Matrices and Determinants
358 Questions
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
The total number of 3 $\times$ 3 matrices A having entries from the set {0, 1, 2, 3} such that the sum of all the diagonal entries of AAT is 9, is equal to _____________.
Correct Answer: 766
Explanation:
$A{A^T} = \left[ {\matrix{
x & y & z \cr
a & b & c \cr
d & e & f \cr
} } \right]\left[ {\matrix{
x & a & d \cr
y & b & e \cr
z & c & f \cr
} } \right]$
$ = \left[ {\matrix{ {{x^2} + {y^2} + {z^2}} & {ax + by + cz} & {dx + ey + fz} \cr {ax + by + cz} & {{a^2} + {b^2} + {c^2}} & {ad + be + cf} \cr {dx + ey + fz} & {ad + be + cf} & {{d^2} + {e^2} + {f^2}} \cr } } \right]$
$Tr(A{A^T}) = {x^2} + {y^2} + {z^2} + {a^2} + {b^2} + {c^2} + {d^2} + {e^2} + {f^2} = 9$
Case-I : Nine ones = 1 case
Case-II : 8 zeroes and one entry is 3 = ${{{9!} \over {8!}} = 9}$ cases
Case-III : Two 2’s, one 1’s and 6 zeroes = ${{9!} \over {2!6!}} = 63 \times 4 = 252$
Case IV : one 2, five 1, rest 0 ${{9!} \over {5!3!}} = 63 \times 8 = 504$
$ \therefore $ Total cases = 9 + 252 + 504 + 1 = 766
$ = \left[ {\matrix{ {{x^2} + {y^2} + {z^2}} & {ax + by + cz} & {dx + ey + fz} \cr {ax + by + cz} & {{a^2} + {b^2} + {c^2}} & {ad + be + cf} \cr {dx + ey + fz} & {ad + be + cf} & {{d^2} + {e^2} + {f^2}} \cr } } \right]$
$Tr(A{A^T}) = {x^2} + {y^2} + {z^2} + {a^2} + {b^2} + {c^2} + {d^2} + {e^2} + {f^2} = 9$
Case-I : Nine ones = 1 case
Case-II : 8 zeroes and one entry is 3 = ${{{9!} \over {8!}} = 9}$ cases
Case-III : Two 2’s, one 1’s and 6 zeroes = ${{9!} \over {2!6!}} = 63 \times 4 = 252$
Case IV : one 2, five 1, rest 0 ${{9!} \over {5!3!}} = 63 \times 8 = 504$
$ \therefore $ Total cases = 9 + 252 + 504 + 1 = 766
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Evening Shift
If the matrix $A = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 2 & 0 \cr
3 & 0 & { - 1} \cr
} } \right]$ satisfies the equation
${A^{20}} + \alpha {A^{19}} + \beta A = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right]$ for some real numbers $\alpha$ and $\beta$, then $\beta$ $-$ $\alpha$ is equal to ___________.
${A^{20}} + \alpha {A^{19}} + \beta A = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right]$ for some real numbers $\alpha$ and $\beta$, then $\beta$ $-$ $\alpha$ is equal to ___________.
Correct Answer: 4
Explanation:
${A^2} = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 2 & 0 \cr
3 & 0 & { - 1} \cr
} } \right]\left[ {\matrix{
1 & 0 & 0 \cr
0 & 2 & 0 \cr
3 & 0 & { - 1} \cr
} } \right] = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 4 & 0 \cr
0 & 0 & 1 \cr
} } \right]$
${A^3} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2 & 0 \cr 3 & 0 & { - 1} \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 8 & 0 \cr 3 & 0 & { - 1} \cr } } \right]$
${A^4} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & {16} & 0 \cr 0 & 0 & 1 \cr } } \right]$
$\eqalign{ & . \cr & . \cr & . \cr & . \cr & . \cr & . \cr} $
${A^{19}} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & {{2^{19}}} & 0 \cr 3 & 0 & { - 1} \cr } } \right],{A^{20}} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & {{2^{20}}} & 0 \cr 0 & 0 & 1 \cr } } \right]$
$L.H.S. = {A^{20}} + \alpha {A^{19}} + \beta A = $
$\left[ {\matrix{ {1 + \alpha + \beta } & 0 & 0 \cr 0 & {{2^{20}} + \alpha {2^{19}} + 2\beta } & 0 \cr {3\alpha + 3\beta } & 0 & {1 - \alpha - \beta } \cr } } \right]$
$R.H.S. = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right] $
$\Rightarrow \alpha + \beta = 0$ and ${2^{20}} + \alpha {2^{19}} + 2\beta = 4$
$ \Rightarrow {2^{20}} + \alpha ({2^{19}} - 2) = 4$
$ \Rightarrow \alpha = {{4 - {2^{20}}} \over {{2^{19}} - 2}} = - 2$
$ \Rightarrow \beta = 2$
$ \therefore $ $\beta - \alpha = 4$
${A^3} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2 & 0 \cr 3 & 0 & { - 1} \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 8 & 0 \cr 3 & 0 & { - 1} \cr } } \right]$
${A^4} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & {16} & 0 \cr 0 & 0 & 1 \cr } } \right]$
$\eqalign{ & . \cr & . \cr & . \cr & . \cr & . \cr & . \cr} $
