cr
} } \right|$$ = 0
<br><br>$$ \Rightarrow $$ 1(35 β 5$$\beta $$) β 4(5 β $$\beta $$) + 1(5 β 7) = 0
<br><br>$$ \Rightarrow $$ 35 - 5$$\beta $$ - 20 + 4$$\beta $$ - 2 = 0
<br><br>$$ \Rightarrow $$ $$\beta $$ = 13
<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = -3 + 13 = 10
|
mcq
|
jee-main-2020-online-9th-january-morning-slot
|
TOdbw5R13QYHwVQqH57k9k2k5hjodld
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
The system of linear equations<br/>
$$\lambda $$x + 2y + 2z = 5<br/>
2$$\lambda $$x + 3y + 5z = 8<br/>
4x + $$\lambda $$y + 6z = 10 has
|
[{"identifier": "A", "content": "a unique solution when $$\\lambda $$ = \u20138"}, {"identifier": "B", "content": "no solution when $$\\lambda $$ = 2"}, {"identifier": "C", "content": "infinitely many solutions when $$\\lambda $$ = 2"}, {"identifier": "D", "content": "no solution when $$\\lambda $$ = 8"}]
|
["B"]
| null |
$$\Delta $$ = $$\left| {\matrix{
\lambda & 2 & 2 \cr
{2\lambda } & 3 & 5 \cr
4 & \lambda & 6 \cr
} } \right|$$
<br><br>= $$\lambda $$ ( 18 β 5$$\lambda $$) β 2(12$$\lambda $$ β 20) + 2(2$$\lambda $$<sup>2</sup>
β 12)
<br><br>= 18$$\lambda $$ β 5$$\lambda $$<sup>2</sup>
β 24$$\lambda $$ + 40 + 4$$\lambda $$<sup>2</sup>
β 24
<br><br>= β $$\lambda $$<sup>2</sup>
β 6$$\lambda $$ + 16
<br><br>= β ($$\lambda $$ + 8)($$\lambda $$ β 2)
<br><br>For no solutions $$\lambda $$ = 0 $$ \Rightarrow $$ $$\lambda $$ = β 8, $$\lambda $$ = 2
<br><br>when $$\lambda $$ = 2
<br><br>$$\Delta $$<sub>x</sub> = $$\left| {\matrix{
5 & 2 & 2 \cr
8 & 3 & 5 \cr
{10} & 2 & 6 \cr
} } \right|$$
<br><br>= 5 (18 β 10) β 2 (48 β 50) + 2 (16 β 30)
<br><br>= 40 + 4 β 28 $$ \ne $$ 0
<br><br>So no solution for $$\lambda $$ = 2
|
mcq
|
jee-main-2020-online-8th-january-evening-slot
|
0qzGq9uNJLFafpg2SS7k9k2k5gz1yx1
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
For which of the following ordered pairs ($$\mu $$, $$\delta $$),
the system of linear equations
<br/>x + 2y + 3z = 1
<br/>3x + 4y + 5z = $$\mu $$
<br/>4x + 4y + 4z = $$\delta $$
<br/>is inconsistent ?
|
[{"identifier": "A", "content": "(1, 0)"}, {"identifier": "B", "content": "(4, 3)"}, {"identifier": "C", "content": "(4, 6)"}, {"identifier": "D", "content": "(3, 4)"}]
|
["B"]
| null |
For inconsistent system we need
<br><br>$$\Delta $$ = 0 and atleast one of $$\Delta $$x, $$\Delta $$y, $$\Delta $$z $$ \ne $$ 0
<br><br>$$ \therefore $$ $$\Delta $$ = $$\left| {\matrix{
1 & 2 & 3 \cr
3 & 4 & 5 \cr
4 & 4 & 4 \cr
} } \right|$$ = 0
<br><br>$$\Delta $$<sub>x</sub> = $$\left| {\matrix{
1 & 2 & 3 \cr
\mu & 4 & 5 \cr
\delta & 4 & 4 \cr
} } \right|$$
<br><br>= (-4) - 2($$\mu $$ - 5$$\delta $$) + 3(4$$\mu $$ - 4$$\delta $$)
<br><br>$$ \Rightarrow $$ 2$$\mu $$ $$ \ne $$ $$\delta $$ + 2 ....(1)
<br><br>Only ($$\mu $$, $$\delta $$) = (4, 3) does satisfy the equation (1).
|
mcq
|
jee-main-2020-online-8th-january-morning-slot
|
Q4MVJjbJtMr0dpsrfw7k9k2k5fopwui
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
If the system of linear equations,
<br/>x + y + z = 6
<br/>x + 2y + 3z = 10
<br/>3x + 2y + $$\lambda $$z = $$\mu $$
<br/>has more than two solutions, then $$\mu $$ - $$\lambda $$<sup>2</sup>
is equal to ______.
|
[]
| null |
13
|
Given system of equation more than
2 solutions.
Hence system of equation has infinite many
solution.
<br><br>$$ \therefore $$ $$\Delta $$ = $$\Delta $$<sub>1</sub> = $$\Delta $$<sub>2</sub> = $$\Delta $$<sub>3</sub> = 0
<br><br>$$\Delta $$ = $$\left| {\matrix{
1 & 1 & 1 \cr
1 & 2 & 3 \cr
3 & 2 & \lambda \cr
} } \right|$$ = 0
<br><br>$$ \Rightarrow $$ 1(2Ξ» β 6) β 1(Ξ» β 9) + 1(β 4) = 0
<br><br>$$ \Rightarrow $$ 2Ξ» β 6 β Ξ» + 9 β 4 = 0
<br><br>$$ \Rightarrow $$ Ξ» = 1
<br><br>$$\Delta $$<sub>1</sub> = $$\left| {\matrix{
6 & 1 & 1 \cr
{10} & 2 & 3 \cr
\mu & 2 & \lambda \cr
} } \right|$$ = 0
<br> 6(2Ξ» β 6) β 1(10Ξ» β 3ΞΌ) + 1(20 β 2ΞΌ) = 0
<br><br>$$ \Rightarrow $$ 12Ξ» β 36 β 10Ξ» + 3ΞΌ + 20 β 2ΞΌ = 0
<br><br>$$ \Rightarrow $$ 2Ξ» + ΞΌ = 16
<br><br>$$ \Rightarrow $$ 2 + ΞΌ = 16
<br><br>$$ \Rightarrow $$ $$\mu $$ = 14
<br><br> $$ \therefore $$ $$\mu $$ - $$\lambda $$<sup>2</sup> = 14 - 1 = 13
|
integer
|
jee-main-2020-online-7th-january-evening-slot
|
ZCZ7Gft603uNCE9sQX7k9k2k5e3cqx9
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
If the system of linear equations<br/>
2x + 2ay + az = 0<br/>
2x + 3by + bz = 0<br/>
2x + 4cy + cz = 0,<br/>
where a, b, c $$ \in $$ R are non-zero distinct; has a non-zero solution, then:
|
[{"identifier": "A", "content": "$${1 \\over a},{1 \\over b},{1 \\over c}$$ are in A.P. "}, {"identifier": "B", "content": "a + b + c = 0"}, {"identifier": "C", "content": "a, b, c are in G.P."}, {"identifier": "D", "content": "a,b,c are in A.P."}]
|
["A"]
| null |
For non-zero solution
<br><br>$$\left| {\matrix{
2 & {2a} & a \cr
2 & {3b} & b \cr
2 & {4c} & c \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
1 & {2a} & a \cr
0 & {3b - 2a} & {b - a} \cr
0 & {4c - 2a} & {c - a} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow $$ (3b β 2a) (c βa) β (b β a) (4c β 2a) = 0
<br><br>$$ \Rightarrow $$ 2ac = bc + ab
<br><br>$$ \Rightarrow $$ $${2 \over b} = {1 \over a} + {1 \over c}$$
<br><br>$$ \therefore $$ $${1 \over a},{1 \over b},{1 \over c}$$ are in A.P.
|
mcq
|
jee-main-2020-online-7th-january-morning-slot
|
lKP9oeIa262XxKtivNjgy2xukf49oomt
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
Let S be the set of all integer solutions, (x, y, z),
of the system of equations
<br/>x β 2y + 5z = 0
<br/>β2x + 4y + z = 0
<br/>β7x + 14y + 9z = 0
<br/>such that 15 $$ \le $$ x<sup>2</sup>
+ y<sup>2</sup>
+ z<sup>2</sup> $$ \le $$ 150. Then, the
number of elements in the set S is equal to
______ .
|
[]
| null |
8
|
$$x - 2y + 5z = 0$$ ....(1)<br><br>$$ - 2x + 4y + z = 0$$ .....(2)<br><br>$$ - 7x + 14y + 9z = 0$$ ....(3)<br><br>2.(1) + (2) we get z = 0, x = 2y<br><br>15 $$ \le $$ 4y<sup>2</sup> + y<sup>2</sup> $$ \le $$ 150<br><br>$$ \Rightarrow $$ 3 $$ \le $$ y<sup>2</sup> $$ \le $$ 30<br><br>$$y \in \left[ { - \sqrt {30} , - \sqrt 3 } \right] \cup \left[ {\sqrt 3 ,\sqrt {30} } \right]$$<br><br>$$y = \pm 2,\, \pm 3,\, \pm 4,\, \pm 5$$<br><br>$$ \therefore $$ no. of integer's in S is 8
|
integer
|
jee-main-2020-online-3rd-september-evening-slot
|
C0QoIvmtVpCzopvrR11klrhinn0
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
The system of linear equations
<br/>3x - 2y - kz = 10
<br/>2x - 4y - 2z = 6
<br/>x+2y - z = 5m
<br/>is inconsistent if :
|
[{"identifier": "A", "content": "k $$ \\ne $$ 3, m $$ \\in $$ <b>R</b>"}, {"identifier": "B", "content": "k = 3, m $$ \\ne $$ $${4 \\over 5}$$"}, {"identifier": "C", "content": "k = 3, m $$ = $$ $${4 \\over 5}$$"}, {"identifier": "D", "content": "k $$ \\ne $$ 3, m $$ \\ne $$ $${4 \\over 5}$$"}]
|
["B"]
| null |
$$\Delta = \left| {\matrix{
3 & { - 2} & { - k} \cr
1 & { - 4} & { - 2} \cr
1 & 2 & { - 1} \cr
} } \right| = 0$$<br><br>$$3(4 + 4) + 2( - 2 + 2) - k(4 + 4) = 0$$<br><br>$$ \Rightarrow k = 3$$<br><br>$${\Delta _x} = \left| {\matrix{
{10} & { - 2} & { - 3} \cr
6 & { - 4} & { - 2} \cr
{5m} & 2 & { - 1} \cr
} } \right| \ne 0$$<br><br>$$10(4 + 4) + 2( - 6 + 10m) - 3(12 + 20m) \ne 0$$<br><br>$$80 - 12 + 20m - 36 - 60m \ne 0$$<br><br>$$40m \ne 32 \Rightarrow m \ne {4 \over 5}$$<br><br>$${\Delta _y} = \left| {\matrix{
3 & {10} & { - 3} \cr
2 & 6 & { - 2} \cr
1 & {5m} & { - 1} \cr
} } \right| \ne 0$$<br><br>$$3( - 6 + 10m) - 10( - 2 + 2) - 3(10m - 6) \ne 0$$<br><br>$$ - 18 + 30m - 30m + 18 \ne 0 \Rightarrow 0$$<br><br>$${\Delta _z} = \left| {\matrix{
3 & { - 2} & {10} \cr
2 & { - 4} & 6 \cr
1 & 2 & {5m} \cr
} } \right| \ne 0$$<br><br>$$3( - 20m - 12) + 2(10m - 6) + 10(4 + 4) - 40m + 32 \ne 0 \Rightarrow m \ne {4 \over 5}$$
|
mcq
|
jee-main-2021-online-24th-february-morning-slot
|
BaRKThiud28DgJukO31klrkt5yt
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
For the system of linear equations:<br/><br/>$$x - 2y = 1,x - y + kz = - 2,ky + 4z = 6,k \in R$$,<br/><br/>consider the following statements :<br/><br/>(A) The system has unique solution if $$k \ne 2,k \ne - 2$$.<br/><br/>(B) The system has unique solution if k = $$-$$2<br/><br/>(C) The system has unique solution if k = 2<br/><br/>(D) The system has no solution if k = 2<br/><br/>(E) The system has infinite number of solutions if k $$ \ne $$ $$-$$2.<br/><br/>Which of the following statements are correct?
|
[{"identifier": "A", "content": "(B) and (E) only"}, {"identifier": "B", "content": "(C) and (D) only"}, {"identifier": "C", "content": "(A) and (E) only"}, {"identifier": "D", "content": "(A) and (D) only"}]
|
["D"]
| null |
$$x - 2y + 0.z = 1$$<br><br>$$x - y + kz = - 2$$<br><br>$$0.x + ky + 4z = 6$$<br><br>$$\Delta = \left| {\matrix{
1 & { - 2} & 0 \cr
1 & { - 1} & k \cr
0 & k & 4 \cr
} } \right| = 4 - {k^2}$$<br><br>For unique solution $$4 - {k^2} \ne 0$$<br><br>$$ \Rightarrow $$ k $$ \ne $$ $$ \pm $$ 2<br><br><b>For k = 2 :</b><br><br>$$x - 2y + 0.z = 1$$<br><br>$$x - y + 2z = - 2$$<br><br>$$0.x + 2y + 4z = 6$$<br><br>$$\Delta x = \left| {\matrix{
1 & { - 2} & 0 \cr
2 & { - 1} & 2 \cr
6 & 2 & 4 \cr
} } \right| = ( - 8) + 2[ - 20]$$<br><br>$$\Delta x = - 48 \ne 0$$<br><br>For k = 2, $$\Delta x \ne 0$$<br><br>$$ \therefore $$ For K = 2; The system has no solution.
|
mcq
|
jee-main-2021-online-24th-february-evening-slot
|
mwbm36x3PoBzX7mXxC1kls5owix
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
If the system of equations<br/><br/>kx + y + 2z = 1<br/><br/>3x $$-$$ y $$-$$ 2z = 2<br/><br/>$$-$$2x $$-$$2y $$-$$4z = 3<br/><br/>has infinitely many solutions, then k is equal to __________.