${A^{19}} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & {{2^{19}}} & 0 \cr 3 & 0 & { - 1} \cr } } \right],{A^{20}} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & {{2^{20}}} & 0 \cr 0 & 0 & 1 \cr } } \right]$
$L.H.S. = {A^{20}} + \alpha {A^{19}} + \beta A = $
$\left[ {\matrix{ {1 + \alpha + \beta } & 0 & 0 \cr 0 & {{2^{20}} + \alpha {2^{19}} + 2\beta } & 0 \cr {3\alpha + 3\beta } & 0 & {1 - \alpha - \beta } \cr } } \right]$
$R.H.S. = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right] $
$\Rightarrow \alpha + \beta = 0$ and ${2^{20}} + \alpha {2^{19}} + 2\beta = 4$
$ \Rightarrow {2^{20}} + \alpha ({2^{19}} - 2) = 4$
$ \Rightarrow \alpha = {{4 - {2^{20}}} \over {{2^{19}} - 2}} = - 2$
$ \Rightarrow \beta = 2$
$ \therefore $ $\beta - \alpha = 4$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Morning Shift
If $A = \left[ {\matrix{
0 & { - \tan \left( {{\theta \over 2}} \right)} \cr
{\tan \left( {{\theta \over 2}} \right)} & 0 \cr
} } \right]$ and
$({I_2} + A){({I_2} - A)^{ - 1}} = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$, then $13({a^2} + {b^2})$ is equal to
$({I_2} + A){({I_2} - A)^{ - 1}} = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$, then $13({a^2} + {b^2})$ is equal to
Correct Answer: 13
Explanation:
$A = \left[ {\matrix{
0 & { - \tan {\theta \over 2}} \cr
{\tan {\theta \over 2}} & 0 \cr
} } \right]$
$ \Rightarrow I + A = \left[ {\matrix{ 1 & { - \tan {\theta \over 2}} \cr {\tan {\theta \over 2}} & 1 \cr } } \right]$
$ \Rightarrow I - A = \left[ {\matrix{ 1 & {\tan {\theta \over 2}} \cr { - \tan {\theta \over 2}} & 1 \cr } } \right]$ { $\therefore$ $\left| {I - A} \right| = {\sec ^2}\theta /2$}
$ \Rightarrow {(I - A)^{ - 1}} = {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{ 1 & { - \tan {\theta \over 2}} \cr {\tan {\theta \over 2}} & 1 \cr } } \right]$
$ \Rightarrow (1 + A){(I - A)^{ - 1}} $
$= {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{ 1 & { - \tan {\theta \over 2}} \cr {\tan {\theta \over 2}} & 1 \cr } } \right]\left[ {\matrix{ 1 & { - \tan {\theta \over 2}} \cr {\tan {\theta \over 2}} & 1 \cr } } \right]$
$ = {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{ {1 - {{\tan }^2}{\theta \over 2}} & { - 2\tan {\theta \over 2}} \cr {2\tan {\theta \over 2}} & {1 - {{\tan }^2}{\theta \over 2}} \cr } } \right]$
$a = {{1 - {{\tan }^2}{\theta \over 2}} \over {{{\sec }^2}{\theta \over 2}}}$
$b = {{2\tan {\theta \over 2}} \over {{{\sec }^2}{\theta \over 2}}}$
$\therefore$ ${a^2} + {b^2} = 1$
$ \Rightarrow $ $13({a^2} + {b^2})$ = 13
$ \Rightarrow I + A = \left[ {\matrix{ 1 & { - \tan {\theta \over 2}} \cr {\tan {\theta \over 2}} & 1 \cr } } \right]$
$ \Rightarrow I - A = \left[ {\matrix{ 1 & {\tan {\theta \over 2}} \cr { - \tan {\theta \over 2}} & 1 \cr } } \right]$ { $\therefore$ $\left| {I - A} \right| = {\sec ^2}\theta /2$}
$ \Rightarrow {(I - A)^{ - 1}} = {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{ 1 & { - \tan {\theta \over 2}} \cr {\tan {\theta \over 2}} & 1 \cr } } \right]$
$ \Rightarrow (1 + A){(I - A)^{ - 1}} $
$= {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{ 1 & { - \tan {\theta \over 2}} \cr {\tan {\theta \over 2}} & 1 \cr } } \right]\left[ {\matrix{ 1 & { - \tan {\theta \over 2}} \cr {\tan {\theta \over 2}} & 1 \cr } } \right]$
$ = {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{ {1 - {{\tan }^2}{\theta \over 2}} & { - 2\tan {\theta \over 2}} \cr {2\tan {\theta \over 2}} & {1 - {{\tan }^2}{\theta \over 2}} \cr } } \right]$
$a = {{1 - {{\tan }^2}{\theta \over 2}} \over {{{\sec }^2}{\theta \over 2}}}$
$b = {{2\tan {\theta \over 2}} \over {{{\sec }^2}{\theta \over 2}}}$
$\therefore$ ${a^2} + {b^2} = 1$
$ \Rightarrow $ $13({a^2} + {b^2})$ = 13
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Morning Shift
Let $A = \left[ {\matrix{
x & y & z \cr
y & z & x \cr
z & x & y \cr
} } \right]$, where x, y and z are real numbers such that x + y + z > 0 and xyz = 2. If ${A^2} = {I_3}$, then the value of ${x^3} + {y^3} + {z^3}$ is ____________.
Correct Answer: 7
Explanation:
$A = \left[ {\matrix{
x & y & z \cr
y & z & x \cr
z & x & y \cr
} } \right]$
$ \therefore $ $|A| = \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)$
Given ${A^2} = {I_3}$
$|{A^2}| = 1$
$ \therefore $ ${({x^3} + {y^3} + {z^3} - 3xyz)^2} = 1$
$ \Rightarrow {x^3} + {y^3} + {z^3} - 3xyz = 1$ only as $(x + y + z > 0)$
$ \Rightarrow {x^3} + {y^3} + {z^3} = 6 + 1 = 7$
$ \therefore $ $|A| = \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)$
Given ${A^2} = {I_3}$
$|{A^2}| = 1$
$ \therefore $ ${({x^3} + {y^3} + {z^3} - 3xyz)^2} = 1$
$ \Rightarrow {x^3} + {y^3} + {z^3} - 3xyz = 1$ only as $(x + y + z > 0)$
$ \Rightarrow {x^3} + {y^3} + {z^3} = 6 + 1 = 7$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Morning Shift
If the system of equations
kx + y + 2z = 1
3x $-$ y $-$ 2z = 2
$-$2x $-$2y $-$4z = 3
has infinitely many solutions, then k is equal to __________.
kx + y + 2z = 1
3x $-$ y $-$ 2z = 2
$-$2x $-$2y $-$4z = 3
has infinitely many solutions, then k is equal to __________.
Correct Answer: 21
Explanation:
D = 0
$ \Rightarrow \left| {\matrix{ k & 1 & 2 \cr 3 & { - 1} & { - 2} \cr { - 2} & { - 2} & { - 4} \cr } } \right| = 0$
$ \Rightarrow $ k (4 $-$ 4) $-$ 1 ($-$ 12 $-$ 4) + 2 ($-$ 6 $-$ 2)
$ \Rightarrow $ 16 $-$ 16 = 0
Also, ${D_1} = {D_2} = {D_3} = 0$
$ \Rightarrow {D_2} = \left| {\matrix{ k & 1 & 2 \cr 3 & 2 & { - 2} \cr { - 2} & 3 & { - 4} \cr } } \right| = 0$
$ \Rightarrow $ k($-$8 + 6) $-$ 1($-$ 12 $-$ 4) + 2(9 + 4) = 0
$ \Rightarrow $ $-$ 2k + 16 + 26 = 0
$ \Rightarrow $ 2k = 42
$ \Rightarrow $ k = 21
$ \Rightarrow \left| {\matrix{ k & 1 & 2 \cr 3 & { - 1} & { - 2} \cr { - 2} & { - 2} & { - 4} \cr } } \right| = 0$
$ \Rightarrow $ k (4 $-$ 4) $-$ 1 ($-$ 12 $-$ 4) + 2 ($-$ 6 $-$ 2)
$ \Rightarrow $ 16 $-$ 16 = 0
Also, ${D_1} = {D_2} = {D_3} = 0$
$ \Rightarrow {D_2} = \left| {\matrix{ k & 1 & 2 \cr 3 & 2 & { - 2} \cr { - 2} & 3 & { - 4} \cr } } \right| = 0$
$ \Rightarrow $ k($-$8 + 6) $-$ 1($-$ 12 $-$ 4) + 2(9 + 4) = 0
$ \Rightarrow $ $-$ 2k + 16 + 26 = 0
$ \Rightarrow $ 2k = 42
$ \Rightarrow $ k = 21
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Morning Shift
Let P = $\left[ {\matrix{
3 & { - 1} & { - 2} \cr
2 & 0 & \alpha \cr
3 & { - 5} & 0 \cr
} } \right]$, where $\alpha $ $ \in $ R. Suppose Q = [ qij] is a matrix satisfying PQ = kl3 for some non-zero k $ \in $ R.
If q23 = $ - {k \over 8}$ and |Q| = ${{{k^2}} \over 2}$, then a2 + k2 is equal to ______.
If q23 = $ - {k \over 8}$ and |Q| = ${{{k^2}} \over 2}$, then a2 + k2 is equal to ______.