|
[]
| null |
21
|
D = 0<br><br>$$ \Rightarrow \left| {\matrix{
k & 1 & 2 \cr
3 & { - 1} & { - 2} \cr
{ - 2} & { - 2} & { - 4} \cr
} } \right| = 0$$<br><br>$$ \Rightarrow $$ k (4 $$-$$ 4) $$-$$ 1 ($$-$$ 12 $$-$$ 4) + 2 ($$-$$ 6 $$-$$ 2)<br><br>$$ \Rightarrow $$ 16 $$-$$ 16 = 0<br><br>Also, $${D_1} = {D_2} = {D_3} = 0$$<br><br>$$ \Rightarrow {D_2} = \left| {\matrix{
k & 1 & 2 \cr
3 & 2 & { - 2} \cr
{ - 2} & 3 & { - 4} \cr
} } \right| = 0$$<br><br>$$ \Rightarrow $$ k($$-$$8 + 6) $$-$$ 1($$-$$ 12 $$-$$ 4) + 2(9 + 4) = 0<br><br>$$ \Rightarrow $$ $$-$$ 2k + 16 + 26 = 0<br><br>$$ \Rightarrow $$ 2k = 42<br><br>$$ \Rightarrow $$ k = 21
|
integer
|
jee-main-2021-online-25th-february-morning-slot
|
7FjcmuhVR7Bug874By1klt9gxr2
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
The following system of linear equations<br/><br/>2x + 3y + 2z = 9<br/><br/>3x + 2y + 2z = 9<br/><br/>x $$-$$ y + 4z = 8
|
[{"identifier": "A", "content": "does not have any solution"}, {"identifier": "B", "content": "has a solution ($$\\alpha$$, $$\\beta$$, $$\\gamma$$) satisfying $$\\alpha$$ + $$\\beta$$<sup>2</sup> + $$\\gamma$$<sup>3</sup> = 12"}, {"identifier": "C", "content": "has a unique solution"}, {"identifier": "D", "content": "has infinitely many solutions"}]
|
["C"]
| null |
$$\Delta = \left| {\matrix{
2 & 3 & 2 \cr
3 & 2 & 2 \cr
1 & { - 1} & 4 \cr
} } \right| = - 20 \ne 0$$ $$ \therefore $$ unique solution<br><br>$${\Delta _x} = \left| {\matrix{
9 & 3 & 2 \cr
9 & 2 & 2 \cr
8 & { - 1} & 4 \cr
} } \right| = 0$$<br><br>$${\Delta _y} = \left| {\matrix{
2 & 9 & 2 \cr
3 & 9 & 2 \cr
1 & 8 & 4 \cr
} } \right| = - 20$$<br><br>$${\Delta _z} = \left| {\matrix{
2 & 3 & 9 \cr
3 & 2 & 9 \cr
1 & { - 1} & 8 \cr
} } \right| = - 40$$<br><br>$$ \therefore $$ $$x = {{{\Delta _x}} \over \Delta } = 0$$<br><br>$$y = {{{\Delta _y}} \over \Delta } = 1$$<br><br>$$z = {{{\Delta _z}} \over \Delta } = 2$$<br><br>Unique solution : (0, 1, 2)
|
mcq
|
jee-main-2021-online-25th-february-evening-slot
|
GHIxtxm7Vb75iE0fjX1kluvwlfb
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
Consider the following system of equations :<br/><br/>x + 2y $$-$$ 3z = a<br/><br/>2x + 6y $$-$$ 11z = b<br/><br/>x $$-$$ 2y + 7z = c,<br/><br/>where a, b and c are real constants. Then the system of equations :
|
[{"identifier": "A", "content": "has no solution for all a, b and c"}, {"identifier": "B", "content": "has a unique solution when 5a = 2b + c"}, {"identifier": "C", "content": "has infinite number of solutions when 5a = 2b + c"}, {"identifier": "D", "content": "has a unique solution for all a, b and c"}]
|
["C"]
| null |
$$D = \left| {\matrix{
1 & 2 & { - 3} \cr
2 & 6 & { - 11} \cr
1 & { - 2} & 7 \cr
} } \right|$$<br><br>= 20 $$-$$ 2(25) $$-$$3($$-$$10)<br><br>= 20 $$-$$ 50 + 30 = 0<br><br>$${D_1} = \left| {\matrix{
a & 2 & { - 3} \cr
b & 6 & { - 11} \cr
c & { - 2} & 7 \cr
} } \right|$$<br><br>= 20a $$-$$ 2(7b + 11c) $$-$$3($$-$$2b $$-$$ 6c)<br><br>= 20a $$-$$ 14b $$-$$ 22c + 6b +18c<br><br>= 20a $$-$$ 8b $$-$$ 4c<br><br>= 4(5a $$-$$ 2b $$-$$ c)<br><br>$${D_2} = \left| {\matrix{
1 & a & { - 3} \cr
2 & b & { - 11} \cr
1 & c & 7 \cr
} } \right|$$<br><br>= 7b + 11c $$-$$ a(25) $$-$$3(2c $$-$$ b)<br><br>= 7b + 11c $$-$$ 25a $$-$$ 6c + 3b<br><br>= $$-$$25a + 10b + 5c<br><br>= $$-$$5(5a $$-$$ 2b $$-$$ c)<br><br>$${D_3} = \left| {\matrix{
1 & 2 & a \cr
2 & 6 & b \cr
1 & { - 2} & c \cr
} } \right|$$<br><br>= 6c + 2b $$-$$ 2(2c $$-$$ b) $$-$$ 10a<br><br>= $$-$$10a + 4b + 2c<br><br>= $$-$$2(5a $$-$$ 2b $$-$$ c)<br><br>for infinite solution <br><br>$$D = {D_1} = {D_2} = {D_3} = 0$$<br><br>$$ \Rightarrow $$ 5a = 2b + c
|
mcq
|
jee-main-2021-online-26th-february-evening-slot
|
C4InkmGv7kT3bDLA8n1kmhx159q
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
Let $$A = \left[ {\matrix{
i & { - i} \cr
{ - i} & i \cr
} } \right],i = \sqrt { - 1} $$. Then, the system of linear equations $${A^8}\left[ {\matrix{
x \cr
y \cr
} } \right] = \left[ {\matrix{
8 \cr
{64} \cr
} } \right]$$ has :
|
[{"identifier": "A", "content": "Exactly two solutions"}, {"identifier": "B", "content": "Infinitely many solutions"}, {"identifier": "C", "content": "A unique solution"}, {"identifier": "D", "content": "No solution"}]
|
["D"]
| null |
$$A = \left[ {\matrix{
i & { - i} \cr
{ - i} & i \cr
} } \right]$$<br><br>$${A^2} = \left[ {\matrix{
i & { - i} \cr
{ - i} & i \cr
} } \right]\left[ {\matrix{
i & { - i} \cr
{ - i} & i \cr
} } \right] = \left[ {\matrix{
{ - 2} & 2 \cr
2 & { - 2} \cr
} } \right] = 2\left[ {\matrix{
{ - 1} & 1 \cr
1 & { - 1} \cr
} } \right]$$<br><br>$${A^4} = 4\left[ {\matrix{
{ - 1} & 1 \cr
1 & { - 1} \cr
} } \right]\left[ {\matrix{
{ - 1} & 1 \cr
1 & { - 1} \cr
} } \right] = 4\left[ {\matrix{
2 & { - 2} \cr
{ - 2} & 2 \cr
} } \right] = 8\left[ {\matrix{
1 & { - 1} \cr
{ - 1} & 1 \cr
} } \right]$$<br><br>$${A^8} = 64\left[ {\matrix{
1 & { - 1} \cr
{ - 1} & 1 \cr
} } \right]\left[ {\matrix{
1 & { - 1} \cr
{ - 1} & 1 \cr
} } \right] = 64\left[ {\matrix{
2 & { - 2} \cr
{ - 2} & 2 \cr
} } \right] = 128\left[ {\matrix{
1 & { - 1} \cr
{ - 1} & 1 \cr
} } \right]$$<br><br>$$128\left[ {\matrix{
1 & { - 1} \cr
{ - 1} & 1 \cr
} } \right]\left[ {\matrix{
x \cr
y \cr
} } \right] = \left[ {\matrix{
8 \cr
{64} \cr
} } \right]$$<br><br>$$128\left[ {\matrix{
{x - y} \cr
{ - x + y} \cr
} } \right] = \left[ {\matrix{
8 \cr
{64} \cr
} } \right] $$
<br><br>$$\Rightarrow 128(x - y) = 8$$<br><br>$$ \Rightarrow x - y = {1 \over {16}}$$ .... (1)<br><br>and $$128( - x + y) = 64 \Rightarrow x - y = {{ - 1} \over 2}$$ .... (2)<br><br>$$ \Rightarrow $$ No solution (from eq. (1) & (2))
|
mcq
|
jee-main-2021-online-16th-march-morning-shift
|
cauxZPLYUekDhfcjXz1kmjb3sh3
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
The system of equations kx + y + z = 1, x + ky + z = k and x + y + zk = k<sup>2</sup> has no solution if k is equal to :
|
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$$-$$1"}, {"identifier": "C", "content": "$$-$$2"}, {"identifier": "D", "content": "1"}]
|
["C"]
| null |
$$D = \left| {\matrix{
k & 1 & 1 \cr
1 & k & 1 \cr
1 & 1 & k \cr
} } \right| = 0$$<br><br>$$ \Rightarrow k({k^2} - 1) - (k - 1) + (1 - k) = 0$$<br><br>$$ \Rightarrow (k - 1)({k^2} + k - 1 - 1) = 0$$<br><br>$$ \Rightarrow (k - 1)({k^2} + k - 2) = 0$$<br><br>$$ \Rightarrow (k - 1)(k - 1)(k + 2) = 0$$<br><br>$$ \Rightarrow k = 1,k = -2$$<br><br>for k = 1 equation identical so k = $$-$$2 for no solution.
|
mcq
|
jee-main-2021-online-17th-march-morning-shift
|
852vLD8DVyX1VNm4W01kmli9o9n
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
Let $$\alpha$$, $$\beta$$, $$\gamma$$ be the real roots of the equation, x<sup>3</sup> + ax<sup>2</sup> + bx + c = 0, (a, b, c $$\in$$ R and a, b $$\ne$$ 0). If the system of equations (in u, v, w) given by $$\alpha$$u + $$\beta$$v + $$\gamma$$w = 0, $$\beta$$u + $$\gamma$$v + $$\alpha$$w = 0; $$\gamma$$u + $$\alpha$$v + $$\beta$$w = 0 has non-trivial solution, then the value of $${{{a^2}} \over b}$$ is
|
[{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "0"}]
|
["B"]
| null |
x<sup>3</sup> + ax<sup>2</sup> + bx + c = 0 <br><br>Roots are $$\alpha$$, $$\beta$$, $$\gamma$$.<br><br>For non-trivial solutions,<br><br>$$\left| {\matrix{
\alpha & \beta & \gamma \cr
\beta & \gamma & \alpha \cr
\gamma & \alpha & \beta \cr
} } \right| = 0$$<br><br>$$ \Rightarrow $$ $${\alpha ^3} + {\beta ^3} + {\gamma ^3} - 3\alpha \beta \gamma = 0$$<br><br>$$ \Rightarrow $$ $$(\alpha + \beta + \gamma )\left[ {{{\left( {\alpha + \beta + \alpha } \right)}^2} - 3\left( {\sum {\alpha \beta } } \right)} \right] = 0$$<br><br>$$ \Rightarrow $$ $$( - a)[{a^2} - 3b] = 0$$<br><br>$$ \Rightarrow $$ $${a^2} = 3b$$ ($$ \because $$ a $$ \ne $$ 0)<br><br>$$ \Rightarrow {{{a^2}} \over b} = 3$$
|
mcq
|
jee-main-2021-online-18th-march-morning-shift
|
ejJWcC5YgNq1NbYw5C1kmm2kpfg
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
Let the system of linear equations <br/><br/>4x + $$\lambda$$y + 2z = 0<br/><br/>2x $$-$$ y + z = 0<br/><br/>$$\mu$$x + 2y + 3z = 0, $$\lambda$$, $$\mu$$$$\in$$R.<br/><br/>has a non-trivial solution. Then which of the following is true?
|
[{"identifier": "A", "content": "$$\\mu$$ = 6, $$\\lambda$$$$\\in$$R"}, {"identifier": "B", "content": "$$\\lambda$$ = 3, $$\\mu$$$$\\in$$R"}, {"identifier": "C", "content": "$$\\mu$$ = $$-$$6, $$\\lambda$$$$\\in$$R"}, {"identifier": "D", "content": "$$\\lambda$$ = 2, $$\\mu$$$$\\in$$R"}]
|
["A"]
| null |
<p>Given, system of linear equations</p>
<p>4x + $$\lambda$$y + 2z = 0</p>
<p>2x $$-$$ y + z = 0</p>
<p>$$\mu$$x + 2y + 3z = 0</p>
<p>For non-trivial solution, $$\Delta$$ = 0</p>
<p>$$\left| {\matrix{
4 & \lambda & 2 \cr
2 & { - 1} & 1 \cr
\mu & 2 & 3 \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow 4( - 3 - 2) - \lambda (6 - \mu ) + 2(4 + \mu ) = 0$$</p>
<p>$$ \Rightarrow - \lambda (6 - \mu ) - 2(6 - \mu ) = 0$$</p>
<p>$$ \Rightarrow (6 - \mu )(\lambda + 2) = 0$$</p>
<p>$$ \Rightarrow \lambda = - 2$$ and $$\mu \in R$$ or $$\mu$$ = 6 and $$\lambda \in R$$.</p>
|
mcq
|
jee-main-2021-online-18th-march-evening-shift
|
1krruoai8
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
The value of k $$\in$$R, for which the following system of linear equations<br/><br/>3x $$-$$ y + 4z = 3,<br/><br/>x + 2y $$-$$ 3z = $$-$$2<br/><br/>6x + 5y + kz = $$-$$3,<br/><br/>has infinitely many solutions, is :
|
[{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$$-$$5"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "$$-$$3"}]
|
["B"]
| null |
$$\left| {\matrix{
3 & { - 1} & 4 \cr
1 & 2 & { - 3} \cr
6 & 5 & k \cr
} } \right| = 0$$<br><br>$$\Rightarrow$$ 3(2k + 15) + K + 18 $$-$$ 28 = 0<br><br>$$\Rightarrow$$ 7k + 35 = 0 <br><br>$$\Rightarrow$$ k = $$-$$ 5
|
mcq
|
jee-main-2021-online-20th-july-evening-shift
|
1krthv2f5
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
The values of $$\lambda$$ and $$\mu$$ such that the system of equations $$x + y + z = 6$$, $$3x + 5y + 5z = 26$$, $$x + 2y + \lambda z = \mu $$ has no solution, are :
|
[{"identifier": "A", "content": "$$\\lambda$$ = 3, $$\\mu$$ = 5"}, {"identifier": "B", "content": "$$\\lambda$$ = 3, $$\\mu$$ $$\\ne$$ 10"}, {"identifier": "C", "content": "$$\\lambda$$ $$\\ne$$ 2, $$\\mu$$ = 10"}, {"identifier": "D", "content": "$$\\lambda$$ = 2, $$\\mu$$ $$\\ne$$ 10"}]
|
["D"]
| null |
$$x + y + z = 6$$ ..... (i)<br><br>$$3x + 5y + 5z = 26$$ .... (ii)<br><br>$$x + 2y + \lambda z = \mu $$ ..... (iii)<br><br>$$5 \times (i) - (ii) \Rightarrow 2x = 4 \Rightarrow x = 2$$<br><br>$$\therefore$$ from (i) and (iii)<br><br>$$y + z = 4$$ ..... (iv)<br><br>$$2y + \lambda z = \mu - 2$$ .....(v)<br><br>$$(v) - 2 \times (iv)$$<br><br>$$ \Rightarrow (\lambda - 2)z = \mu - 10$$<br><br>$$ \Rightarrow z = {{\mu - 10} \over {\lambda - 2}}$$ & $$y = 4 - {{\mu - 10} \over {\lambda - 2}}$$<br><br>$$\therefore$$ For no solution $$\lambda$$ = 2 and $$\mu$$ $$\ne$$ 10.
|
mcq
|
jee-main-2021-online-22th-july-evening-shift
|
1krw0ttza
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
The values of a and b, for which the system of equations <br/><br/>2x + 3y + 6z = 8<br/><br/>x + 2y + az = 5<br/><br/>3x + 5y + 9z = b<br/><br/>has no solution, are :
|
[{"identifier": "A", "content": "a = 3, b $$\\ne$$ 13"}, {"identifier": "B", "content": "a $$\\ne$$ 3, b $$\\ne$$ 13"}, {"identifier": "C", "content": "a $$\\ne$$ 3, b = 3"}, {"identifier": "D", "content": "a = 3, b = 13"}]
|
["A"]
| null |
$$D = \left| {\matrix{
2 & 3 & 6 \cr
1 & 2 & a \cr
3 & 5 & 9 \cr
} } \right| = 3 - a$$<br><br>$$D = \left| {\matrix{
2 & 3 & 8 \cr
1 & 2 & 5 \cr
3 & 5 & b \cr
} } \right| = b - 13$$<br><br>If a = 3, b $$\ne$$ 13, no solution.
|
mcq
|
jee-main-2021-online-25th-july-morning-shift
|
1ks0bjb5g
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
For real numbers $$\alpha$$ and $$\beta$$, consider the following system of linear equations :<br/><br/>x + y $$-$$ z = 2, x + 2y + $$\alpha$$z = 1, 2x $$-$$ y + z = $$\beta$$. If the system has infinite solutions, then $$\alpha$$ + $$\beta$$ is equal to ______________.
|
[]
| null |
5
|
For infinite solutions<br><br>$$\Delta$$ = $$\Delta$$<sub>1</sub> = $$\Delta$$<sub>2</sub> = $$\Delta$$<sub>3</sub> = 0<br><br>$$\Delta$$ = $$\left| {\matrix{
1 & 1 & { - 1} \cr
1 & 2 & \alpha \cr
2 & { - 1} & 1 \cr
} } \right| = 0$$<br><br>$$\Delta = \left| {\matrix{
3 & 0 & 0 \cr
1 & 2 & \alpha \cr
2 & { - 1} & 1 \cr
} } \right| = 0$$<br><br>$$\Delta$$ = 3(2 + $$\alpha$$) = 0<br><br>$$\Rightarrow$$ $$\alpha$$ = $$-$$2<br><br>$${\Delta _2} = \left| {\matrix{
1 & 2 & { - 1} \cr
1 & 1 & { - 2} \cr
2 & \beta & 1 \cr
} } \right| = 0$$<br><br>1(1 + 2$$\beta$$) $$-$$2(1 + 4) $$-$$ ($$\beta$$ $$-$$ 2) = 0<br><br> $$\beta$$ $$-$$ 7 = 0<br><br>$$\beta$$ = 7<br><br>$$\therefore$$ $$\alpha$$ + $$\beta$$ = 5
|
integer
|
jee-main-2021-online-27th-july-morning-shift
|
1ktbc2w2e
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
Let $$\theta \in \left( {0,{\pi \over 2}} \right)$$. If the system of linear equations<br/><br/>$$(1 + {\cos ^2}\theta )x + {\sin ^2}\theta y + 4\sin 3\,\theta z = 0$$<br/><br/>$${\cos ^2}\theta x + (1 + {\sin ^2}\theta )y + 4\sin 3\,\theta z = 0$$<br/><br/>$${\cos ^2}\theta x + {\sin ^2}\theta y + (1 + 4\sin 3\,\theta )z = 0$$<br/><br/>has a non-trivial solution, then the value of $$\theta$$ is :
|
[{"identifier": "A", "content": "$${{4\\pi } \\over 9}$$"}, {"identifier": "B", "content": "$${{7\\pi } \\over {18}}$$"}, {"identifier": "C", "content": "$${\\pi \\over {18}}$$"}, {"identifier": "D", "content": "$${{5\\pi } \\over {18}}$$"}]
|
["B"]
| null |
$$\left| {\matrix{
{1 + {{\cos }^2}\theta } & {{{\sin }^2}\theta } & {4\sin 3\,\theta } \cr
{{{\cos }^2}\theta } & {1 + {{\sin }^2}\theta } & {4\sin 3\,\theta } \cr
{{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\sin 3\,\theta } \cr
} } \right| = 0$$<br><br>$${C_1} \to {C_1} + {C_2}$$<br><br>$$\left| {\matrix{
2 & {{{\sin }^2}\theta } & {4\sin 3\,\theta } \cr
2 & {1 + {{\sin }^2}\theta } & {4\sin 3\,\theta } \cr
1 & {{{\sin }^2}\theta } & {1 + 4\sin 3\,\theta } \cr
} } \right| = 0$$<br><br>$${R_1} \to {R_1} - {R_2},{R_2} \to {R_2} - {R_3}$$<br><br>$$\left| {\matrix{
0 & { - 1} & 0 \cr
1 & 1 & { - 1} \cr
1 & {{{\sin }^2}\theta } & {1 + 4\sin 3\,\theta } \cr
} } \right| = 0$$<br><br>or $$4\sin 3\theta = - 2$$<br><br>$$\sin 3\theta = - {1 \over 2}$$<br><br>$$\theta = {{7\pi } \over {18}}$$
|
mcq
|
jee-main-2021-online-26th-august-morning-shift
|
1ktepe5b4
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
If the system of linear equations<br/><br/>2x + y $$-$$ z = 3<br/><br/>x $$-$$ y $$-$$ z = $$\alpha$$<br/><br/>3x + 3y + $$\beta$$z = 3<br/><br/>has infinitely many solution, then $$\alpha$$ + $$\beta$$ $$-$$ $$\alpha$$$$\beta$$ is equal to _____________.
|
[]
| null |
5
|
2 $$\times$$ (i) $$-$$ (ii) $$-$$ (iii) gives :<br><br>$$-$$ (1 + $$\beta$$)z = 3 $$-$$ $$\alpha$$<br><br>For infinitely many solution<br><br>$$\beta$$ + 1 = 0 = 3 $$-$$ $$\alpha$$ $$\Rightarrow$$ ($$\alpha$$, $$\beta$$) = (3, $$-$$1)<br><br>Hence, $$\alpha$$ + $$\beta$$ $$-$$ $$\alpha$$$$\beta$$ = 5
|
integer
|
jee-main-2021-online-27th-august-morning-shift
|
1ktg0cvqr
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
Let [$$\lambda$$] be the greatest integer less than or equal to $$\lambda$$. The set of all values of $$\lambda$$ for which the system of linear equations <br/>x + y + z = 4, <br/>3x + 2y + 5z = 3, <br/>9x + 4y + (28 + [$$\lambda$$])z = [$$\lambda$$] has a solution is :
|
[{"identifier": "A", "content": "R"}, {"identifier": "B", "content": "($$-$$$$\\infty$$, $$-$$9) $$\\cup$$ ($$-$$9, $$\\infty$$)"}, {"identifier": "C", "content": "[$$-$$9, $$-$$8)"}, {"identifier": "D", "content": "($$-$$$$\\infty$$, $$-$$9) $$\\cup$$ [$$-$$8, $$\\infty$$)"}]
|
["A"]
| null |
$$D = \left| {\matrix{
1 & 1 & 1 \cr
3 & 2 & 5 \cr
9 & 4 & {28 + [\lambda ]} \cr
} } \right| = - 24 - [\lambda ] + 15 = - [\lambda ] - 9$$<br><br>if $$[\lambda ] + 9 \ne 0$$ then unique solution<br><br>if $$[\lambda ] + 9 = 0$$ then D<sub>1</sub> = D<sub>2</sub> = D<sub>3</sub> = 0<br><br>so infinite solutions<br><br>Hence, $$\lambda$$ can be any red number.
|
mcq
|
jee-main-2021-online-27th-august-evening-shift
|
1ktiqe5l2
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
If the following system of linear equations<br/><br/>2x + y + z = 5<br/><br/>x $$-$$ y + z = 3<br/><br/>x + y + az = b<br/><br/>has no solution, then :
|
[{"identifier": "A", "content": "$$a = - {1 \\over 3},b \\ne {7 \\over 3}$$"}, {"identifier": "B", "content": "$$a \\ne {1 \\over 3},b = {7 \\over 3}$$"}, {"identifier": "C", "content": "$$a \\ne - {1 \\over 3},b = {7 \\over 3}$$"}, {"identifier": "D", "content": "$$a = {1 \\over 3},b \\ne {7 \\over 3}$$"}]
|
["D"]
| null |
Here $$D = \left| {\matrix{
2 & 1 & 1 \cr
1 & { - 1} & 1 \cr
1 & 1 & a \cr
} } \right|\matrix{
{ = 2(a - 1) - 1(a - 1) + 1 + 1} \cr
{ = 1 - 3a} \cr
} $$<br><br>$${D_3} = \left| {\matrix{
2 & 1 & 5 \cr
1 & { - 1} & 3 \cr
1 & 1 & b \cr
} } \right|\matrix{
{ = 2( - b - 3) - 1(b - 3) + 5(1 + 1)} \cr
{ = 7 - 3b} \cr
} $$<br><br>for $$a = {1 \over 3},b \ne {7 \over 3}$$, system has no solutions.
|
mcq
|
jee-main-2021-online-31st-august-morning-shift
|
1ktk3cn98
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
If $$\alpha$$ + $$\beta$$ + $$\gamma$$ = 2$$\pi$$, then the system of equations <br/><br/>x + (cos $$\gamma$$)y + (cos $$\beta$$)z = 0<br/><br/>(cos $$\gamma$$)x + y + (cos $$\alpha$$)z = 0<br/><br/>(cos $$\beta$$)x + (cos $$\alpha$$)y + z = 0<br/><br/>has :
|
[{"identifier": "A", "content": "no solution"}, {"identifier": "B", "content": "infinitely many solution"}, {"identifier": "C", "content": "exactly two solutions"}, {"identifier": "D", "content": "a unique solution"}]
|
["B"]
| null |
<p>Given $$\alpha$$ + $$\beta$$ + $$\gamma$$ = 2$$\pi$$</p>
<p>$$\Delta = \left| {\matrix{
1 & {\cos \gamma } & {\cos \beta } \cr
{\cos \gamma } & 1 & {\cos \alpha } \cr
{\cos \beta } & {\cos \alpha } & 1 \cr
} } \right|$$</p>
<p>$$ = 1 - {\cos ^2}\alpha - \cos \gamma (\cos \gamma - \cos \alpha \cos \beta ) + \cos \beta (\cos \alpha \cos \gamma - \cos \beta )$$</p>
<p>$$ = 1 - {\cos ^2}\alpha - {\cos ^2}\beta - {\cos ^2}\gamma + 2\cos \alpha \cos \beta \cos \gamma $$</p>
<p>$$ = {\sin ^2}\alpha - {\cos ^2}\beta - \cos \gamma (\cos \gamma - 2\cos \alpha \cos \beta )$$</p>
<p>$$ = - \cos (\alpha + \beta )\cos (\alpha - \beta ) - \cos \gamma (\cos (2\pi - (\alpha - \beta )) - 2\cos \alpha \cos \beta )$$</p>
<p>$$ = - \cos (2\pi - \gamma )\cos (\alpha - \beta ) - \cos \gamma (\cos (\alpha + \beta ) - 2\cos \alpha \cos \beta )$$</p>
<p>$$ = - \cos \gamma \cos (\alpha - \beta ) + \cos \gamma (\cos \alpha \cos \beta + \sin \alpha \sin \beta )$$</p>
<p>$$ = - \cos \gamma \cos (\alpha - \beta ) + \cos \gamma \cos (\alpha - \beta )$$</p>
<p>$$ = 0$$</p>
<p>So, the system of equation has infinitely many solutions.</p>
|
mcq
|
jee-main-2021-online-31st-august-evening-shift
|
1ktnzxwbq
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
Consider the system of linear equations<br/><br/>$$-$$x + y + 2z = 0<br/><br/>3x $$-$$ ay + 5z = 1<br/><br/>2x $$-$$ 2y $$-$$ az = 7<br/><br/>Let S<sub>1</sub> be the set of all a$$\in$$R for which the system is inconsistent and S<sub>2</sub> be the set of all a$$\in$$R for which the system has infinitely many solutions. If n(S<sub>1</sub>) and n(S<sub>2</sub>) denote the number of elements in S<sub>1</sub> and S<sub>2</sub> respectively, then
|
[{"identifier": "A", "content": "n(S<sub>1</sub>) = 2, n(S<sub>2</sub>) = 2"}, {"identifier": "B", "content": "n(S<sub>1</sub>) = 1, n(S<sub>2</sub>) = 0"}, {"identifier": "C", "content": "n(S<sub>1</sub>) = 2, n(S<sub>2</sub>) = 0"}, {"identifier": "D", "content": "n(S<sub>1</sub>) = 0, n(S<sub>2</sub>) = 2"}]
|
["C"]
| null |
$$\Delta = \left| {\matrix{
{ - 1} & 1 & 2 \cr
3 & { - a} & 5 \cr
2 & { - 2} & { - a} \cr
} } \right|$$<br><br>$$ = - 1({a^2} + 10) - 1( - 3a - 10) + 2( - 6 + 2a)$$<br><br>$$ = - {a^2} - 10 + 3a + 10 - 12 + 4a$$<br><br>$$\Delta = - {a^2} + 7a - 12$$<br><br>$$\Delta = - [{a^2} - 7a + 12]$$<br><br>$$\Delta = - [(a - 3)(a - 4)]$$<br><br>$${\Delta _1} = \left| {\matrix{
0 & 1 & 2 \cr
1 & { - a} & 5 \cr
7 & { - 2} & { - a} \cr
} } \right|$$<br><br><br>$$ = a + 35 - 4 + 14a$$<br><br>= $$15a + 31$$<br><br>Now, $${\Delta _1} = 15a + 31$$<br><br>For inconsistent $$\Delta$$ = 0 $$\therefore$$ a = 3, a = 4 and for a = 3 and 4, $$\Delta$$<sub>1</sub> $$\ne$$ 0<br><br>n(S<sub>1</sub>) = 2<br><br>For infinite solution : $$\Delta$$ = 0 and $$\Delta$$<sub>1</sub> = $$\Delta$$<sub>2</sub> = $$\Delta$$<sub>3</sub> = 0<br><br>Not possible<br><br>$$\therefore$$ n(S<sub>2</sub>) = 0
|
mcq
|
jee-main-2021-online-1st-september-evening-shift
|
1l544wdkc
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>If the system of linear equations</p>
<p>2x + y $$-$$ z = 7</p>
<p>x $$-$$ 3y + 2z = 1</p>
<p>x + 4y + $$\delta$$z = k, where $$\delta$$, k $$\in$$ R has infinitely many solutions, then $$\delta$$ + k is equal to:</p>
|
[{"identifier": "A", "content": "$$-$$3"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "9"}]
|
["B"]
| null |
<p>$$2x + y - z = 7$$</p>
<p>$$x - 3y + 2z = 1$$</p>
<p>$$x + 4y + \delta z = k$$</p>
<p>$$\Delta = \left| {\matrix{
2 & 1 & { - 1} \cr
1 & { - 3} & 2 \cr
1 & 4 & \delta \cr
} } \right| = - 7\delta - 21 = 0$$</p>
<p>$$\delta = - 3$$</p>
<p>$${\Delta _1} = \left| {\matrix{
7 & 1 & { - 1} \cr
1 & { - 3} & 2 \cr
k & 4 & { - 3} \cr
} } \right|$$</p>
<p>$$ \Rightarrow 6 - k = 0 \Rightarrow k = 6$$</p>
<p>$$\delta + k = - 3 + 6 = 3$$</p>
|
mcq
|
jee-main-2022-online-29th-june-morning-shift
|
1l55j68za
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>If the system of linear equations <br/>$$2x - 3y = \gamma + 5$$, <br/>$$\alpha x + 5y = \beta + 1$$, where $$\alpha$$, $$\beta$$, $$\gamma$$ $$\in$$ R has infinitely many solutions then the value <br/>of | 9$$\alpha$$ + 3$$\beta$$ + 5$$\gamma$$ | is equal to ____________.</p>
|
[]
| null |
58
|
<p>If 2x $$-$$ 3y = $$\gamma$$ + 5 and $$\alpha$$x + 5y = $$\beta$$ + 1 have infinitely many solutions then</p>
<p>$${2 \over \alpha } = {{ - 3} \over 5} = {{\gamma + 5} \over {\beta + 1}}$$</p>
<p>$$ \Rightarrow \alpha = - {{10} \over 3}$$ and $$3\beta + 5\gamma = - 28$$</p>
<p>So $$|9\alpha + 3\beta + 5\gamma | = | - 30 - 28| = 58$$</p>
|
integer
|
jee-main-2022-online-28th-june-evening-shift
|
1l56668zy
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>If the system of linear equations</p>
<p>$$2x + 3y - z = - 2$$</p>
<p>$$x + y + z = 4$$</p>
<p>$$x - y + |\lambda |z = 4\lambda - 4$$</p>
<p>where, $$\lambda$$ $$\in$$ R, has no solution, then</p>
|
[{"identifier": "A", "content": "$$\\lambda$$ = 7"}, {"identifier": "B", "content": "$$\\lambda$$ = $$-$$7"}, {"identifier": "C", "content": "$$\\lambda$$ = 8"}, {"identifier": "D", "content": "$$\\lambda$$<sup>2</sup> = 1"}]
|
["B"]
| null |
<p>$$\Delta = \left| {\matrix{
2 & 3 & { - 1} \cr
1 & 1 & 1 \cr
1 & { - 1} & {|\lambda |} \cr
} } \right| = 0 \Rightarrow |\lambda | = 7$$</p>
<p>But at $$\lambda = 7,\,{D_x} = {D_y} = {D_z} = 0$$</p>
<p>$${P_1}:2x + 3y - z = - 2$$</p>
<p>$${P_2}:x + y + z = 4$$</p>
<p>$${P_3}:x - y + |\lambda |z = 4\lambda - 4$$</p>
<p>So clearly $$5{P_2} - 2{P_1} = {P_3}$$, so at $$\lambda = 7$$, system of equation is having infinite solutions.</p>
<p>So $$\lambda = - 7$$ is correct answer.</p>
|
mcq
|
jee-main-2022-online-28th-june-morning-shift
|
1l57nuni9
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>Let the system of linear equations <br/>$$x + 2y + z = 2$$, <br/>$$\alpha x + 3y - z = \alpha $$, <br/>$$ - \alpha x + y + 2z = - \alpha $$ <br/>be inconsistent. Then $$\alpha$$ is equal to :</p>
|
[{"identifier": "A", "content": "$${5 \\over 2}$$"}, {"identifier": "B", "content": "$$-$$$${5 \\over 2}$$"}, {"identifier": "C", "content": "$${7 \\over 2}$$"}, {"identifier": "D", "content": "$$-$$$${7 \\over 2}$$"}]
|
["D"]
| null |
<p>$$x + 2y + z = 2$$</p>
<p>$$\alpha x + 3y - z = \alpha $$</p>
<p>$$ - \alpha x + y + 2z = - \alpha $$</p>
<p>$$\Delta = \left| {\matrix{
1 & 2 & 1 \cr
\alpha & 3 & { - 1} \cr
{ - \alpha } & 1 & 2 \cr
} } \right| = 1(6 + 1) - 2(2\alpha - \alpha ) + 1(\alpha + 3\alpha )$$</p>
<p>$$ = 7 + 2\alpha $$</p>
<p>$$\Delta = 0 \Rightarrow \alpha = - {7 \over 2}$$</p>
<p>$${\Delta _1} = \left| {\matrix{
2 & 2 & 1 \cr
\alpha & 3 & { - 1} \cr
{ - \alpha } & 1 & 2 \cr
} } \right| = 14 + 2\alpha \ne 0$$ for $$\alpha = - {7 \over 2}$$</p>
<p>$$\therefore$$ For no solution $$\alpha = - {7 \over 2}$$</p>
|
mcq
|
jee-main-2022-online-27th-june-morning-shift
|
1l587hfiu
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>The ordered pair (a, b), for which the system of linear equations</p>
<p>3x $$-$$ 2y + z = b</p>
<p>5x $$-$$ 8y + 9z = 3</p>
<p>2x + y + az = $$-$$1</p>
<p>has no solution, is :</p>
|
[{"identifier": "A", "content": "$$\\left( {3,{1 \\over 3}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - 3,{1 \\over 3}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - 3, - {1 \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {3, - {1 \\over 3}} \\right)$$"}]
|
["C"]
| null |
<p>$$\left| {\matrix{
3 & { - 2} & 1 \cr
5 & { - 8} & 9 \cr
2 & 1 & a \cr
} } \right| = 0 \Rightarrow - 14a - 42 = 0 \Rightarrow a = - 3$$</p>
<p>Now 3 (equation (1)) $$-$$ (equation (2)) $$-$$ 2 (equation (3)) is</p>
<p>$$3(3x - 2y + z - b) - (5x - 8y + 9z - 3) - 2(2x + y + az + 1) = 0$$</p>
<p>$$ \Rightarrow - 3b + 3 - 2 = 0 \Rightarrow b = {1 \over 3}$$</p>
<p>So for no solution $$a = - 3$$ and $$b \ne {1 \over 3}$$
|
mcq
|
jee-main-2022-online-26th-june-morning-shift
|
1l58ez9zk
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>If the system of equations</p>
<p>$$\alpha$$x + y + z = 5, x + 2y + 3z = 4, x + 3y + 5z = $$\beta$$</p>
<p>has infinitely many solutions, then the ordered pair ($$\alpha$$, $$\beta$$) is equal to :</p>
|
[{"identifier": "A", "content": "(1, $$-$$3)"}, {"identifier": "B", "content": "($$-$$1, 3)"}, {"identifier": "C", "content": "(1, 3)"}, {"identifier": "D", "content": "($$-$$1, $$-$$3)"}]
|
["C"]
| null |
<p>Given system of equations</p>
<p>$$\alpha x + y + z = 5$$</p>
<p>$$x + 2y + 3z = 4$$, has infinite solution</p>
<p>$$x + 3y + 5z = \beta $$</p>
<p>$$\therefore$$ $$\Delta = \left| {\matrix{
\alpha & 1 & 1 \cr
1 & 2 & 3 \cr
1 & 3 & 5 \cr
} } \right| = 0 \Rightarrow \alpha (1) - 1(2) + 1(1) = 0$$</p>
<p>$$ \Rightarrow \alpha = 1$$</p>
<p>and $${\Delta _1} = \left| {\matrix{
5 & 1 & 1 \cr
4 & 2 & 3 \cr
\beta & 3 & 5 \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow 5(1) - 1(20 - 3\beta ) + 1(12 - 2\beta ) = 0$$</p>
<p>$$ \Rightarrow \beta = 3$$</p>
<p>and $${\Delta _2} = \left| {\matrix{
1 & 5 & 1 \cr
1 & 4 & 3 \cr
1 & \beta & 5 \cr
} } \right| = 0 \Rightarrow (20 - 3\beta ) - 5(2) + 1(\beta - 4) = 0$$</p>
<p>$$ \Rightarrow - 2\beta + 6 = 0$$</p>
<p>$$ \Rightarrow \beta = 3$$</p>
<p>Similarly,</p>
<p>$$ \Rightarrow {\Delta _3} = \left| {\matrix{
1 & 1 & 5 \cr
1 & 2 & 4 \cr
1 & 3 & \beta \cr
} } \right| = 0 \Rightarrow \beta = 3$$</p>
<p>$$\therefore$$ ($$\alpha$$, $$\beta$$) = (1, 3)</p>
|
mcq
|
jee-main-2022-online-26th-june-evening-shift
|
1l59js96c
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>The system of equations</p>
<p>$$ - kx + 3y - 14z = 25$$</p>
<p>$$ - 15x + 4y - kz = 3$$</p>
<p>$$ - 4x + y + 3z = 4$$</p>
<p>is consistent for all k in the set</p>
|
[{"identifier": "A", "content": "R"}, {"identifier": "B", "content": "R $$-$$ {$$-$$11, 13}"}, {"identifier": "C", "content": "R $$-$$ {13}"}, {"identifier": "D", "content": "R $$-$$ {$$-$$11, 11}"}]
|
["D"]
| null |
<p>The system may be inconsistent if</p>
<p>$$\left| {\matrix{
{ - k} & 3 & { - 14} \cr
{ - 15} & 4 & { - k} \cr
{ - 4} & 1 & 3 \cr
} } \right| = 0 \Rightarrow k = \, \pm \,11$$</p>
<p>Hence if system is consistent then $$k \in R - \{ 11, - 11\} $$.</p>
|
mcq
|
jee-main-2022-online-25th-june-evening-shift
|
1l5ahzzd3
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>Let A be a 3 $$\times$$ 3 real matrix such that</p>
<p>$$A\left( {\matrix{
1 \cr
1 \cr
0 \cr
} } \right) = \left( {\matrix{
1 \cr
1 \cr
0 \cr
} } \right);A\left( {\matrix{
1 \cr
0 \cr
1 \cr
} } \right) = \left( {\matrix{
{ - 1} \cr
0 \cr
1 \cr
} } \right)$$ and $$A\left( {\matrix{
0 \cr
0 \cr
1 \cr
} } \right) = \left( {\matrix{
1 \cr
1 \cr
2 \cr
} } \right)$$.</p>
<p>If $$X = {({x_1},{x_2},{x_3})^T}$$ and I is an identity matrix of order 3, then the system $$(A - 2I)X = \left( {\matrix{
4 \cr
1 \cr
1 \cr
} } \right)$$ has :</p>
|
[{"identifier": "A", "content": "no solution"}, {"identifier": "B", "content": "infinitely many solutions"}, {"identifier": "C", "content": "unique solution"}, {"identifier": "D", "content": "exactly two solutions"}]
|
["B"]
| null |
<p>Let $$A = \left[ {\matrix{
a & b & c \cr
d & e & f \cr
g & h & i \cr
} } \right]$$</p>