Correct Answer: 17
Explanation:
As $PQ = kI \Rightarrow Q = k{P^{ - 1}}I$
now $Q = {k \over {|P|}}(adjP)I $
$\Rightarrow Q = {k \over {(20 + 12\alpha )}}\left[ {\matrix{ - & - & - \cr - & - & {( - 3\alpha - 4)} \cr - & - & - \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$
$ \because $ ${q_{23}} = {{ - k} \over 8} $
$\Rightarrow {k \over {(20 + 12\alpha )}}( - 3\alpha - 4) = {{ - k} \over 8} $
$\Rightarrow 2(3\alpha + 4) = 5 + 3\alpha $
$3\alpha = - 3 \Rightarrow \alpha = - 1$
also $|Q| = {{{k^3}|I|} \over {|P|}} \Rightarrow {{{k^2}} \over 2} = {{{k^3}} \over {(20 + 12\alpha )}}$
$ \Rightarrow $ $(20 + 12\alpha ) = 2k \Rightarrow 8 = 2k \Rightarrow k = 4$
now $Q = {k \over {|P|}}(adjP)I $
$\Rightarrow Q = {k \over {(20 + 12\alpha )}}\left[ {\matrix{ - & - & - \cr - & - & {( - 3\alpha - 4)} \cr - & - & - \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$
$ \because $ ${q_{23}} = {{ - k} \over 8} $
$\Rightarrow {k \over {(20 + 12\alpha )}}( - 3\alpha - 4) = {{ - k} \over 8} $
$\Rightarrow 2(3\alpha + 4) = 5 + 3\alpha $
$3\alpha = - 3 \Rightarrow \alpha = - 1$
also $|Q| = {{{k^3}|I|} \over {|P|}} \Rightarrow {{{k^2}} \over 2} = {{{k^3}} \over {(20 + 12\alpha )}}$
$ \Rightarrow $ $(20 + 12\alpha ) = 2k \Rightarrow 8 = 2k \Rightarrow k = 4$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Morning Shift
Let M be any 3 $ \times $ 3 matrix with entries from the set {0, 1, 2}. The maximum number of such matrices, for which the sum of diagonal elements of MTM is seven, is ________.
Correct Answer: 540
Explanation:
$\left[ {\matrix{
a & b & c \cr
d & e & f \cr
g & h & i \cr
} } \right]\left[ {\matrix{
a & d & g \cr
b & e & h \cr
c & f & i \cr
} } \right]$
${a^2} + {b^2} + {c^2} + {d^2} + {e^2} + {f^2} + {g^2} + {h^2} + {i^2} = 7$
Case I : Seven (1's) and two (0's)
Number of such matrices = ${}^9{C_2} = 36$
Case II : One (2) and three (1's) and five (0's)
Number of such matrices = ${{9!} \over {5!3!}} = 504$
$ \therefore $ Total = 540
${a^2} + {b^2} + {c^2} + {d^2} + {e^2} + {f^2} + {g^2} + {h^2} + {i^2} = 7$
Case I : Seven (1's) and two (0's)
Number of such matrices = ${}^9{C_2} = 36$
Case II : One (2) and three (1's) and five (0's)
Number of such matrices = ${{9!} \over {5!3!}} = 504$
$ \therefore $ Total = 540
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 1st September Evening Shift
Consider the system of linear equations
$-$x + y + 2z = 0
3x $-$ ay + 5z = 1
2x $-$ 2y $-$ az = 7
Let S1 be the set of all a$\in$R for which the system is inconsistent and S2 be the set of all a$\in$R for which the system has infinitely many solutions. If n(S1) and n(S2) denote the number of elements in S1 and S2 respectively, then
$-$x + y + 2z = 0
3x $-$ ay + 5z = 1
2x $-$ 2y $-$ az = 7
Let S1 be the set of all a$\in$R for which the system is inconsistent and S2 be the set of all a$\in$R for which the system has infinitely many solutions. If n(S1) and n(S2) denote the number of elements in S1 and S2 respectively, then
A.
n(S1) = 2, n(S2) = 2
B.
n(S1) = 1, n(S2) = 0
C.
n(S1) = 2, n(S2) = 0
D.
n(S1) = 0, n(S2) = 2
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Evening Shift
If $\alpha$ + $\beta$ + $\gamma$ = 2$\pi$, then the system of equations
x + (cos $\gamma$)y + (cos $\beta$)z = 0
(cos $\gamma$)x + y + (cos $\alpha$)z = 0
(cos $\beta$)x + (cos $\alpha$)y + z = 0
has :
x + (cos $\gamma$)y + (cos $\beta$)z = 0
(cos $\gamma$)x + y + (cos $\alpha$)z = 0
(cos $\beta$)x + (cos $\alpha$)y + z = 0
has :
A.
no solution
B.
infinitely many solution
C.
exactly two solutions
D.
a unique solution
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Morning Shift
If the following system of linear equations
2x + y + z = 5
x $-$ y + z = 3
x + y + az = b
has no solution, then :
2x + y + z = 5
x $-$ y + z = 3
x + y + az = b
has no solution, then :
A.
$a = - {1 \over 3},b \ne {7 \over 3}$
B.
$a \ne {1 \over 3},b = {7 \over 3}$
C.