<p>$$A = \left[ {\matrix{
1 \cr
1 \cr
0 \cr
} } \right] = \left[ {\matrix{
1 \cr
1 \cr
0 \cr
} } \right] \Rightarrow \left[ {\matrix{
a & b & c \cr
d & e & f \cr
g & h & i \cr
} } \right]\left[ {\matrix{
1 \cr
1 \cr
0 \cr
} } \right] = \left[ {\matrix{
1 \cr
1 \cr
0 \cr
} } \right] \Rightarrow \matrix{
{a + b = 1} \cr
{d + e = 1} \cr
{g + h = 0} \cr
} $$</p>
<p>$$A = \left[ {\matrix{
1 \cr
0 \cr
1 \cr
} } \right] = \left[ {\matrix{
{ - 1} \cr
0 \cr
1 \cr
} } \right] \Rightarrow \left[ {\matrix{
a & b & c \cr
d & e & f \cr
g & h & i \cr
} } \right]\left[ {\matrix{
1 \cr
0 \cr
1 \cr
} } \right] = \left[ {\matrix{
{ - 1} \cr
0 \cr
1 \cr
} } \right] \Rightarrow \matrix{
{a + c = - 1} \cr
{d + f = 1} \cr
{g + i = 0} \cr
} $$</p>
<p>$$A = \left[ {\matrix{
0 \cr
0 \cr
1 \cr
} } \right] = \left[ {\matrix{
1 \cr
1 \cr
2 \cr
} } \right] \Rightarrow \left[ {\matrix{
a & b & c \cr
d & e & f \cr
g & h & i \cr
} } \right]\left[ {\matrix{
0 \cr
0 \cr
1 \cr
} } \right] = \left[ {\matrix{
1 \cr
1 \cr
2 \cr
} } \right] \Rightarrow \matrix{
{c = 1} \cr
{f = 1} \cr
{i = 2} \cr
} $$</p>
<p>Solving will get</p>
<p>$$a = - 2,\,b = 3,\,c = 1,\,d = - 1,\,e = 2,\,f = 1,\,g = - 1,\,h = 1,\,i = 2$$</p>
<p>$$A = \left[ {\matrix{
{ - 2} & 3 & 1 \cr
{ - 1} & 2 & 1 \cr
{ - 1} & 1 & 2 \cr
} } \right] \Rightarrow A = 2I = \left[ {\matrix{
{ - 4} & 3 & 1 \cr
{ - 1} & 0 & 1 \cr
{ - 1} & 1 & 0 \cr
} } \right]$$</p>
<p>$$(A - 2I)x = \left[ {\matrix{
4 \cr
1 \cr
1 \cr
} } \right]$$</p>
<p>$$ \Rightarrow - 4{x_1} + 3{x_2} + {x_3} = 4$$ ..... (i)</p>
<p>$$ - {x_1} + {x_3} = 1$$ ...... (ii)</p>
<p>$$ - {x_1} + {x_2} = 1$$ ...... (iii)</p>
<p>So 3(iii) + (ii) = (i)</p>
<p>$$\therefore$$ Infinite solution</p>
|
mcq
|
jee-main-2022-online-25th-june-morning-shift
|
1l5b7v78i
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>Let the system of linear equations</p>
<p>x + y + $$\alpha$$z = 2</p>
<p>3x + y + z = 4</p>
<p>x + 2z = 1</p>
<p>have a unique solution (x$$^ * $$, y$$^ * $$, z$$^ * $$). If ($$\alpha$$, x$$^ * $$), (y$$^ * $$, $$\alpha$$) and (x$$^ * $$, $$-$$y$$^ * $$) are collinear points, then the sum of absolute values of all possible values of $$\alpha$$ is</p>
|
[{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}]
|
["C"]
| null |
<p>Given system of equations</p>
<p>$$x + y + az = 2$$ ..... (i)</p>
<p>$$3x + y + z = 4$$ ..... (ii)</p>
<p>$$x + 2z = 1$$ ..... (iii)</p>
<p>Solving (i), (ii) and (iii), we get</p>
<p>x = 1, y = 1, z = 0 (and for unique solution a $$\ne$$ $$-$$3)</p>
<p>Now, ($$\alpha$$, 1), (1, $$\alpha$$) and (1, $$-$$1) are collinear</p>
<p>$$\therefore$$ $$\left| {\matrix{
\alpha & 1 & 1 \cr
1 & \alpha & 1 \cr
1 & { - 1} & 1 \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow \alpha (\alpha + 1) - 1(0) + 1( - 1 - \alpha ) = 0$$</p>
<p>$$ \Rightarrow {\alpha ^2} - 1 = 0$$</p>
<p>$$\therefore$$ $$\alpha = \, \pm \,1$$</p>
<p>$$\therefore$$ Sum of absolute values of $$\alpha = 1 + 1 = 2$$</p>
|
mcq
|
jee-main-2022-online-24th-june-evening-shift
|
1l5c0yfg6
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>The number of values of $$\alpha$$ for which the system of equations :</p>
<p>x + y + z = $$\alpha$$</p>
<p>$$\alpha$$x + 2$$\alpha$$y + 3z = $$-$$1</p>
<p>x + 3$$\alpha$$y + 5z = 4</p>
<p>is inconsistent, is</p>
|
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}]
|
["B"]
| null |
$\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ \alpha & 2 \alpha & 3 \\ 1 & 3 \alpha & 5\end{array}\right|$
<br/><br/>
$$
\begin{aligned}
&=1(10 \alpha-9 \alpha)-1(5 \alpha-3)+1\left(3 \alpha^{2}-2 \alpha\right) \\\\
&=\alpha-5 \alpha+3+3 \alpha^{2}-2 \alpha \\\\
&=3 \alpha^{2}-6 \alpha+3
\end{aligned}
$$
<br/><br/>
For inconsistency $\Delta=0$ i.e. $\alpha=1$
<br/><br/>
Now check for $\alpha=1$
<br/><br/>
$$
x+y+z=1\quad\quad...(i)
$$
<br/><br/>
$x+2 y+3 z=-1\quad\quad...(ii)$
<br/><br/>
$x+3 y+5 z=4\quad\quad...(iii)$
<br/><br/>
By (ii) $\times 2-$ (i) $\times 1$
<br/><br/>
$$
x+3 y+5 z=-3
$$
<br/><br/>
so equations are inconsistent for $\alpha=1$
|
mcq
|
jee-main-2022-online-24th-june-morning-shift
|
1l6dv2rsx
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>The number of $$\theta \in(0,4 \pi)$$ for which the system of linear equations
</p>
<p>$$
\begin{aligned}
&3(\sin 3 \theta) x-y+z=2 \\\\
&3(\cos 2 \theta) x+4 y+3 z=3 \\\\
&6 x+7 y+7 z=9
\end{aligned}
$$</p>
<p>has no solution, is :</p>
|
[{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "9"}]
|
["B"]
| null |
<p>Given,</p>
<p>$$3(\sin 3\theta )x - y + z = 2$$</p>
<p>$$3(\cos 2\theta )x + 4y + 3z = 3$$</p>
<p>$$6x + 7y + 7z = 9$$</p>
<p>For no solutions determinant of coefficient will be = 0</p>
<p>$$\therefore$$ $$D = \left| {\matrix{
{3\sin 3\theta } & { - 1} & 1 \cr
{3\cos 2\theta } & 4 & 3 \cr
6 & 7 & 7 \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow 3\sin 3\theta (28 - 21) + 1(21\cos 2\theta - 18) + 1(21\cos 2\theta - 24) = 0$$</p>
<p>$$ \Rightarrow 21\sin 3\theta + 42\cos 2\theta - 42 = 0$$</p>
<p>$$ \Rightarrow \sin 3\theta + 2\cos 2\theta - 2 = 0$$</p>
<p>$$ \Rightarrow 3\sin \theta - 4{\sin ^3}\theta + 2(1 - 2{\sin ^2}\theta ) - 2 = 0$$</p>
<p>$$ \Rightarrow 3\sin \theta - 4{\sin ^3}\theta - 4{\sin ^2}\theta = 0$$</p>
<p>$$ \Rightarrow 4{\sin ^3}\theta + 4{\sin ^2}\theta - 3\sin \theta = 0$$</p>
<p>$$ \Rightarrow \sin \theta (4{\sin ^2}\theta + 4\sin \theta - 3) = 0$$</p>
<p>$$\therefore$$ $$\sin \theta = 0$$</p>
<p>$$ \Rightarrow \theta = \pi ,2\pi ,3\pi $$ when $$\theta \in (0,4\pi )$$</p>
<p>or,</p>
<p>$$4{\sin ^2}\theta + 4\sin \theta - 3 = 0$$</p>
<p>$$ \Rightarrow 4{\sin ^2}\theta + 6\sin \theta - 2\sin \theta - 3 = 0$$</p>
<p>$$ \Rightarrow 2\sin \theta (2\sin \theta + 3) - 1(2\sin \theta + 3) = 0$$</p>
<p>$$ \Rightarrow (2\sin \theta - 1)(2\sin \theta + 3) = 0$$</p>
<p>$$\therefore$$ $$\sin \theta = {1 \over 2}$$</p>
<p>or,</p>
<p>$$\sin \theta = - {3 \over 2}$$ [not possible as $$\sin \in [ - 1,1]$$]</p>
<p>$$\therefore$$ $$\sin \theta = {1 \over 2}$$</p>
<p>$$ \Rightarrow \theta = {\pi \over 6},{{5\pi } \over 6},{{13\pi } \over 6},{{17\pi } \over 6}$$</p>
<p>$$\therefore$$ Possible values of $$\theta = \pi ,2\pi ,3\pi ,{\pi \over 6},{{5\pi } \over 6},{{13\pi } \over 6},{{17\pi } \over 6}$$</p>
<p>$$\therefore$$ Total 7 values of $$\theta$$ possible.</p>
|
mcq
|
jee-main-2022-online-25th-july-morning-shift
|
1l6f0ne7j
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>The number of real values of $$\lambda$$, such that the system of linear equations</p>
<p>2x $$-$$ 3y + 5z = 9</p>
<p>x + 3y $$-$$ z = $$-$$18</p>
<p>3x $$-$$ y + ($$\lambda$$<sup>2</sup> $$-$$ | $$\lambda$$ |)z = 16</p>
<p>has no solutions, is</p>
|
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "4"}]
|
["C"]
| null |
<p>$$\Delta = \left| {\matrix{
2 & { - 3} & 5 \cr
1 & 3 & { - 1} \cr
3 & { - 1} & {{\lambda ^2} - |\lambda |} \cr
} } \right| = 2\left( {3{\lambda ^2} - 3|\lambda | - 1} \right) + 3\left( {{\lambda ^2} - |\lambda | + 3} \right) + 5( - 1 - 9)$$</p>
<p>$$ = 9{\lambda ^2} - 9|\lambda | - 43$$</p>
<p>$$ = 9|\lambda {|^2} - 9|\lambda | - 43$$</p>
<p>$$\Delta = 0$$ for 2 values of $$|\lambda |$$ out of which one is $$-\mathrm{ve}$$ and other is $$+\mathrm{ve}$$</p>
<p>So, 2 values of $$\lambda$$ satisfy the system of equations to obtain no solution.</p>
|
mcq
|
jee-main-2022-online-25th-july-evening-shift
|
1l6ggh9jg
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>If the system of linear equations.</p>
<p>$$8x + y + 4z = - 2$$</p>
<p>$$x + y + z = 0$$</p>
<p>$$\lambda x - 3y = \mu $$</p>
<p>has infinitely many solutions, then the distance of the point $$\left( {\lambda ,\mu , - {1 \over 2}} \right)$$ from the plane $$8x + y + 4z + 2 = 0$$ is :</p>
|
[{"identifier": "A", "content": "$$3\\sqrt 5 $$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$${{26} \\over 9}$$"}, {"identifier": "D", "content": "$${{10} \\over 3}$$"}]
|
["D"]
| null |
<p>$$\Delta = \left| {\matrix{
8 & 1 & 4 \cr
1 & 1 & 1 \cr
\lambda & { - 3} & 0 \cr
} } \right|$$</p>
<p>$$ = 8(3) - 1( - \lambda ) + 4( - 3 - \lambda )$$</p>
<p>$$ = 24 + \lambda - 12 - 4\lambda $$</p>
<p>$$ = 12 - 3\lambda $$</p>
<p>So for $$\lambda = 4$$, it is having infinitely many solutions.</p>
<p>$${\Delta _x} = \left| {\matrix{
{ - 2} & 1 & 4 \cr
0 & 1 & 1 \cr
\mu & { - 3} & 0 \cr
} } \right|$$</p>
<p>$$ = - 2(3) - 1( - \mu )+4( - \mu )$$</p>
<p>$$ = - 6 - 3\mu = 0$$</p>
<p>For $$\mu = - 2$$</p>
<p>Distance of $$(4, - 2,{{ - 1} \over 2})$$ from $$8x + y + 4z + 2 = 0$$</p>
<p>$$ = {{32 - 2 - 2 + 2} \over {\sqrt {64 + 1 + 16} }} = {{10} \over 3}$$ units</p>
|
mcq
|
jee-main-2022-online-26th-july-morning-shift
|
1l6p0ni83
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>Let A and B be two $$3 \times 3$$ non-zero real matrices such that AB is a zero matrix. Then</p>
|
[{"identifier": "A", "content": "the system of linear equations $$A X=0$$ has a unique solution"}, {"identifier": "B", "content": "the system of linear equations $$A X=0$$ has infinitely many solutions"}, {"identifier": "C", "content": "B is an invertible matrix"}, {"identifier": "D", "content": "$$\\operatorname{adj}(\\mathrm{A})$$ is an invertible matrix"}]
|
["B"]
| null |
<p>AB is zero matrix</p>
<p>$$ \Rightarrow |A| = |B| = 0$$</p>
<p>So neither A nor B is invertible</p>
<p>If $$|A| = 0$$</p>
<p>$$ \Rightarrow |\mathrm{adj}\,A| = 0$$ so $$\mathrm{adj}\,A$$</p>
<p>$$AX = 0$$ is homogeneous system and $$|A| = 0$$</p>
<p>So, it is having infinitely many solutions</p>
|
mcq
|
jee-main-2022-online-29th-july-morning-shift
|
1l6rdwkbl
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>If the system of equations</p>
<p>$$
\begin{aligned}
&x+y+z=6 \\
&2 x+5 y+\alpha z=\beta \\
&x+2 y+3 z=14
\end{aligned}
$$</p>
<p>has infinitely many solutions, then $$\alpha+\beta$$ is equal to</p>
|
[{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "36"}, {"identifier": "C", "content": "44"}, {"identifier": "D", "content": "48"}]
|
["C"]
| null |
<p>Given,</p>
<p>$$x + y + z = 6$$ ...... (1)</p>
<p>$$2x + 5y + \alpha z = \beta $$ ..... (2)</p>
<p>$$x + 2y + 3z = 14$$ ...... (3)</p>
<p>System of equation have infinite many solutions.</p>
<p>$$\therefore$$ $${\Delta _x} = {\Delta _y} = {\Delta _z} = 0$$ and $$\Delta = 0$$</p>
<p>Now, $$\Delta = \left| {\matrix{
1 & 1 & 1 \cr
2 & 5 & \alpha \cr
1 & 2 & 3 \cr
} } \right| = 0$$</p>
<p>$${C_1} \to {C_1} - {C_3}$$</p>
<p>$${C_2} \to {C_2} - {C_3}$$</p>
<p>$$ \Rightarrow \left| {\matrix{
0 & 0 & 1 \cr
{2 - \alpha } & {5 - \alpha } & \alpha \cr
{ - 2} & { - 1} & 3 \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow - 2 + \alpha + 10 - 2\alpha = 0$$</p>
<p>$$ \Rightarrow 8 - \alpha = 0$$</p>
<p>$$ \Rightarrow \alpha = 8$$</p>
<p>Now, $$x + y + z = 6$$</p>
<p>$$2x + 5y + 8z = \beta $$</p>
<p>$$x + 2y + 3z = 14$$</p>
<p>$$\therefore$$ $${\Delta _x} = \left| {\matrix{
6 & 1 & 1 \cr
\beta & 5 & 8 \cr
{14} & 2 & 3 \cr
} } \right| = 0$$</p>
<p>$${C_1} \to {C_1} - 6{C_3}$$</p>
<p>$${C_2} \to {C_2} - {C_3}$$</p>
<p>$$ \Rightarrow \left| {\matrix{
0 & 0 & 1 \cr
{\beta - 48} & { - 3} & 8 \cr
{ - 4} & { - 1} & 3 \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow - \beta + 48 - 12 = 0$$</p>
<p>$$ \Rightarrow \beta = 36$$</p>
<p>$$\therefore$$ $$\alpha + \beta = 8 + 36 = 44$$</p>
|
mcq
|
jee-main-2022-online-29th-july-evening-shift
|
1ldo5ix8v
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>For the system of linear equations $$\alpha x+y+z=1,x+\alpha y+z=1,x+y+\alpha z=\beta$$, which one of the following statements is <b>NOT</b> correct?</p>
|
[{"identifier": "A", "content": "It has infinitely many solutions if $$\\alpha=1$$ and $$\\beta=1$$"}, {"identifier": "B", "content": "It has infinitely many solutions if $$\\alpha=2$$ and $$\\beta=-1$$"}, {"identifier": "C", "content": "$$x+y+z=\\frac{3}{4}$$ if $$\\alpha=2$$ and $$\\beta=1$$"}, {"identifier": "D", "content": "It has no solution if $$\\alpha=-2$$ and $$\\beta=1$$"}]
|
["B"]
| null |
For infinite solution $\Delta=\Delta_x=\Delta_y=\Delta_z=0$
<br/><br/>$$
\Delta=\left|\begin{array}{lll}
\alpha & 1 & 1 \\
1 & \alpha & 1 \\
1 & 1 & \alpha
\end{array}\right|=0 \Rightarrow\left(\alpha^3-3 \alpha+2\right)=0 \Rightarrow \alpha=1,-2
$$
<br/><br/>If $\beta=1$, then all planes are overlapping
<br/><br/>$\therefore$ Option (A) is correct.
<br/><br/><b>Option (B) :</b>
<br/><br/>If $\alpha=2 \Rightarrow \Delta \neq 0$
<br/><br/>$\therefore $ Unique solution exist
<br/><br/>$\therefore$ Option (B) is incorrect.
<br/><br/><b>Option (C) :</b>
<br/><br/>$$
\begin{aligned}
& \alpha=2, \beta=1 \\\\
& 2 x+y+z=1 \\\\
& x+2 y+z=1 \\\\
& x+y+2 z=1
\end{aligned}
$$
<br/><br/>Adding all three equations,
<br/><br/>$$
x+y+z=\frac{3}{4}
$$
<br/><br/>$\therefore$ option (C) is correct.
<br/><br/><b>Option (D) :</b>
<br/><br/>If $\alpha=-2$ and $\beta=1$, then $\Delta=0, \Delta_x \neq 0$
<br/><br/>$\therefore$ No solution.
<br/><br/>$\therefore$ Option (D) is correct.
|
mcq
|
jee-main-2023-online-1st-february-evening-shift
|
1ldomfjmn
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>Let $$S$$ denote the set of all real values of $$\lambda$$ such that the system of equations</p>
<p>$$\lambda x+y+z=1$$</p>
<p>$$x+\lambda y+z=1$$</p>
<p>$$x+y+\lambda z=1$$</p>
<p>is inconsistent, then $$\sum_\limits{\lambda \in S}\left(|\lambda|^{2}+|\lambda|\right)$$ is equal to</p>
|
[{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "6"}]
|
["D"]
| null |
$\left|\begin{array}{lll}\lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda\end{array}\right|=0$
<br/><br/>$$
\begin{aligned}
& \lambda\left(\lambda^{2}-1\right)-1(\lambda-1)+1(1-\lambda)=0 \\\\
& \Rightarrow \lambda^{3}-\lambda-\lambda+1+1-\lambda=0 \\\\
& \Rightarrow \lambda^{3}-3 \lambda+2=0 \\\\
& \Rightarrow (\lambda-1)\left(\lambda^{2}+\lambda-2\right)=0
\end{aligned}
$$
<br/><br/>$\Rightarrow$ $\lambda=1,-2$
<br/><br/>For $\lambda=1 \Rightarrow \infty$ solution
<br/><br/>$\lambda=-2 \Rightarrow$ no solution
<br/><br/>$\sum\limits_{\lambda \in S}|\lambda|^{2}+|\lambda|=6$
|
mcq
|
jee-main-2023-online-1st-february-morning-shift
|
1ldprqy0n
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>For the system of linear equations</p>
<p>$$x+y+z=6$$</p>
<p>$$\alpha x+\beta y+7 z=3$$</p>
<p>$$x+2 y+3 z=14$$</p>
<p>which of the following is <b>NOT</b> true ?</p>
|
[{"identifier": "A", "content": "If $$\\alpha=\\beta=7$$, then the system has no solution"}, {"identifier": "B", "content": "For every point $$(\\alpha, \\beta) \\neq(7,7)$$ on the line $$x-2 y+7=0$$, the system has infinitely many solutions"}, {"identifier": "C", "content": "There is a unique point $$(\\alpha, \\beta)$$ on the line $$x+2 y+18=0$$ for which the system has infinitely many solutions"}, {"identifier": "D", "content": "If $$\\alpha=\\beta$$ and $$\\alpha \\neq 7$$, then the system has a unique solution"}]
|
["B"]
| null |
$\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ \alpha & \beta & 7 \\ 1 & 2 & 3\end{array}\right|$
<br/><br/>$$
\begin{aligned}
& =1(3 \beta-14)-1(3 \alpha-7)+1(2 \alpha-\beta) \\\\
& =3 \beta-14+7-3 \alpha+2 \alpha-\beta \\\\
& =2 \beta-\alpha-7
\end{aligned}
$$
<br/><br/>So, for $\alpha=\beta \neq 7, \Delta \neq 0$ so unique solution.
<br/><br/>$\alpha=\beta=7$, equation (i) and (ii) represent 2 parallel planes so no solution.
<br/><br/>If $\alpha-2 \beta+7=0$, but $(\alpha, \beta) \neq(7,7)$, then no solution.
|
mcq
|
jee-main-2023-online-31st-january-morning-shift
|
ldqv7zxq
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
For $\alpha, \beta \in \mathbb{R}$, suppose the system of linear equations
<br/><br/>$$
\begin{aligned}
& x-y+z=5 \\
& 2 x+2 y+\alpha z=8 \\
& 3 x-y+4 z=\beta
\end{aligned}
$$
<br/><br/>has infinitely many solutions. Then $\alpha$ and $\beta$ are the roots of :
|
[{"identifier": "A", "content": "$x^2+18 x+56=0$"}, {"identifier": "B", "content": "$x^2-10 x+16=0$"}, {"identifier": "C", "content": "$x^2+14 x+24=0$"}, {"identifier": "D", "content": "$x^2-18 x+56=0$"}]
|
["D"]
| null |
<p>$$\Delta = \left| {\matrix{
1 & { - 1} & 1 \cr
2 & 2 & \alpha \cr
3 & { - 1} & 4 \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow \alpha = 4$$</p>
<p>$${\Delta _3} = 0$$</p>
<p>$$ = \left| {\matrix{
1 & { - 1} & 5 \cr
2 & 2 & 8 \cr
3 & { - 1} & \beta \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow \beta = 14$$</p>
<p>$$\therefore$$ $${x^2} - 18x + 56 = 0$$</p>
|
mcq
|
jee-main-2023-online-30th-january-evening-shift
|
1ldr6t8zm
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>Let the system of linear equations</p>
<p>$$x+y+kz=2$$</p>
<p>$$2x+3y-z=1$$</p>
<p>$$3x+4y+2z=k$$</p>
<p>have infinitely many solutions. Then the system</p>
<p>$$(k+1)x+(2k-1)y=7$$</p>
<p>$$(2k+1)x+(k+5)y=10$$</p>
<p>has :</p>
|
[{"identifier": "A", "content": "unique solution satisfying $$x-y=1$$"}, {"identifier": "B", "content": "infinitely many solutions"}, {"identifier": "C", "content": "no solution"}, {"identifier": "D", "content": "unique solution satisfying $$x+y=1$$"}]
|
["D"]
| null |
<p>$$x + y + kz = 2$$ ............(i)</p>
<p>$$2x + 3y - z = 1$$ ..........(ii)</p>
<p>$$3x + 4y + 2z = k$$ ......(iii)</p>
<p>(1) + (2)</p>
<p>$$3x + 4y + z(k - 1) = 3$$</p>
<p>Comparing with (3)</p>
<p>$$k = 3$$</p>
<p>Now, $$4x + 5y = 7$$</p>
<p>$$ \Rightarrow 3x + 3y = 3$$</p>
<p>$$7x + 8y = 10$$</p>
<p>as $${4 \over 7} \ne {5 \over 8}$$</p>
<p>$$\therefore$$ Unique solution satisfying $$x + y = 1$$</p>
|
mcq
|
jee-main-2023-online-30th-january-morning-shift
|
1ldswccvm
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>Consider the following system of equations</p>
<p>$$\alpha x+2y+z=1$$</p>
<p>$$2\alpha x+3y+z=1$$</p>
<p>$$3x+\alpha y+2z=\beta$$</p>
<p>for some $$\alpha,\beta\in \mathbb{R}$$. Then which of the following is NOT correct.</p>
|
[{"identifier": "A", "content": "It has a solution for all $$\\alpha\\ne-1$$ and $$\\beta=2$$"}, {"identifier": "B", "content": "It has no solution if $$\\alpha=-1$$ and $$\\beta\\ne2$$"}, {"identifier": "C", "content": "It has no solution for $$\\alpha=-1$$ and for all $$\\beta \\in \\mathbb{R}$$"}, {"identifier": "D", "content": "It has no solution for $$\\alpha=3$$ and for all $$\\beta\\ne2$$"}]
|
["C"]
| null |
$D=\left|\begin{array}{ccc}\alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & 2\end{array}\right|=0 \Rightarrow \alpha=-1,3$
<br/><br/>
$D_{x}=\left|\begin{array}{ccc}2 & 1 & 1 \\ 3 & 1 & 1 \\ \alpha & 2 & \beta\end{array}\right|=0 \Rightarrow \beta=2$
<br/><br/>
$D_{y}=\left|\begin{array}{ccc}\alpha & 1 & 1 \\ 2 \alpha & 1 & 1 \\ 3 & 2 & \beta\end{array}\right|=0$
<br/><br/>
$D_{z}=\left|\begin{array}{ccc}\alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & \beta\end{array}\right|=0$
<br/><br/>
$\beta=2, \alpha=-1$
<br/><br/>
$\alpha=-1, \beta=2$ Infinite solution
|
mcq
|
jee-main-2023-online-29th-january-morning-shift
|
1ldv2dqa1
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>Let S$$_1$$ and S$$_2$$ be respectively the sets of all $$a \in \mathbb{R} - \{ 0\} $$ for which the system of linear equations</p>
<p>$$ax + 2ay - 3az = 1$$</p>
<p>$$(2a + 1)x + (2a + 3)y + (a + 1)z = 2$$</p>
<p>$$(3a + 5)x + (a + 5)y + (a + 2)z = 3$$</p>
<p>has unique solution and infinitely many solutions. Then</p>
|
[{"identifier": "A", "content": "$$\\mathrm{n({S_1}) = 2}$$ and S$$_2$$ is an infinite set"}, {"identifier": "B", "content": "$$\\mathrm{{S_1} = \\Phi} $$ and $$\\mathrm{{S_2} = \\mathbb{R} - \\{ 0\\}}$$"}, {"identifier": "C", "content": "$$\\mathrm{{S_1} = \\mathbb{R} - \\{ 0\\}}$$ and $$\\mathrm{{S_2} = \\Phi} $$"}, {"identifier": "D", "content": "S$$_1$$ is an infinite set and n(S$$_2$$) = 2"}]
|
["C"]
| null |
Given system of equations
<br/><br/>
$$
\begin{aligned}
& a x+2 a y-3 a z=1 \\\\
& (2 a+1) x+(2 a+3) y+(a+1) z=2 \\\\
& (3 a+5) x+(a+5) y+(a+2) z=3 \\\\
& \text { Let } A=\left|\begin{array}{ccc}
a & 2 a & -3 a \\\\
2 a+1 & 2 a+3 & a+1 \\\\
3 a+5 & a+5 & a+2
\end{array}\right| \\\\
& =a\left|\begin{array}{ccc}
1 & 0 & 0 \\\\
2 a+1 & 1-2 a & 7 a+4 \\\\
3 a+5 & -5 a-5 & 10 a+17
\end{array}\right| \\\\
& =a\left(15 a^{2}+31 a+37\right) \\\\
& \text { Now } A=0 \\\\
& \Rightarrow a=0
\end{aligned}
$$
<br/><br/>
So, $S_{1}=R-\{0\}$ and at $a=0$
<br/><br/>
System has infinite solution but $a \in R-\{0\}$
<br/><br/>
$\therefore S_{2}=\Phi$
|
mcq
|
jee-main-2023-online-25th-january-morning-shift
|
1ldwwf3hd
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>If the system of equations</p>
<p>$$x+2y+3z=3$$</p>
<p>$$4x+3y-4z=4$$</p>
<p>$$8x+4y-\lambda z=9+\mu$$</p>
<p>has infinitely many solutions, then the ordered pair ($$\lambda,\mu$$) is equal to :</p>
|
[{"identifier": "A", "content": "$$\\left( {{{72} \\over 5},{{21} \\over 5}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - {{72} \\over 5}, - {{21} \\over 5}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - {{72} \\over 5},{{21} \\over 5}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{{72} \\over 5}, - {{21} \\over 5}} \\right)$$"}]
|
["D"]
| null |
For infinite many solution, $\Delta=0$ and $\Delta_x=0$
<br/><br/>$$
\begin{aligned}
& \Delta=\left|\begin{array}{ccc}
1 & 2 & 3 \\
4 & 3 & -4 \\
8 & 4 & -\lambda
\end{array}\right|=0 \\\\
& \Rightarrow 1(-3 \lambda+16)-2(-4 \lambda+32)+3(16-24)=0 \\\\
& \Rightarrow 16-3 \lambda+8 \lambda-64-24=0 \Rightarrow 5 \lambda=72 \\\\
& \therefore \lambda=\frac{72}{5} \\\\
& \Delta_x=\left|\begin{array}{ccc}
3 & 2 & 3 \\
4 & 3 & -4 \\
9+\mu & 4 & -\lambda
\end{array}\right|=0
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \quad 3(-3 \lambda+16)-2(-4 \lambda+36+4 \mu)+3(16-27-3 \mu)=0 \\\\
& \Rightarrow-9 \lambda+48+8 \lambda-72-8 \mu-33-9 \mu=0 \\\\
& \Rightarrow-\lambda-17 \mu=57 \\\\
& \Rightarrow-17 \mu=57+\lambda \\\\
& \therefore -\mu=\frac{57+\frac{72}{5}}{17}\\\\
& \Rightarrow \mu=\frac{-357}{85}=\frac{-21}{5}
\end{aligned}
$$
<br/><br/>$$
\text { Thus, }(\lambda, \mu) \equiv\left(\frac{72}{5}, \frac{-21}{5}\right)
$$
|
mcq
|
jee-main-2023-online-24th-january-evening-shift
|
1lgoxgbmt
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>If the system of equations</p>
<p>$$2 x+y-z=5$$</p>
<p>$$2 x-5 y+\lambda z=\mu$$</p>
<p>$$x+2 y-5 z=7$$</p>
<p>has infinitely many solutions, then $$(\lambda+\mu)^{2}+(\lambda-\mu)^{2}$$ is equal to</p>
|
[{"identifier": "A", "content": "916"}, {"identifier": "B", "content": "912"}, {"identifier": "C", "content": "920"}, {"identifier": "D", "content": "904"}]
|
["A"]
| null |
$$
\begin{aligned}
& 2 x+y-z=5 \\
& 2 x-5 y+\lambda z=\mu \\
& x+2 y-5 z=7
\end{aligned}
$$
<br/><br/>For infinite solution $\Delta=0=\Delta_1=\Delta_2=\Delta_3$
<br/><br/>$$
\Delta=\left|\begin{array}{ccc}
2 & 1 & -1 \\
2 & -5 & \lambda \\
1 & 2 & -5
\end{array}\right|=0
$$
<br/><br/>$$
\begin{aligned}
\Rightarrow& 2(25-2 \lambda)-(-10-\lambda)-(4+5)=0 \\\\
\Rightarrow& 50-4 \lambda+10+\lambda-9=0 \\\\
\Rightarrow& 51=3 \lambda \Rightarrow \lambda=17
\end{aligned}
$$
<br/><br/>$$
\Delta_3=\left|\begin{array}{ccc}
5 & 2 & 1 \\
\mu & 2 & -5 \\
7 & 1 & 2
\end{array}\right|=0
$$
<br/><br/>$$
\begin{aligned}
\Rightarrow & 2(-35-2 \mu)-(14-\mu)+5(4+5)=0 \\\\
\Rightarrow & -70-4 \mu-14+\mu+45=0 \\\\
\Rightarrow & -3 \mu=39 \\\\
\Rightarrow & -\mu=13
\end{aligned}
$$
<br/><br/>Now $(\lambda+\mu)^2+(\lambda-\mu)^2$
<br/><br/>$$
\begin{aligned}
& (17+13)^2+(17-13)^2 \\\\
& 900+16 \\\\
& =916
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-13th-april-evening-shift
|
1lgpxsxhd
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>For the system of linear equations</p>
<p>$$2 x+4 y+2 a z=b$$</p>
<p>$$x+2 y+3 z=4$$</p>
<p>$$2 x-5 y+2 z=8$$</p>
<p>which of the following is NOT correct?</p>
|
[{"identifier": "A", "content": "It has infinitely many solutions if $$a=3, b=8$$"}, {"identifier": "B", "content": "It has infinitely many solutions if $$a=3, b=6$$"}, {"identifier": "C", "content": "It has unique solution if $$a=b=8$$"}, {"identifier": "D", "content": "It has unique solution if $$a=b=6$$"}]
|
["B"]
| null |
The given system of equations is :
<br/><br/>1. $$2x + 4y + 2az = b$$
<br/><br/>2. $$x + 2y + 3z = 4$$
<br/><br/>3. $$2x - 5y + 2z = 8$$
<br/><br/>We can write this in matrix form :
<br/><br/>$$
\begin{bmatrix}
2 & 4 & 2a \\
1 & 2 & 3 \\
2 & -5 & 2
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
=
\begin{bmatrix}
b \\
4 \\
8
\end{bmatrix}
$$
<br/><br/>First, we need to find the determinant of the coefficient matrix, which we'll call Ξ. The coefficient matrix is :
<br/><br/>$$
\begin{bmatrix}
2 & 4 & 2a \\
1 & 2 & 3 \\
2 & -5 & 2
\end{bmatrix}
$$
<br/><br/>We find its determinant for 3 $$ \times $$ 3 matrices :
<br/><br/>$$
\Delta = 2(2)(2) + 4(3)(2) + 2a(1)(-5) - 2a(2)(2) - 4(1)(2) - 2(3)(-5)
$$
<br/><br/>$$
\Delta = 8 + 24 - 10a - 8a - 8 + 30
$$
<br/><br/>$$
\Delta = -18a + 54
$$
<br/><br/>$$
\Delta = -18(a - 3)
$$
<br/><br/>Next, we substitute the third column of our matrix with the column of constants (b, 4, 8) and calculate the determinant Ξβ :
<br/><br/>$$
\begin{bmatrix}
2 & 4 & b \\
1 & 2 & 4 \\
2 & -5 & 8
\end{bmatrix}
$$
<br/><br/>We find its determinant :
<br/><br/>$$
\Delta_3 = 2(2)(8) + 4(4)(2) + b(1)(-5) - b(2)(2) - 4(1)(8) - 2(4)(-5)
$$
<br/><br/>$$
\Delta_3 = 32 + 32 - 5b - 4b - 32 + 40
$$
<br/><br/>$$
\Delta_3 = -9b + 72
$$
<br/><br/>$$
\Delta_3 = 9(8 - b)
$$
<br/><br/>Now, let's analyze the options :
<br/><br/>1. For a=3 and b=8, we have Ξ = 0 and Ξβ = 0, which indicates an infinite number of solutions.
<br/><br/>2. For a=3 and b=6, we have Ξ = 0 and Ξβ β 0, which indicates no solution.
<br/><br/>3. For a=8 and b=8, we have Ξ β 0, which indicates a unique solution.
<br/><br/>4. For a=6 and b=6, we have Ξ β 0, which indicates a unique solution.