$a \ne - {1 \over 3},b = {7 \over 3}$
D.
$a = {1 \over 3},b \ne {7 \over 3}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Morning Shift
If ${a_r} = \cos {{2r\pi } \over 9} + i\sin {{2r\pi } \over 9}$, r = 1, 2, 3, ....., i = $\sqrt { - 1} $, then
the determinant $\left| {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right|$ is equal to :
the determinant $\left| {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right|$ is equal to :
A.
a2a6 $-$ a4a8
B.
a9
C.
a1a9 $-$ a3a7
D.
a5
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th August Evening Shift
Let $A = \left( {\matrix{
{[x + 1]} & {[x + 2]} & {[x + 3]} \cr
{[x]} & {[x + 3]} & {[x + 3]} \cr
{[x]} & {[x + 2]} & {[x + 4]} \cr
} } \right)$, where [t] denotes the greatest integer less than or equal to t. If det(A) = 192, then the set of values of x is the interval :
A.
[68, 69)
B.
[62, 63)
C.
[65, 66)
D.
[60, 61)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th August Evening Shift
Let A(a, 0), B(b, 2b + 1) and C(0, b), b $\ne$ 0, |b| $\ne$ 1, be points such that the area of triangle ABC is 1 sq. unit, then the sum of all possible values of a is :
A.
${{ - 2b} \over {b + 1}}$
B.
${{2b} \over {b + 1}}$
C.
${{2{b^2}} \over {b + 1}}$
D.
${{ - 2{b^2}} \over {b + 1}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th August Evening Shift
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
x + y + z = 4,
3x + 2y + 5z = 3,
9x + 4y + (28 + [$\lambda$])z = [$\lambda$] has a solution is :
x + y + z = 4,
3x + 2y + 5z = 3,
9x + 4y + (28 + [$\lambda$])z = [$\lambda$] has a solution is :
A.
R
B.
($-$$\infty$, $-$9) $\cup$ ($-$9, $\infty$)
C.
[$-$9, $-$8)
D.
($-$$\infty$, $-$9) $\cup$ [$-$8, $\infty$)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th August Morning Shift
If the matrix $A = \left( {\matrix{
0 & 2 \cr
K & { - 1} \cr
} } \right)$ satisfies $A({A^3} + 3I) = 2I$, then the value of K is :
A.
${1 \over 2}$
B.
$-$${1 \over 2}$
C.
$-$1
D.
1
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Evening Shift
Let $A = \left( {\matrix{
1 & 0 & 0 \cr
0 & 1 & 1 \cr
1 & 0 & 0 \cr
} } \right)$. Then A2025 $-$ A2020 is equal to :
A.
A6 $-$ A
B.
A5
C.
A5 $-$ A
D.
A6
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Morning Shift
Let $\theta \in \left( {0,{\pi \over 2}} \right)$. If the system of linear equations
$(1 + {\cos ^2}\theta )x + {\sin ^2}\theta y + 4\sin 3\,\theta z = 0$
${\cos ^2}\theta x + (1 + {\sin ^2}\theta )y + 4\sin 3\,\theta z = 0$
${\cos ^2}\theta x + {\sin ^2}\theta y + (1 + 4\sin 3\,\theta )z = 0$
has a non-trivial solution, then the value of $\theta$ is :
$(1 + {\cos ^2}\theta )x + {\sin ^2}\theta y + 4\sin 3\,\theta z = 0$
${\cos ^2}\theta x + (1 + {\sin ^2}\theta )y + 4\sin 3\,\theta z = 0$
${\cos ^2}\theta x + {\sin ^2}\theta y + (1 + 4\sin 3\,\theta )z = 0$
has a non-trivial solution, then the value of $\theta$ is :
A.
${{4\pi } \over 9}$
B.
${{7\pi } \over {18}}$
C.
${\pi \over {18}}$
D.
${{5\pi } \over {18}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Morning Shift
If $A = \left( {\matrix{
{{1 \over {\sqrt 5 }}} & {{2 \over {\sqrt 5 }}} \cr
{{{ - 2} \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \cr
} } \right)$, $B = \left( {\matrix{
1 & 0 \cr
i & 1 \cr
} } \right)$, $i = \sqrt { - 1} $, and Q = ATBA, then the inverse of the matrix A Q2021 AT is equal to :
A.
$\left( {\matrix{
{{1 \over {\sqrt 5 }}} & { - 2021} \cr
{2021} & {{1 \over {\sqrt 5 }}} \cr
} } \right)$
B.
$\left( {\matrix{
1 & 0 \cr
{ - 2021i} & 1 \cr
} } \right)$
C.
$\left( {\matrix{
1 & 0 \cr
{2021i} & 1 \cr
} } \right)$
D.
$\left( {\matrix{
1 & { - 2021i} \cr
0 & 1 \cr
} } \right)$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Evening Shift
Let A and B be two 3 $\times$ 3 real matrices such that (A2 $-$ B2) is invertible matrix. If A5 = B5 and A3B2 = A2B3, then the value of the determinant of the matrix A3 + B3 is equal to :
A.