<br/><br/>Therefore, the statement that is NOT correct is Option B: "It has infinitely many solutions if a=3, b=6", because in this case the system actually has no solution.
|
mcq
|
jee-main-2023-online-13th-april-morning-shift
|
1lgsuczop
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>If the system of linear equations</p>
<p>$$
\begin{aligned}
& 7 x+11 y+\alpha z=13 \\\\
& 5 x+4 y+7 z=\beta \\\\
& 175 x+194 y+57 z=361
\end{aligned}
$$</p>
<p>has infinitely many solutions, then $$\alpha+\beta+2$$ is equal to :</p>
|
[{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "3"}]
|
["B"]
| null |
Given,
<br/><br/>$$
\begin{aligned}
& 7 x+11 y+\alpha z=13 \\\\
& 5 x+4 y+7 z=\beta \\\\
& 175 x+194 y+57 z=361
\end{aligned}
$$
<br/><br/>$$
\text { For infinite solution, }\left|\begin{array}{ccc}
7 & 11 & \alpha \\
5 & 4 & 7 \\
175 & 194 & 57
\end{array}\right|=0
$$
<br/><br/>$$
\Rightarrow\left|\begin{array}{ccc}
7 & 11 & \alpha \\
5 & 4 & 7 \\
0 & -81 & 57-25 \alpha
\end{array}\right|=0
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow 81(49-5 \alpha)+(57-25 \alpha)(-27)=0 \\\\
& \Rightarrow 270 \alpha=-2430 \Rightarrow \alpha=-9
\end{aligned}
$$
<br/><br/>And $\Delta_1=0$
<br/><br/>$$
\begin{aligned}
& \left|\begin{array}{ccc}
13 & 11 & -9 \\
\beta & 4 & 7 \\
361 & 194 & 57
\end{array}\right|=0 \\\\
& \Rightarrow \beta=11
\end{aligned}
$$
<br/><br/>$$
\therefore \alpha+\beta+2=4
$$
|
mcq
|
jee-main-2023-online-11th-april-evening-shift
|
1lgvqob1i
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>Let $$\mathrm{S}$$ be the set of values of $$\lambda$$, for which the system of equations <br/><br/>$$6 \lambda x-3 y+3 z=4 \lambda^{2}$$, <br/><br/>$$2 x+6 \lambda y+4 z=1$$, <br/><br/>$$3 x+2 y+3 \lambda z=\lambda$$ has no solution. Then $$12 \sum_\limits{i \in S}|\lambda|$$ is equal to ___________.</p>
|
[]
| null |
24
|
Given that $S$ be the set of values of $\lambda$ for which given system of equations has no solution.
<br/><br/>Therefore for the given set of equations
<br/><br/>$$
\Delta=\left|\begin{array}{ccc}
6 \lambda & -3 & 3 \\
2 & 6 \lambda & 4 \\
3 & 2 & 3 \lambda
\end{array}\right|=0
$$
<br/><br/>$$
\begin{aligned}
&\Rightarrow6 \lambda\left(18 \lambda^2-8\right)+3(6 \lambda-12)+3(4-18 \lambda)=0 \\\\
&\Rightarrow18 \lambda^3-14 \lambda-4=0 \\\\
&\Rightarrow(\lambda-1)(3 \lambda+1)(3 \lambda+2)=0 \\\\
&\Rightarrow \lambda=1,-\frac{1}{3},-\frac{2}{3}
\end{aligned}
$$
<br/><br/>Also for each values of $\lambda=1, \frac{-1}{3}, \frac{-2}{3}$, we have
<br/><br/>$$
\left|\begin{array}{ccc}
6 \lambda & -3 & 4 \lambda^2 \\
2 & 6 \lambda & 1 \\
3 & 2 & \lambda
\end{array}\right| \neq 0
$$
<br/><br/>which implies that, for each values of $\lambda$, the given system of equations has no solution.
<br/><br/>$$
\begin{aligned}
& \text { Therefore } S \in\left\{1, \frac{-1}{3}, \frac{-2}{3}\right\} \text { and } \\\\
&12 \sum_{\lambda \in S}|\lambda| \\\\
& =12\left(|1|+\left|\frac{-1}{3}\right|+\left|\frac{-2}{3}\right|\right) \\\\
& =12\left(1+\frac{1}{3}+\frac{2}{3}\right)=12\left(\frac{6}{3}\right)=24
\end{aligned}
$$
|
integer
|
jee-main-2023-online-10th-april-evening-shift
|
1lgxvxhlk
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>For the system of linear equations</p>
<p>$$2x - y + 3z = 5$$</p>
<p>$$3x + 2y - z = 7$$</p>
<p>$$4x + 5y + \alpha z = \beta $$,</p>
<p>which of the following is <b>NOT</b> correct?</p>
|
[{"identifier": "A", "content": "The system has infinitely many solutions for $$\\alpha=-6$$ and $$\\beta=9$$"}, {"identifier": "B", "content": "The system has a unique solution for $$\\alpha$$ $$ \\ne $$ $$-5$$ and $$\\beta=8$$"}, {"identifier": "C", "content": "The system is inconsistent for $$\\alpha=-5$$ and $$\\beta=8$$"}, {"identifier": "D", "content": "The system has infinitely many solutions for $$\\alpha=-5$$ and $$\\beta=9$$"}]
|
["A"]
| null |
Given system of linear equation is
<br/><br/>$$
\begin{gathered}
2 x-y+3 z=5 \\\\
3 x+2 y-z=7 \\\\
4 x+5 y+\alpha z=\beta \\\\
\text { Now, } \Delta=\left|\begin{array}{ccc}
2 & -1 & 3 \\
3 & 2 & -1 \\
4 & 5 & \alpha
\end{array}\right|=7(\alpha+5)
\end{gathered}
$$
<br/><br/>So, this system of equation has unique solution, if $\alpha \neq-5$
<br/><br/>and $\Delta_1=\left|\begin{array}{ccc}5 & -1 & 3 \\ 7 & 2 & -1 \\ \beta & 5 & \alpha\end{array}\right|=17 \alpha-5 \beta+30$
<br/><br/>and $\Delta_2=\left|\begin{array}{ccc}2 & 5 & 3 \\ 3 & 7 & -1 \\ 4 & \beta & \alpha\end{array}\right|=-11 \beta+\alpha+104$
<br/><br/>and $\Delta_3=\left|\begin{array}{ccc}2 & -1 & 5 \\ 3 & 2 & 7 \\ 4 & 5 & \beta\end{array}\right|=7(\beta-9)$
<br/><br/>For infinitely many solutions,
<br/><br/>$$
\Delta=\Delta_1=\Delta_2=\Delta_3=0
$$
<br/><br/>$$
\begin{aligned}
& \text { So, } \Delta=0 \\\\
& \Rightarrow 7(\alpha+5)=0 \\\\
& \Rightarrow \alpha=-5
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Delta_3 =0 \\\\
& \Rightarrow 7(\beta-9) =0 \\\\
& \Rightarrow \beta =9
\end{aligned}
$$
<br/><br/>If $\alpha=-5$ and $\beta=8$, then $\Delta$ equals zero but $\Delta_3$ does not, which would imply the system is inconsistent for $\alpha=-5$ and $\beta=8$.
<br/><br/>Therefore, the option "The system is inconsistent for $\alpha=-5$ and $\beta=8$ " is correct.
|
mcq
|
jee-main-2023-online-10th-april-morning-shift
|
1lgym2kp0
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>Let S be the set of all values of $$\theta \in[-\pi, \pi]$$ for which the system of linear equations</p>
<p>$$x+y+\sqrt{3} z=0$$</p>
<p>$$-x+(\tan \theta) y+\sqrt{7} z=0$$</p>
<p>$$x+y+(\tan \theta) z=0$$</p>
<p>has non-trivial solution. Then $$\frac{120}{\pi} \sum_\limits{\theta \in \mathrm{s}} \theta$$ is equal to :</p>
|
[{"identifier": "A", "content": "40"}, {"identifier": "B", "content": "30"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "20"}]
|
["D"]
| null |
Since, the given system has a non trivial solution,
<br/><br/>$$
\text { So, } \Delta=0
$$
<br/><br/>$$
\Rightarrow \Delta=\left|\begin{array}{ccc}
1 & 1 & \sqrt{3} \\
-1 & \tan \theta & \sqrt{7} \\
1 & 1 & \tan \theta
\end{array}\right|=0
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow 1\left(\tan ^2 \theta-\sqrt{7}\right)-1(-\tan \theta-\sqrt{7})+\sqrt{3}(-1-\tan \theta)=0 \\\\
& \Rightarrow \tan ^2 \theta-\sqrt{7}+\tan \theta+\sqrt{7}-\sqrt{3}-\sqrt{3} \tan \theta=0 \\\\
& \Rightarrow \tan \theta(\tan \theta-\sqrt{3})+1(\tan \theta-\sqrt{3})=0 \\\\
& \Rightarrow \tan \theta=\sqrt{3} \text { or } \tan \theta=-1 \\\\
& \therefore \theta=\left\{\frac{\pi}{3}, \frac{-2 \pi}{3}, \frac{-\pi}{4}, \frac{3 \pi}{4}\right\} \\\\
& \text { So, } \frac{120}{\pi} \sum \theta=\frac{120}{\pi}\left\{\frac{4 \pi-8 \pi-3 \pi+9 \pi}{12}\right\} \\\\
& =\frac{120}{\pi}\left[\frac{2 \pi}{12}\right]=20
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-8th-april-evening-shift
|
1lh204tzs
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>If the system of equations</p>
<p>$$x+y+a z=b$$</p>
<p>$$2 x+5 y+2 z=6$$</p>
<p>$$x+2 y+3 z=3$$</p>
<p>has infinitely many solutions, then $$2 a+3 b$$ is equal to :</p>
|
[{"identifier": "A", "content": "28"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "23"}]
|
["D"]
| null |
Given system of equations,
<br/><br/>$$
\text { and } \quad \begin{aligned}
x+y+a z & =b \\
2 x+5 y+2 z & =6 \\
x+2 y+3 z & =3
\end{aligned}
$$
<br/><br/>Since, given system of equation has infinitely many solutions
<br/><br/>$$
\therefore D=0 \text { and } D_1=D_2=D_3=0
$$
<br/><br/>$$
\begin{aligned}
& \text { Here, } D=\left|\begin{array}{lll}
1 & 1 & a \\
2 & 5 & 2 \\
1 & 2 & 3
\end{array}\right|=0 \\\\
& \Rightarrow 1(15-4)-1(6-2)+a(4-5)=0 \\\\
& \Rightarrow -a+11-4=0 \\\\
& \Rightarrow a=7
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { and } D_1=\left|\begin{array}{ccc}
b & 1 & a \\
6 & 5 & 2 \\
3 & 2 & 3
\end{array}\right|=0 \\\\
& \Rightarrow b(15-4)-1(18-6)+a(12-15)=0 \\\\
& \Rightarrow 11 b-12-3 a=0 \\\\
& \Rightarrow 11 b-12-21=0 ~~~~~~~(\because a=7)\\\\
& \Rightarrow 11 b-33=0 \\\\
& \Rightarrow b=3 \\\\
& \therefore 2 a+3 b=2(7)+3(3)=14+9=23
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-6th-april-morning-shift
|
1lh2yj42z
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>For the system of equations</p>
<p>$$x+y+z=6$$</p>
<p>$$x+2 y+\alpha z=10$$</p>
<p>$$x+3 y+5 z=\beta$$, which one of the following is <b>NOT</b> true?</p>
|
[{"identifier": "A", "content": "System has a unique solution for $$\\alpha=3,\\beta\\ne14$$."}, {"identifier": "B", "content": "System has infinitely many solutions for $$\\alpha=3, \\beta=14$$."}, {"identifier": "C", "content": "System has no solution for $$\\alpha=3, \\beta=24$$."}, {"identifier": "D", "content": "System has a unique solution for $$\\alpha=-3, \\beta=14$$."}]
|
["A"]
| null |
Given system of equations,
<br/><br/>$$
\begin{aligned}
x+y+z & =6 ........(i)\\\\
x+2 y+\alpha z & =10 ........(ii)\\\\
x+3 y+5 z & =\beta ........(iii)
\end{aligned}
$$
<br/><br/>Here,
<br/><br/>$$
\begin{aligned}
\Delta & =\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & \alpha \\
1 & 3 & 5
\end{array}\right| \\\\
& =1(10-3 \alpha)-1(5-\alpha)+1(3-2) \\\\
& =10-3 \alpha-5+\alpha+1 \\\\
& =6-2 \alpha
\end{aligned}
$$
<br/><br/>For unique solution, $\Delta \neq 0$
<br/><br/>$$
\Rightarrow 6-2 \alpha \neq 0 \Rightarrow \alpha \neq 3
$$
<br/><br/>When, $\alpha=3$
<br/><br/>$$
\begin{aligned}
\Delta_1 & =\left|\begin{array}{ccc}
6 & 1 & 1 \\
10 & 2 & 3 \\
\beta & 3 & 5
\end{array}\right| \\\\
& =\left|\begin{array}{ccc}
0 & 1 & 0 \\
-2 & 2 & 1 \\
\beta-18 & 3 & 2
\end{array}\right|=-1(-4-\beta+18) \\\\
& =\beta-14
\end{aligned}
$$
<br/><br/>and
<br/><br/>$\begin{aligned} \Delta_2 & =\left|\begin{array}{ccc}1 & 6 & 1 \\ 1 & 10 & 3 \\ 1 & \beta & 5\end{array}\right| \\\\ & =1(50-3 \beta)-6(5-3)+1(\beta-10) \\\\ & =50-3 \beta-12+\beta-10 \\\\ & =28-2 \beta=2(14-\beta)\end{aligned}$
<br/><br/>
and
<br/><br/>$$
\begin{aligned}
\Delta_3 & =\left|\begin{array}{ccc}
1 & 1 & 6 \\
1 & 2 & 10 \\
1 & 3 & \beta
\end{array}\right| \\
& =1(2 \beta-30)-1(\beta-10)+6(3-2) \\\\
& =2 \beta-30-\beta+10+6 \\\\
& =\beta-14
\end{aligned}
$$
<br/><br/>Thus, at $\beta=14, \Delta_1=\Delta_2=\Delta_3=0$
<br/><br/>$$
\Rightarrow \alpha=3, \beta=14$$
<br/><br/>So, system has infinite solutions.