2
B.
4
C.
1
D.
0
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Morning Shift
Let $A = \left[ {\matrix{
1 & 2 \cr
{ - 1} & 4 \cr
} } \right]$. If A$-$1 = $\alpha$I + $\beta$A, $\alpha$, $\beta$ $\in$ R, I is a 2 $\times$ 2 identity matrix then 4($\alpha$ $-$ $\beta$) is equal to :
A.
5
B.
${8 \over 3}$
C.
2
D.
4
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Evening Shift
The number of distinct real roots
of $\left| {\matrix{ {\sin x} & {\cos x} & {\cos x} \cr {\cos x} & {\sin x} & {\cos x} \cr {\cos x} & {\cos x} & {\sin x} \cr } } \right| = 0$ in the interval $ - {\pi \over 4} \le x \le {\pi \over 4}$ is :
of $\left| {\matrix{ {\sin x} & {\cos x} & {\cos x} \cr {\cos x} & {\sin x} & {\cos x} \cr {\cos x} & {\cos x} & {\sin x} \cr } } \right| = 0$ in the interval $ - {\pi \over 4} \le x \le {\pi \over 4}$ is :
A.
4
B.
1
C.
2
D.
3
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Evening Shift
If $P = \left[ {\matrix{
1 & 0 \cr
{{1 \over 2}} & 1 \cr
} } \right]$, then P50 is :
A.
$\left[ {\matrix{
1 & 0 \cr
{25} & 1 \cr
} } \right]$
B.
$\left[ {\matrix{
1 & {50} \cr
0 & 1 \cr
} } \right]$
C.
$\left[ {\matrix{
1 & {25} \cr
0 & 1 \cr
} } \right]$
D.
$\left[ {\matrix{
1 & 0 \cr
{50} & 1 \cr
} } \right]$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Morning Shift
The values of a and b, for which the system of equations
2x + 3y + 6z = 8
x + 2y + az = 5
3x + 5y + 9z = b
has no solution, are :
2x + 3y + 6z = 8
x + 2y + az = 5
3x + 5y + 9z = b
has no solution, are :
A.
a = 3, b $\ne$ 13
B.
a $\ne$ 3, b $\ne$ 13
C.
a $\ne$ 3, b = 3
D.
a = 3, b = 13
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 22th July Evening Shift
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 :
A.
$\lambda$ = 3, $\mu$ = 5
B.
$\lambda$ = 3, $\mu$ $\ne$ 10
C.
$\lambda$ $\ne$ 2, $\mu$ = 10
D.
$\lambda$ = 2, $\mu$ $\ne$ 10
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 22th July Evening Shift
Let A = [aij] be a real matrix of order 3 $\times$ 3, such that ai1 + ai2 + ai3 = 1, for i = 1, 2, 3. Then, the sum of all the entries of the matrix A3 is equal to :
A.
2
B.
1
C.
3
D.
9
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Evening Shift
The value of k $\in$R, for which the following system of linear equations
3x $-$ y + 4z = 3,
x + 2y $-$ 3z = $-$2
6x + 5y + kz = $-$3,
has infinitely many solutions, is :
3x $-$ y + 4z = 3,
x + 2y $-$ 3z = $-$2
6x + 5y + kz = $-$3,
has infinitely many solutions, is :
A.
3
B.
$-$5
C.
5
D.
$-$3
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Morning Shift
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 :
A.
36
B.
24
C.
45
D.
18
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Evening Shift
Let the system of linear equations
4x + $\lambda$y + 2z = 0
2x $-$ y + z = 0
$\mu$x + 2y + 3z = 0, $\lambda$, $\mu$$\in$R.
has a non-trivial solution. Then which of the following is true?
4x + $\lambda$y + 2z = 0
2x $-$ y + z = 0
$\mu$x + 2y + 3z = 0, $\lambda$, $\mu$$\in$R.
has a non-trivial solution. Then which of the following is true?
A.
$\mu$ = 6, $\lambda$$\in$R
B.
$\lambda$ = 3, $\mu$$\in$R
C.
$\mu$ = $-$6, $\lambda$$\in$R
D.
$\lambda$ = 2, $\mu$$\in$R
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
The solutions of the equation $\left| {\matrix{
{1 + {{\sin }^2}x} & {{{\sin }^2}x} & {{{\sin }^2}x} \cr
{{{\cos }^2}x} & {1 + {{\cos }^2}x} & {{{\cos }^2}x} \cr
{4\sin 2x} & {4\sin 2x} & {1 + 4\sin 2x} \cr
} } \right| = 0,(0 < x < \pi )$, are
A.
${\pi \over {12}},{\pi \over 6}$
B.
${\pi \over 6},{{5\pi } \over 6}$
C.
${{5\pi } \over {12}},{{7\pi } \over {12}}$
D.
${{7\pi } \over {12}},{{11\pi } \over {12}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
Let $\alpha$, $\beta$, $\gamma$ be the real roots of the equation, x3 + ax2 + 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
A.
5
B.
3
C.
1
D.
0
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
Let $A + 2B = \left[ {\matrix{
1 & 2 & 0 \cr
6 & { - 3} & 3 \cr
{ - 5} & 3 & 1 \cr
} } \right]$ and $2A - B = \left[ {\matrix{
2 & { - 1} & 5 \cr
2 & { - 1} & 6 \cr
0 & 1 & 2 \cr
} } \right]$. If Tr(A) denotes the sum of all diagonal elements of the matrix A, then Tr(A) $-$ Tr(B) has value equal to
A.
1
B.
2
C.
0
D.
3
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Evening Shift
If x, y, z are in arithmetic progression with common difference d, x $\ne$ 3d, and the determinant of the matrix $\left[ {\matrix{
3 & {4\sqrt 2 } & x \cr
4 & {5\sqrt 2 } & y \cr
5 & k & z \cr
} } \right]$ is zero, then the value of k2 is :
A.
72
B.
12
C.
36
D.
6
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Morning Shift
The system of equations kx + y + z = 1, x + ky + z = k and x + y + zk = k2 has no solution if k is equal to :
A.
0
B.
$-$1
C.
$-$2
D.
1
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Morning Shift
If $A = \left( {\matrix{
0 & {\sin \alpha } \cr
{\sin \alpha } & 0 \cr
} } \right)$ and $\det \left( {{A^2} - {1 \over 2}I} \right) = 0$, then a possible value of $\alpha$ is :
A.
${\pi \over 4}$
B.
${\pi \over 6}$
C.
${\pi \over 2}$
D.
${\pi \over 3}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Morning Shift
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 :
A.
Exactly two solutions
B.
Infinitely many solutions
C.
A unique solution
D.
No solution
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Evening Shift
Consider the following system of equations :
x + 2y $-$ 3z = a
2x + 6y $-$ 11z = b
x $-$ 2y + 7z = c,
where a, b and c are real constants. Then the system of equations :
x + 2y $-$ 3z = a
2x + 6y $-$ 11z = b
x $-$ 2y + 7z = c,
where a, b and c are real constants. Then the system of equations :
A.
has no solution for all a, b and c
B.
has a unique solution when 5a = 2b + c
C.
has infinite number of solutions when 5a = 2b + c
D.
has a unique solution for all a, b and c
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Morning Shift
Let A be a symmetric matrix of order 2 with integer entries. If the sum of the diagonal elements of A2 is 1, then the possible number of such matrices is :
A.
6
B.
4
C.
1
D.
12
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Morning Shift
The value of $\left| {\matrix{
{(a + 1)(a + 2)} & {a + 2} & 1 \cr
{(a + 2)(a + 3)} & {a + 3} & 1 \cr
{(a + 3)(a + 4)} & {a + 4} & 1 \cr
} } \right|$ is :
A.
$-$2
B.
0
C.
(a + 2)(a + 3)(a + 4)
D.
(a + 1)(a + 2)(a + 3)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
Let A be a 3 $\times$ 3 matrix with det(A) = 4. Let Ri denote the ith row of A. If a matrix B is obtained by performing the operation R2 $ \to $ 2R2 + 5R3 on 2A, then det(B) is equal to :
A.
64
B.
16
C.
128
D.
80
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
If for the matrix, $A = \left[ {\matrix{
1 & { - \alpha } \cr
\alpha & \beta \cr
} } \right]$, $A{A^T} = {I_2}$, then the value of ${\alpha ^4} + {\beta ^4}$ is :
A.
3
B.
2
C.
1
D.
4
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
The following system of linear equations
2x + 3y + 2z = 9
3x + 2y + 2z = 9
x $-$ y + 4z = 8
2x + 3y + 2z = 9
3x + 2y + 2z = 9
x $-$ y + 4z = 8
A.
does not have any solution
B.
has a solution ($\alpha$, $\beta$, $\gamma$) satisfying $\alpha$ + $\beta$2 + $\gamma$3 = 12
C.
has a unique solution
D.
has infinitely many solutions
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Evening Shift
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 (A2B2 $-$ B2A2) X = O, where X is a 3 $\times$ 1 column matrix of unknown variables and O is a 3 $\times$ 1 null matrix, has :
A.
no solution
B.
exactly two solutions
C.
infinitely many solutions
D.
a unique solution
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Evening Shift
For the system of linear equations:
$x - 2y = 1,x - y + kz = - 2,ky + 4z = 6,k \in R$,
consider the following statements :
(A) The system has unique solution if $k \ne 2,k \ne - 2$.
(B) The system has unique solution if k = $-$2
(C) The system has unique solution if k = 2
(D) The system has no solution if k = 2
(E) The system has infinite number of solutions if k $ \ne $ $-$2.
Which of the following statements are correct?
$x - 2y = 1,x - y + kz = - 2,ky + 4z = 6,k \in R$,
consider the following statements :
(A) The system has unique solution if $k \ne 2,k \ne - 2$.
(B) The system has unique solution if k = $-$2
(C) The system has unique solution if k = 2
(D) The system has no solution if k = 2
(E) The system has infinite number of solutions if k $ \ne $ $-$2.