|
mcq
|
jee-main-2023-online-6th-april-evening-shift
|
lsam7nu4
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
Let the system of equations $x+2 y+3 z=5,2 x+3 y+z=9,4 x+3 y+\lambda z=\mu$ have infinite number of solutions. Then $\lambda+2 \mu$ is equal to :
|
[{"identifier": "A", "content": "22"}, {"identifier": "B", "content": "17"}, {"identifier": "C", "content": "15"}, {"identifier": "D", "content": "28"}]
|
["B"]
| null |
$$
\begin{aligned}
& x+2 y+3 z=5 \\\\
& 2 x+3 y+z=9 \\\\
& 4 x+3 y+\lambda z=\mu
\end{aligned}
$$
<br/><br/>For infinite following $\Delta=\Delta_1=\Delta_2=\Delta_3=0$
<br/><br/>$\begin{aligned} & \Delta=\left|\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1 \\ 4 & 3 & \lambda\end{array}\right|=0 \Rightarrow \lambda=-13 \\\\ & \Delta_1=\left|\begin{array}{llc}5 & 2 & 3 \\ 9 & 3 & 1 \\ \mu & 3 & -13\end{array}\right|=0 \Rightarrow \mu=15 \\\\ & \Delta_2=\left|\begin{array}{ccc}1 & 5 & 3 \\ 2 & 9 & 1 \\ 4 & 15 & -13\end{array}\right|=0\end{aligned}$
<br/><br/>$$
\Delta_3=\left|\begin{array}{ccc}
1 & 2 & 5 \\
2 & 3 & 9 \\
4 & 3 & 15
\end{array}\right|=0
$$
<br/><br/>For $\lambda=-13, \mu=15$ system of equation has infinite solution hence $\lambda+2 \mu=17$
|
mcq
|
jee-main-2024-online-1st-february-evening-shift
|
lsaorfpq
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
If the system of equations
<br/><br/>$$
\begin{aligned}
& 2 x+3 y-z=5 \\\\
& x+\alpha y+3 z=-4 \\\\
& 3 x-y+\beta z=7
\end{aligned}
$$
<br/><br/>has infinitely many solutions, then $13 \alpha \beta$ is equal to :
|
[{"identifier": "A", "content": "1110"}, {"identifier": "B", "content": "1120"}, {"identifier": "C", "content": "1210"}, {"identifier": "D", "content": "1220"}]
|
["B"]
| null |
$\begin{aligned} & \text { Given } 2 x+3 y-z=5 \\\\ & x+\alpha y+3 z=-4 \\\\ & 3 x-y+\beta z=7 \\\\ & \Delta_2=\left|\begin{array}{ccc}2 & -1 & 5 \\ 1 & 3 & -4 \\ 3 & \beta & 7\end{array}\right| \\\\ & \Delta_2=2(21+4 \beta)+1(7+12)+5(\beta-9) \\\\& \Delta_2=42+8 \beta+19+5 \beta-45 \\\\ & \Delta_2=13 \beta+16 \\\\ & \Delta_2=0\end{aligned}$
<br/><br/>$\begin{aligned} & \therefore \beta=-\frac{16}{13} \\\\ & \Delta_3=\left|\begin{array}{lll}2 & 3 & 5 \\ 1 & \alpha & -4 \\ 3 & -1 & 7\end{array}\right| \\\\ & \Delta_3=2(7 \alpha-4)-3(7+12)+5(-1-3 \alpha) \\\\ & \Delta_3=14 \alpha-8-57-5-15 \alpha \\\\ & \Delta_3=-\alpha-70\end{aligned}$
<br/><br/>$\begin{aligned} & \Delta_3=0 \\\\ & \alpha=-70 \\\\ & 13 \alpha \beta=(13)(-70)\left(-\frac{16}{13}\right) \\\\ & =+1120\end{aligned}$
|
mcq
|
jee-main-2024-online-1st-february-morning-shift
|
jaoe38c1lsd3de46
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>Let $$A$$ be a $$3 \times 3$$ real matrix such that</p>
<p>$$A\left(\begin{array}{l}
1 \\
0 \\
1
\end{array}\right)=2\left(\begin{array}{l}
1 \\
0 \\
1
\end{array}\right), A\left(\begin{array}{l}
-1 \\
0 \\
1
\end{array}\right)=4\left(\begin{array}{l}
-1 \\
0 \\
1
\end{array}\right), A\left(\begin{array}{l}
0 \\
1 \\
0
\end{array}\right)=2\left(\begin{array}{l}
0 \\
1 \\
0
\end{array}\right) \text {. }$$</p>
<p>Then, the system $$(A-3 I)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)$$ has :</p>
|
[{"identifier": "A", "content": "exactly two solutions\n"}, {"identifier": "B", "content": "infinitely many solutions\n"}, {"identifier": "C", "content": "unique solution\n"}, {"identifier": "D", "content": "no solution"}]
|
["C"]
| null |
<p>$$\text { Let } A=\left[\begin{array}{lll}
x_1 & y_1 & z_1 \\
x_2 & y_2 & z_2 \\
x_3 & y_3 & z_3
\end{array}\right]$$</p>
<p>$$\text { Given } A\left[\begin{array}{l}
1 \\
0 \\
1
\end{array}\right]=\left[\begin{array}{l}
2 \\
0 \\
2
\end{array}\right] \quad \text{ ..... (1)}$$</p>
<p>$$\therefore\left[\begin{array}{l}
\mathrm{x}_1+\mathrm{z}_1 \\
\mathrm{x}_2+\mathrm{z}_2 \\
\mathrm{x}_3+\mathrm{z}_3
\end{array}\right]=\left[\begin{array}{l}
2 \\
0 \\
2
\end{array}\right]$$</p>
<p>$$\begin{aligned}
\therefore \mathrm{x}_1+\mathrm{z}_1 & =2 \quad \text{.... (2)}\\
\mathrm{x}_2+\mathrm{z}_2 & =0 \quad \text{.... (3)}\\
\mathrm{x}_3+\mathrm{z}_3 & =0 \quad \text{.... (4)}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { Given } A\left[\begin{array}{l}
-1 \\
0 \\
1
\end{array}\right]=\left[\begin{array}{l}
-4 \\
0 \\
4
\end{array}\right] \\
& \therefore\left[\begin{array}{l}
-\mathrm{x}_1+\mathrm{z}_1 \\
-\mathrm{x}_2+\mathrm{z}_2 \\
-\mathrm{x}_3+\mathrm{z}_3
\end{array}\right]=\left[\begin{array}{l}
4 \\
0 \\
4
\end{array}\right]
\end{aligned}$$</p>
<p>$$\begin{array}{r}
\Rightarrow-\mathrm{x}_1+\mathrm{z}_1=-4 \quad \text{... (5)}\\
-\mathrm{x}_2+\mathrm{x}_2=0 \quad \text{... (6)}\\
-\mathrm{x}_3+\mathrm{z}_3=4
\end{array}$$</p>
<p>$$\begin{aligned}
& \text { Given } A\left[\begin{array}{l}
0 \\
1 \\
0
\end{array}\right]=\left[\begin{array}{l}
0 \\
2 \\
0
\end{array}\right] \\
& \therefore\left[\begin{array}{l}
\mathrm{y}_1 \\
\mathrm{y}_2 \\
\mathrm{y}_3
\end{array}\right]=\left[\begin{array}{l}
0 \\
2 \\
0
\end{array}\right]
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \therefore \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0 \\
& \therefore \text { from }(2),(3),(4),(5),(6) \text { and }(7) \\
& \mathrm{x}_1=3 \mathrm{x}, \mathrm{x}_2=0, \mathrm{x}_3=-1 \\
& \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0 \\
& \mathrm{z}_1=-1, \mathrm{z}_2=0, \mathrm{z}_3=3
\end{aligned}$$</p>
<p>$$\begin{gathered}
\therefore A=\left[\begin{array}{ccc}
3 & 0 & -1 \\
0 & 2 & 0 \\
-1 & 0 & 3
\end{array}\right] \\
\therefore \text { Now }(A-31)\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
-1 \\
2 \\
3
\end{array}\right] \\
\therefore\left[\begin{array}{ccc}
0 & 0 & -1 \\
0 & -1 & 0 \\
-1 & 0 & 0
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
-1 \\
2 \\
3
\end{array}\right] \\
{\left[\begin{array}{c}
-z \\
-y \\
-x
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]} \\
{[z=-1],[y=-2],[x=-3]}
\end{gathered}$$</p>
|
mcq
|
jee-main-2024-online-31st-january-evening-shift
|
jaoe38c1lse5aryy
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>If the system of linear equations</p>
<p>$$\begin{aligned}
& x-2 y+z=-4 \\
& 2 x+\alpha y+3 z=5 \\
& 3 x-y+\beta z=3
\end{aligned}$$</p>
<p>has infinitely many solutions, then $$12 \alpha+13 \beta$$ is equal to</p>
|
[{"identifier": "A", "content": "60"}, {"identifier": "B", "content": "54"}, {"identifier": "C", "content": "64"}, {"identifier": "D", "content": "58"}]
|
["D"]
| null |
<p>$$\begin{aligned}
& D=\left|\begin{array}{ccc}
1 & -2 & 1 \\
2 & \alpha & 3 \\
3 & -1 & \beta
\end{array}\right| \\
& =1(\alpha \beta+3)+2(2 \beta-9)+1(-2-3 \alpha) \\
& =\alpha \beta+3+4 \beta-18-2-3 \alpha
\end{aligned}$$</p>
<p>For infinite solutions $$\mathrm{D}=0, \mathrm{D}_1=0, \mathrm{D}_2=0$$ and</p>
<p>$$\begin{aligned}
& \mathrm{D}_3=0 \\
& \mathrm{D}=0
\end{aligned}$$</p>
<p>$$\alpha \beta-3 \alpha+4 \beta=17$$ ..... (1)</p>
<p>$$\begin{aligned}
& \mathrm{D}_1=\left|\begin{array}{ccc}
-4 & -2 & 1 \\
5 & \alpha & 3 \\
3 & -1 & \beta
\end{array}\right|=0 \\
& \mathrm{D}_2=\left|\begin{array}{ccc}
1 & -4 & 1 \\
2 & 5 & 3 \\
3 & 3 & \beta
\end{array}\right|=0 \\
& \Rightarrow 1(5 \beta-9)+4(2 \beta-9)+1(6-15)=0 \\
& 13 \beta-9-36-9=0
\end{aligned}$$</p>
<p>$$13 \beta=54, \beta=\frac{54}{13}$$ put in (1)</p>
<p>$$\frac{54}{13} \alpha-3 \alpha+4\left(\frac{54}{13}\right)=17$$</p>
<p>$$54 \alpha-39 \alpha+216=221$$</p>
<p>$$15 \alpha=5 \quad \alpha=\frac{1}{3}$$</p>
<p>Now, $$12 \alpha+13 \beta=12 \cdot \frac{1}{3}+13 \cdot \frac{54}{13}$$</p> <p>$$=4+54=58$$</p>
|
mcq
|
jee-main-2024-online-31st-january-morning-shift
|
jaoe38c1lsfl56cy
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>Let for any three distinct consecutive terms $$a, b, c$$ of an A.P, the lines $$a x+b y+c=0$$ be concurrent at the point $$P$$ and $$Q(\alpha, \beta)$$ be a point such that the system of equations</p>
<p>$$\begin{aligned}
& x+y+z=6, \\
& 2 x+5 y+\alpha z=\beta \text { and }
\end{aligned}$$</p>
<p>$$x+2 y+3 z=4$$, has infinitely many solutions. Then $$(P Q)^2$$ is equal to _________.</p>
|
[]
| null |
113
|
<p>$$\because \mathrm{a}, \mathrm{b}, \mathrm{c}$$ and in A.P</p>
<p>$$\Rightarrow 2 b=a+c \Rightarrow a-2 b+c=0$$</p>
<p>$$\therefore \mathrm{ax}+\mathrm{by}+\mathrm{c}$$ passes through fixed point $$(1,-2)$$</p>
<p>$$\therefore \mathrm{P}=(1,-2)$$</p>
<p>For infinite solution,</p>
<p>$$\begin{aligned}
& D=D_1=D_2=D_3=0 \\
& D:\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 5 & \alpha \\
1 & 2 & 3
\end{array}\right|=0 \\
& \Rightarrow \alpha=8 \\
& D_1:\left|\begin{array}{lll}
6 & 1 & 1 \\
\beta & 5 & \alpha \\
4 & 2 & 3
\end{array}\right|=0 \Rightarrow \beta=6 \\
& \therefore Q=(8,6) \\
& \therefore Q^2=113
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-29th-january-evening-shift
|
1lsg4izdm
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>Consider the system of linear equations $$x+y+z=5, x+2 y+\lambda^2 z=9, x+3 y+\lambda z=\mu$$, where $$\lambda, \mu \in \mathbb{R}$$. Then, which of the following statement is NOT correct?</p>
|
[{"identifier": "A", "content": "System is consistent if $$\\lambda \\neq 1$$ and $$\\mu=13$$\n"}, {"identifier": "B", "content": "System is inconsistent if $$\\lambda=1$$ and $$\\mu \\neq 13$$\n"}, {"identifier": "C", "content": "System has unique solution if $$\\lambda \\neq 1$$ and $$\\mu \\neq 13$$\n"}, {"identifier": "D", "content": "System has infinite number of solutions if $$\\lambda=1$$ and $$\\mu=13$$"}]
|
["C"]
| null |
<p>$$\begin{aligned}
& \left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & \lambda^2 \\
1 & 3 & \lambda
\end{array}\right|=0 \\
& \Rightarrow 2 \lambda^2-\lambda-1=0 \\
& \lambda=1,-\frac{1}{2} \\
& \left|\begin{array}{ccc}
1 & 1 & 5 \\
2 & \lambda^2 & 9 \\
3 & \lambda & \mu
\end{array}\right|=0 \Rightarrow \mu=13
\end{aligned}$$</p>
<p>Infinite solution $$\lambda=1 \& \mu=13$$</p>
<p>For unique $$\operatorname{sol}^{\mathrm{n}} \lambda \neq 1$$</p>
<p>For no $$\operatorname{sol}^{\mathrm{n}} \lambda=1 \& \mu \neq 13$$</p>
<p>If $$\lambda \neq 1$$ and $$\mu \neq 13$$</p>
<p>Considering the case when $$\lambda=-\frac{1}{2}$$ and $$\mu \neq 13$$ this will generate no solution case</p>
|
mcq
|
jee-main-2024-online-30th-january-evening-shift
|
1lsga8e3h
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>Consider the system of linear equations $$x+y+z=4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda^2 z=\mu^2+15$$ where $$\lambda, \mu \in \mathbf{R}$$. Which one of the following statements is NOT correct ?</p>
|
[{"identifier": "A", "content": "The system has unique solution if $$\\lambda \\neq \\frac{1}{2}$$ and $$\\mu \\neq 1,15$$"}, {"identifier": "B", "content": "The system has infinite number of solutions if $$\\lambda=\\frac{1}{2}$$ and $$\\mu=15$$\n"}, {"identifier": "C", "content": "The system is consistent if $$\\lambda \\neq \\frac{1}{2}$$\n"}, {"identifier": "D", "content": "The system is inconsistent if $$\\lambda=\\frac{1}{2}$$ and $$\\mu \\neq 1$$"}]
|
["D"]
| null |
<p>$$x+y+z=4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda{ }^2 z=\mu^2+15$$,</p>
<p>$$\Delta=\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & 2 \lambda \\
1 & 3 & 4 \lambda^2
\end{array}\right|=(2 \lambda-1)^2$$</p>
<p>For unique solution $$\Delta \neq 0,2 \lambda-1 \neq 0,\left(\lambda \neq \frac{1}{2}\right)$$</p>
<p>Let $$\Delta=0, \lambda=\frac{1}{2}$$</p>
<p>$$\begin{aligned}
& \Delta_y=0, \Delta_x=\Delta_z=\left|\begin{array}{ccc}
4 \mu & 1 & 1 \\
10 \mu & 2 & 1 \\
\mu^2+15 & 3 & 1
\end{array}\right| \\
& =(\mu-15)(\mu-1)
\end{aligned}$$</p>
<p>For infinite solution $$\lambda=\frac{1}{2}, \mu=1$$ or 15</p>
|
mcq
|
jee-main-2024-online-30th-january-morning-shift
|
luxwcbqo
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>Consider the matrices : $$A=\left[\begin{array}{cc}2 & -5 \\ 3 & m\end{array}\right], B=\left[\begin{array}{l}20 \\ m\end{array}\right]$$ and $$X=\left[\begin{array}{l}x \\ y\end{array}\right]$$. Let the set of all $$m$$, for which the system of equations $$A X=B$$ has a negative solution (i.e., $$x<0$$ and $$y<0$$), be the interval $$(a, b)$$. Then $$8 \int_\limits a^b|A| d m$$ is equal to _________.</p>
|
[]
| null |
450
|
<p>$$\begin{aligned}
& A X=B \\
& 2 x-5 y=20 \\
& 3 x+m y=m \\
& \Rightarrow 3\left(\frac{20+5 y}{2}\right)+m y=m
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow 30+\frac{15}{2} y+m y=m \\
& \Rightarrow y\left(\frac{15}{2}+m\right)=m-30 \\
& \Rightarrow y=\frac{m-30}{\frac{15}{2}+m}<0 \Rightarrow m \in\left(-\frac{15}{2}, 30\right)
\end{aligned}$$</p>
<p>Similarly : $$3 x+m\left(\frac{2 x-20}{5}\right)=m$$</p>
<p>$$\begin{aligned}
\Rightarrow & 3 x+\frac{2 m x}{5}-\frac{20 m}{5}=m \\
\Rightarrow & \frac{15 x+2 m x}{5}=5 m \Rightarrow x=\frac{25 m}{15+2 m} \\
& x<0 \Rightarrow \frac{25 m}{15+2 m}<0 \Rightarrow m \in\left(-\frac{15}{2}, 0\right) \\
\therefore \quad & m \in\left(-\frac{15}{2}, 0\right) \\
& a=-\frac{15}{2}, b=0 \\
& 8 \int_\limits{-\frac{15}{2}}^0(2 m+15) d m=450 \\
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-9th-april-evening-shift
|
luy6z5o7
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>Let $$\lambda, \mu \in \mathbf{R}$$. If the system of equations</p>
<p>$$\begin{aligned}
& 3 x+5 y+\lambda z=3 \\
& 7 x+11 y-9 z=2 \\
& 97 x+155 y-189 z=\mu
\end{aligned}$$</p>
<p>has infinitely many solutions, then $$\mu+2 \lambda$$ is equal to :</p>
|
[{"identifier": "A", "content": "24"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "27"}, {"identifier": "D", "content": "22"}]
|
["B"]
| null |
<p>$$\begin{aligned}
& 3 x+5 y+\lambda z=3 \\
& 7 x+11 y-9 z=2 \\
& 97 x+155 y-189 z=\mu
\end{aligned}$$</p>
<p>$$\begin{aligned}
& {\left[\begin{array}{ccc}
3 & 5 & \lambda \\
7 & 11 & -9 \\
97 & 155 & -189
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
3 \\
2 \\
\mu
\end{array}\right]} \\
& A X=B \\
& X=A^{-1} B \\
& X=\frac{\operatorname{adj} A}{|A|} B
\end{aligned}$$</p>
<p>For Infinitely many solution</p>
<p>$$\begin{aligned}
& |A|=0 \text { and }(\operatorname{adj} A) B=0 \\
& \left|\begin{array}{ccc}
3 & 5 & \lambda \\
7 & 11 & -9 \\
97 & 155 & -189
\end{array}\right|=0 \\
& -2052+2250+18 \lambda=0 \\
& \Rightarrow \lambda=-11 \\
& \operatorname{adj} A=\left[\begin{array}{ccc}
-684 & -760 & 76 \\
450 & 500 & -50 \\
18 & 20 & -2
\end{array}\right] \\
& (\operatorname{adj} A) B=\left[\begin{array}{ccc}
684 & -760 & 76 \\
450 & 500 & -50 \\
18 & 20 & -2
\end{array}\right]\left[\begin{array}{l}
3 \\
2 \\
\mu
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right] \\
& \Rightarrow 54+40-2 \mu=0 \\
& \Rightarrow 2 \mu=94 \\
& \Rightarrow \mu=47 \\
& \Rightarrow \mu+2 \lambda=25
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-9th-april-morning-shift
|
lv0vxd86
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>If the system of equations</p>
<p>$$\begin{aligned}
& x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\
& x+(\cos \alpha) y+(\sin \alpha) z=0 \\
& x+(\sin \alpha) y-(\cos \alpha) z=0
\end{aligned}$$</p>
<p>has a non-trivial solution, then $$\alpha \in\left(0, \frac{\pi}{2}\right)$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$\\frac{5 \\pi}{24}$$\n"}, {"identifier": "B", "content": "$$\\frac{11 \\pi}{24}$$\n"}, {"identifier": "C", "content": "$$\\frac{7 \\pi}{24}$$\n"}, {"identifier": "D", "content": "$$\\frac{3 \\pi}{4}$$"}]
|
["A"]
| null |
<p>$$\begin{aligned}
& x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\
& x+(\cos \alpha) y+(\sin \alpha) z=0 \\
& x+(\sin \alpha) y-(\cos \alpha) z=0
\end{aligned}$$</p>
<p>$$\because$$ Non-trivial solution</p>
<p>$$\Rightarrow D=0$$</p>
<p>$$\begin{aligned}
& \left|\begin{array}{ccc}
1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \\
1 & \cos \alpha & \sin \alpha \\
1 & \sin \alpha & -\cos \alpha
\end{array}\right|=0 \\
& 1{\left[ { - {{\cos }^2}\alpha - \sin \alpha } \right]^{ - 1}}\left[ { - \sqrt 2 \sin \alpha \cos \alpha - \sqrt 2 \sin \alpha \cos \alpha } \right] + 1\left[ {\sqrt 2 {{\sin }^2}\alpha - \sqrt 2 {{\cos }^2}\alpha } \right] = 0 \\
\end{aligned}$$</p>
<p>$$\begin{aligned}
& -1+2 \sqrt{2} \sin \alpha \cos \alpha+\sqrt{2}\left(\sin ^2 \alpha-\cos ^2 \alpha\right)=0 \\
& \sqrt{2} \sin 2 \alpha-\sqrt{2} \cos 2 \alpha=1 \\
& \frac{\sin 2 \alpha}{\sqrt{2}}-\frac{\cos 2 \alpha}{\sqrt{2}}=\frac{1}{2} \\
& \sin \left(2 \alpha-\frac{\pi}{4}\right)=\sin \frac{\pi}{6} \\
& \Rightarrow 2 \alpha-\frac{\pi}{4}=n \pi+(-1)^n \frac{\pi}{6} \text { for } n=0 \\
& \Rightarrow \alpha=\frac{5 \pi}{24}
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-4th-april-morning-shift
|
lv3ve3w9
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>If the system of equations $$x+4 y-z=\lambda, 7 x+9 y+\mu z=-3,5 x+y+2 z=-1$$ has infinitely many solutions, then $$(2 \mu+3 \lambda)$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$-2$$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$$-3$$"}]
|
["D"]
| null |
<p>$$\begin{aligned}
& x+4 y-z=\lambda \\
& 7 x+9 y+\mu z=-3 \\
& 5 x+y+2 z=-1 \\
& {\left[\begin{array}{ccc}
1 & 4 & -1 \\
7 & 9 & \mu \\
5 & 1 & 2
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
\lambda \\
-3 \\
-1
\end{array}\right]} \\
& A=\left[\begin{array}{lll}
1 & 4 & -1 \\
7 & 9 & \mu \\
5 & 1 & 2
\end{array}\right], B=\left[\begin{array}{c}
\lambda \\
-3 \\
-1
\end{array}\right] \\
& A X=B \\
& X=A^{-1} B \\
& \quad=\frac{\operatorname{adj} A}{|A|} B
\end{aligned}$$</p>
<p>If $$|A|=0$$ and $$(\operatorname{adj} A) \cdot B=0$$, system has infinitely many solutions.</p>
<p>$$\begin{aligned}
& |A|=18-\mu-4(14-5 \mu)-1(7-45)=0 \\
& \Rightarrow 18-\mu-56+20 \mu+38=0 \\
& \Rightarrow 19 \mu=0 \\
& \Rightarrow \mu=0 \\
& \text { Also adjA }=\left[\begin{array}{ccc}
18 & -9 & 9 \\
-14 & 7 & -7 \\
-38 & 19 & -19
\end{array}\right] \\
& (\text { adj } A) \cdot B=0 \\
& {\left[\begin{array}{ccc}
18 & -9 & 9 \\
-14 & 7 & -7 \\
-38 & 19 & -19
\end{array}\right]\left[\begin{array}{c}
\lambda \\
-3 \\
-1
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]} \\
& 18 \lambda+27-9=0 \\
& \Rightarrow 18 \lambda=-18 \\
& \Rightarrow \lambda=-1 \\
& \Rightarrow 2 \mu+3 \lambda=3(-1)=-3
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-8th-april-evening-shift
|
lv7v3k8u
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>If the system of equations</p>
<p>$$\begin{array}{r}
11 x+y+\lambda z=-5 \\
2 x+3 y+5 z=3 \\
8 x-19 y-39 z=\mu
\end{array}$$</p>
<p>has infinitely many solutions, then $$\lambda^4-\mu$$ is equal to :</p>
|
[{"identifier": "A", "content": "51"}, {"identifier": "B", "content": "45"}, {"identifier": "C", "content": "47"}, {"identifier": "D", "content": "49"}]
|
["C"]
| null |
<p>$$\begin{aligned}
& 11 x+y+\lambda z=-5 \\
& 2 x+3 y+5 z=3 \\
& 8 x-19 y-39 z=\mu \\
& \Delta=0 \Rightarrow\left|\begin{array}{ccc}
11 & 1 & \lambda \\
2 & 3 & 5 \\
8 & -19 & -39
\end{array}\right|=0 \\
& 11(-39.3+19.5)-1(-39.2-40)+\lambda(-38-24)=0 \\
& =11(-117+95)-1(-118)-62 \lambda=0 \\
& =-242+118=62 \lambda \\
& \Rightarrow \lambda=-2 \\
& \Delta 2=0 \\
& \Rightarrow\left|\begin{array}{ccc}
11 & 1 & -5 \\
2 & 3 & 3 \\
8 & -19 & \mu
\end{array}\right|=0
\end{aligned}$$</p>
<p>$$\begin{aligned}
& 11(3 \mu+57)-1(2 \mu-24)-5(-38-24)=0 \\
& 33 \mu+627-2 \mu+24+310=0 \\
& \mu=-31 \\
& \Rightarrow \lambda^4-31 \\
& =16+31 \\
& =47
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-5th-april-morning-shift
|
lv9s2037
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>The values of $$m, n$$, for which the system of equations</p>
<p>$$\begin{aligned}
& x+y+z=4, \\
& 2 x+5 y+5 z=17, \\
& x+2 y+\mathrm{m} z=\mathrm{n}
\end{aligned}$$</p>
<p>has infinitely many solutions, satisfy the equation :</p>
|
[{"identifier": "A", "content": "$$\\mathrm{m}^2+\\mathrm{n}^2-\\mathrm{m}-\\mathrm{n}=46$$\n"}, {"identifier": "B", "content": "$$\\mathrm{m}^2+\\mathrm{n}^2+\\mathrm{mn}=68$$\n"}, {"identifier": "C", "content": "$$\\mathrm{m}^2+\\mathrm{n}^2-\\mathrm{mn}=39$$\n"}, {"identifier": "D", "content": "$$\\mathrm{m}^2+\\mathrm{n}^2+\\mathrm{m}+\\mathrm{n}=64$$"}]
|
["C"]
| null |
<p>The given system of linear equations can be represented as,</p>
<p>$$\begin{aligned}
& \left(\begin{array}{ccc|c}
1 & 1 & 1 & 4 \\
2 & 5 & 5 & 17 \\
1 & 2 & m & n
\end{array}\right) \\
& \sim\left(\begin{array}{ccc|c}
1 & 1 & 1 & 4 \\
0 & 3 & 3 & 9 \\
0 & 1 & m-1 & n-4
\end{array}\right) \\
& \sim\left(\begin{array}{ccc|c}
1 & 1 & 1 & 4 \\
0 & 1 & 1 & 3 \\
0 & 0 & m-2 & n-7
\end{array}\right)
\end{aligned}$$</p>
<p>$$\because$$ System of equations has infinitely many solutions</p>
<p>$$\therefore m=2 \& n=7$$</p>
<p>Which satisfy equation given in option (1).</p>
<p>$$\text { (i.e., } 2^2+7^2-14=39 \text { ) }$$</p>
|
mcq
|
jee-main-2024-online-5th-april-evening-shift
|
lvb29535
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>If the system of equations</p>
<p>$$\begin{aligned}
& 2 x+7 y+\lambda z=3 \\
& 3 x+2 y+5 z=4 \\
& x+\mu y+32 z=-1
\end{aligned}$$</p>
<p>has infinitely many solutions, then $$(\lambda-\mu)$$ is equal to ______ :</p>
|
[]
| null |
38
|
<p>To determine if the system of equations:</p>
<p>$$\begin{aligned} 2x + 7y + \lambda z = 3 \\ 3x + 2y + 5z = 4 \\ x + \mu y + 32z = -1 \end{aligned}$$</p>
<p>has infinitely many solutions, we must use Cramer's rule.</p>
<p>The determinants are calculated as follows:</p>
<p>$$\begin{aligned} \Delta &= -2\lambda + 3\lambda\mu - 10\mu - 509 \\ \Delta_1 &= 2\lambda + 3\lambda\mu - 15\mu - 739 \\ \Delta_2 &= -7\lambda - 7 \\ \Delta_3 &= \mu + 39 \end{aligned}$$</p>
<p>To have infinitely many solutions, the determinants must satisfy:</p>
<p>$$\begin{aligned} \Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0 \end{aligned}$$</p>
<p>Solving these equations, we find:</p>
<p>$$\lambda = -1, \mu = -39$$</p>
<p>Thus, the value of $ \lambda - \mu $ is:</p>
<p>$$ \lambda - \mu = 38 $$</p>
|
integer
|
jee-main-2024-online-6th-april-evening-shift
|
lvc57b1q
|
maths
|
matrices-and-determinants
|
solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices
|
<p>Let $$\alpha \beta \gamma=45 ; \alpha, \beta, \gamma \in \mathbb{R}$$. If $$x(\alpha, 1,2)+y(1, \beta, 2)+z(2,3, \gamma)=(0,0,0)$$ for some $$x, y, z \in \mathbb{R}, x y z \neq 0$$, then $$6 \alpha+4 \beta+\gamma$$ is equal to _________.</p>
|
[]
| null |
55
|
<p>Given that $\alpha \beta \gamma = 45$ and $\alpha, \beta, \gamma \in \mathbb{R}$, consider the equation $x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0)$ for some $x, y, z \in \mathbb{R}$ where $x y z \neq 0$. To find the value of $6 \alpha + 4 \beta + \gamma$, follow these steps:</p>
<ol>
<li>Express the given equation in matrix form:</li>
</ol>
<p>$ \begin{aligned} & \alpha x + y + 2 z = 0 \\ & x + \beta y + 3 z = 0 \\ & 2 x + 2 y + \gamma z = 0 \end{aligned} $</p>
<ol>
<li>Since $x, y, z \neq 0$, the determinant of the coefficients matrix must be zero:</li>
</ol>
<p>$ \left|\begin{array}{ccc} \alpha & 1 & 2 \\ 1 & \beta & 3 \\ 2 & 2 & \gamma \end{array}\right| = 0 $</p>
<ol>
<li>Calculate the determinant of the matrix:</li>
</ol>
<p>$ \alpha \beta \gamma - 6 \alpha - 4 \beta - \gamma + 10 = 0 $</p>
<ol>
<li>Given $\alpha \beta \gamma = 45$, substitute this value into the equation:</li>
</ol>
<p>$ 45 - 6 \alpha - 4 \beta - \gamma + 10 = 0 $</p>
<ol>
<li>Simplify the equation:</li>
</ol>
<p>$ 45 + 10 = 6 \alpha + 4 \beta + \gamma $</p>
<ol>
<li>Thus,</li>
</ol>
<p>$ 6 \alpha + 4 \beta + \gamma = 55 $</p>
<p>So, the value of $6 \alpha + 4 \beta + \gamma$ is 55.</p>
|
integer
|
jee-main-2024-online-6th-april-morning-shift
|
C0L4Ct0WYGS8JsmI
|
maths
|
matrices-and-determinants
|
symmetric-and-skew-symmetric-matrices
|
Let $$A$$ and $$B$$ be two symmetric matrices of order $$3$$.
<br/><br><b>Statement - 1 :</b> $$A(BA)$$ and $$(AB)$$$$A$$ are symmetric matrices.
<br/><br><b>Statement - 2 :</b> $$AB$$ is symmetric matrix if matrix multiplication of $$A$$ with $$B$$ is commutative.</br></br>
|
[{"identifier": "A", "content": "statement - 1 is true, statement - 2 is true; statement - 2 is <b>not</b> a correct explanation for statement - 1. "}, {"identifier": "B", "content": "statement - 1 is true, statement - 2 is false. "}, {"identifier": "C", "content": "statement - 1 is false, statement -2 is true "}, {"identifier": "D", "content": "statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1. "}]
|
["A"]
| null |
$$\therefore$$ $$A' = A,B' = B$$
<br><br>Now $$\,\,\,\left( {A\left( {BA} \right)} \right)' = \left( {BA} \right)'A'$$
<br><br>$$ = \left( {A'B'} \right)A' = \left( {AB} \right)A = A\left( {BA} \right)$$
<br><br>Similarly $$\left( {\left( {AB} \right)A} \right)' = \left( {AB} \right)A$$
<br><br>So, $$A\left( {BA} \right)\,\,\,\,$$ and $$A\left( {BA} \right)\,\,\,\,$$ are symmetric matrices.
<br><br>Again $$\left( {AB} \right)' = B'A' = BA$$
<br><br>Now if $$BA=AB$$, then $$AB$$ is symmetric matrix.
|
mcq
|
aieee-2011
|
rfc0afjOYGFlvHdYbD3rsa0w2w9jx6595rt
|
maths
|
matrices-and-determinants
|
symmetric-and-skew-symmetric-matrices
|
If A is a symmetric matrix and B is a skew-symmetric matrix such that A + B = $$\left[ {\matrix{
2 & 3 \cr
5 & { - 1} \cr
} } \right]$$, then AB is equal
to :
|
[{"identifier": "A", "content": "$$\\left[ {\\matrix{\n 4 & { - 2} \\cr \n 1 & { - 4} \\cr \n\n } } \\right]$$"}, {"identifier": "B", "content": "$$\\left[ {\\matrix{\n { - 4} & { - 2} \\cr \n { - 1} & 4 \\cr \n\n } } \\right]$$"}, {"identifier": "C", "content": "$$\\left[ {\\matrix{\n { - 4} & 2 \\cr \n 1 & 4 \\cr \n\n } } \\right]$$"}, {"identifier": "D", "content": "$$\\left[ {\\matrix{\n 4 & { - 2} \\cr \n { - 1} & { - 4} \\cr \n\n } } \\right]$$"}]
|
["D"]
| null |
$$A + B = \left[ {\matrix{
2 & 3 \cr
5 & { - 1} \cr
} } \right] = P(say)$$<br><br>
Now $$A = {{P + {P^T}} \over 2}\& B = {{P - {P^T}} \over 2}$$<br><br>
So $$A = {1 \over 2}\left( {\left[ {\matrix{
2 & 3 \cr
5 & { - 1} \cr
} } \right] + \left[ {\matrix{
2 & 5 \cr
3 & { - 1} \cr
} } \right]} \right) = \left[ {\matrix{
2 & 4 \cr
4 & { - 1} \cr
} } \right]$$<br><br>
$$B = {1 \over 2}\left( {\left[ {\matrix{
2 & 3 \cr
5 & { - 1} \cr
} } \right] - \left[ {\matrix{
2 & 5 \cr
3 & { - 1} \cr
} } \right]} \right) = \left[ {\matrix{
0 & { - 1} \cr
1 & 0 \cr
} } \right]$$<br><br>
So $$AB = \left( {\left[ {\matrix{
2 & 4 \cr
4 & { - 1} \cr
} } \right]\left[ {\matrix{
0 & { - 1} \cr
1 & 0 \cr
} } \right]} \right) = \left[ {\matrix{
4 & { - 2} \cr
{ - 1} & { - 4} \cr
} } \right]$$
|
mcq
|
jee-main-2019-online-12th-april-morning-slot
|
0EINDVkT81wWvriaR41klrkbb9t
|
maths
|
matrices-and-determinants
|
symmetric-and-skew-symmetric-matrices
|
Let A and B be 3 $$\times$$ 3 real matrices such that A is symmetric matrix and B is skew-symmetric matrix. Then the system of linear equations (A<sup>2</sup>B<sup>2</sup> $$-$$ B<sup>2</sup>A<sup>2</sup>) X = O, where X is a 3 $$\times$$ 1 column matrix of unknown variables and O is a 3 $$\times$$ 1 null matrix, has :
|
[{"identifier": "A", "content": "no solution"}, {"identifier": "B", "content": "exactly two solutions"}, {"identifier": "C", "content": "infinitely many solutions"}, {"identifier": "D", "content": "a unique solution"}]
|
["C"]
| null |
A<sup>T</sup> = A, B<sup>T</sup> = $$-$$B<br><br>Let A<sup>2</sup>B<sup>2</sup> $$-$$ B<sup>2</sup>A<sup>2</sup> = P<br><br>P<sup>T</sup> = (A<sup>2</sup>B<sup>2</sup> $$-$$ B<sup>2</sup>A<sup>2</sup>)<sup>T</sup> = (A<sup>2</sup>B<sup>2</sup>)<sup>T</sup> $$-$$ (B<sup>2</sup>A<sup>2</sup>)<sup>T</sup><br><br>= (B<sup>2</sup>)<sup>T</sup> (A<sup>2</sup>)<sup>T</sup> $$-$$ (A<sup>2</sup>)<sup>T</sup> (B<sup>2</sup>)<sup>T</sup><br><br>= B<sup>2</sup>A<sup>2</sup> $$-$$ A<sup>2</sup>B<sup>2</sup><br><br>$$ \Rightarrow $$ P is skew-symmetric matrix<br><br>$$\left[ {\matrix{
0 & a & b \cr
{ - a} & 0 & c \cr
{ - b} & { - c} & 0 \cr
} } \right]\left[ {\matrix{
x \cr
y \cr
z \cr
} } \right] = \left[ {\matrix{
0 \cr
0 \cr
0 \cr
} } \right]$$<br><br>$$ \therefore $$ ay + bz = 0 ..... (1)<br><br>$$-$$ax + cz = 0 .... (2)<br><br>$$-$$bx $$-$$cy = 0 ..... (3)<br><br>From equation 1, 2, 3<br><br>$$\Delta$$ = 0 & $$\Delta$$<sub>1</sub> = $$\Delta$$<sub>2</sub> = $$\Delta$$<sub>3</sub> = 0<br><br>$$ \therefore $$ equation have infinite number of solution
|
mcq
|
jee-main-2021-online-24th-february-evening-slot
|
BTVrgRhS5ZberuUFEM1klug7r15
|
maths
|
matrices-and-determinants
|
symmetric-and-skew-symmetric-matrices
|
Let A be a symmetric matrix of order 2 with integer entries. If the sum of the diagonal elements of A<sup>2</sup> is 1, then the possible number of such matrices is :
|
[{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "12"}]
|
["B"]
| null |
Let $$A = \left[ {\matrix{
a & b \cr
b & c \cr
} } \right]$$<br><br>$${A^2} = \left[ {\matrix{
a & b \cr
b & c \cr
} } \right]\left[ {\matrix{
a & b \cr
b & c \cr
} } \right] = \left[ {\matrix{
{{a^2} + {b^2}} & {ab + bc} \cr
{ab + bc} & {{c^2} + {b^2}} \cr
} } \right]$$<br><br>$$ = {a^2} + 2{b^2} + {c^2} = 1$$<br><br>$$a = 1,b = 0,c = 0$$<br><br>$$a = 0,b = 0,c = 1$$<br><br>$$a = - 1,b = 0,c = 0$$<br><br>$$c = - 1,b = 0,a = 0$$
|
mcq
|
jee-main-2021-online-26th-february-morning-slot
|
1krptnjcb
|
maths
|
matrices-and-determinants
|
symmetric-and-skew-symmetric-matrices
|
Let $$A = \left[ {\matrix{
2 & 3 \cr
a & 0 \cr
} } \right]$$, a$$\in$$R be written as P + Q where P is a symmetric matrix and Q is skew symmetric matrix. If det(Q) = 9, then the modulus of the sum of all possible values of determinant of P is equal to :
|
[{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "45"}, {"identifier": "D", "content": "18"}]
|
["A"]
| null |
$$A = \left[ {\matrix{
2 & 3 \cr
a & 0 \cr
} } \right]$$, $${A^T} = \left[ {\matrix{
2 & a \cr
3 & 0 \cr
} } \right]$$<br/><br/>$$A = {{A + {A^T}} \over 2} + {{A - {A^T}} \over 2}$$<br/><br/>Let $$P = {{A + {A^T}} \over 2}$$ and $$Q = {{A - {A^T}} \over 2}$$<br/><br/>$$Q = \left( {\matrix{
0 & {{{3 - a} \over 2}} \cr
{{{a - 3} \over 2}} & 0 \cr
} } \right)$$<br/><br/>Det (Q) = 9<br/><br/>$$0 - \left( {{{3 - a} \over 2}} \right)\left( {{{a - 3} \over 2}} \right) = 9$$<br/><br/>$$ \Rightarrow {\left( {{{a - 3} \over 2}} \right)^2} = 9 \Rightarrow {(a - 3)^2} = 36$$<br/><br/>$$a - 3 = \pm \,6 \Rightarrow a = 9, - 3$$<br/><br/>$$P = \left[ {\matrix{
2 & {{{a + 3} \over 2}} \cr
{{{a + 3} \over 2}} & 0 \cr
} } \right]$$<br/><br/>$$P = \left[ {\matrix{
2 & 6 \cr
6 & 0 \cr
} } \right]$$ or $$\left[ {\matrix{
2 & 0 \cr
0 & 0 \cr
} } \right]$$
<br/><br/> | P | = - 36 or 0
<br/><br/>$$\therefore$$ | $$-$$36 + 0 | = 36
|
mcq
|
jee-main-2021-online-20th-july-morning-shift
|
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