Which of the following statements are correct?
A.
(B) and (E) only
B.
(C) and (D) only
C.
(A) and (E) only
D.
(A) and (D) only
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Morning Shift
The system of linear equations
3x - 2y - kz = 10
2x - 4y - 2z = 6
x+2y - z = 5m
is inconsistent if :
3x - 2y - kz = 10
2x - 4y - 2z = 6
x+2y - z = 5m
is inconsistent if :
A.
k $ \ne $ 3, m $ \in $ R
B.
k = 3, m $ \ne $ ${4 \over 5}$
C.
k = 3, m $ = $ ${4 \over 5}$
D.
k $ \ne $ 3, m $ \ne $ ${4 \over 5}$
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 6th September Evening Slot
The sum of distinct values of $\lambda $ for which the
system of equations
$\left( {\lambda - 1} \right)x + \left( {3\lambda + 1} \right)y + 2\lambda z = 0$
$\left( {\lambda - 1} \right)x + \left( {4\lambda - 2} \right)y + \left( {\lambda + 3} \right)z = 0$
$2x + \left( {3\lambda + 1} \right)y + 3\left( {\lambda - 1} \right)z = 0$
has non-zero solutions, is ________ .
$\left( {\lambda - 1} \right)x + \left( {3\lambda + 1} \right)y + 2\lambda z = 0$
$\left( {\lambda - 1} \right)x + \left( {4\lambda - 2} \right)y + \left( {\lambda + 3} \right)z = 0$
$2x + \left( {3\lambda + 1} \right)y + 3\left( {\lambda - 1} \right)z = 0$
has non-zero solutions, is ________ .
Correct Answer: 3
Explanation:
$\left| {\matrix{
{\lambda - 1} & {3\lambda + 1} & {2\lambda } \cr
{\lambda - 1} & {4\lambda - 2} & {\lambda + 3} \cr
2 & {3\lambda + 1} & {3\left( {\lambda - 1} \right)} \cr
} } \right|$ = 0
R2 $ \to $ R2 – R1
R3 $ \to $ R3 – R1
$\left| {\matrix{ {\lambda - 1} & {3\lambda + 1} & {2\lambda } \cr 0 & {\lambda - 3} & { - \lambda + 3} \cr {3 - \lambda } & 0 & {\lambda - 3} \cr } } \right| = 0$
C1 $ \to $ C1 + C3
$\left| {\matrix{ {3\lambda - 1} & {3\lambda + 1} & {2\lambda } \cr { - \lambda + 3} & {\lambda - 3} & { - \lambda + 3} \cr 0 & 0 & {\lambda - 3} \cr } } \right| = 0$
$ \Rightarrow $ ($\lambda $ - 3) [(3$\lambda $ - 1) ($\lambda $ - 3) – (3 – $\lambda $) (3$\lambda $ + 1)] = 0
$ \Rightarrow $ ($\lambda $ – 3) [3$\lambda $2 – 10$\lambda $ + 3 –(8$\lambda $ –3$\lambda $2 + 3)] = 0
$ \Rightarrow $ ($\lambda $ – 3) (6$\lambda $2 – 18$\lambda $) = 0
$ \Rightarrow $ (6$\lambda $) ($\lambda $ – 3)2 = 0
$ \Rightarrow $ $\lambda $ = 0, 3
$ \therefore $ sum of values of $\lambda $ = 0 + 3 = 3
R2 $ \to $ R2 – R1
R3 $ \to $ R3 – R1
$\left| {\matrix{ {\lambda - 1} & {3\lambda + 1} & {2\lambda } \cr 0 & {\lambda - 3} & { - \lambda + 3} \cr {3 - \lambda } & 0 & {\lambda - 3} \cr } } \right| = 0$
C1 $ \to $ C1 + C3
$\left| {\matrix{ {3\lambda - 1} & {3\lambda + 1} & {2\lambda } \cr { - \lambda + 3} & {\lambda - 3} & { - \lambda + 3} \cr 0 & 0 & {\lambda - 3} \cr } } \right| = 0$
$ \Rightarrow $ ($\lambda $ - 3) [(3$\lambda $ - 1) ($\lambda $ - 3) – (3 – $\lambda $) (3$\lambda $ + 1)] = 0
$ \Rightarrow $ ($\lambda $ – 3) [3$\lambda $2 – 10$\lambda $ + 3 –(8$\lambda $ –3$\lambda $2 + 3)] = 0
$ \Rightarrow $ ($\lambda $ – 3) (6$\lambda $2 – 18$\lambda $) = 0
$ \Rightarrow $ (6$\lambda $) ($\lambda $ – 3)2 = 0
$ \Rightarrow $ $\lambda $ = 0, 3
$ \therefore $ sum of values of $\lambda $ = 0 + 3 = 3
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Morning Slot
If the system of equations
x - 2y + 3z = 9
2x + y + z = b
x - 7y + az = 24,
has infinitely many solutions, then a - b is equal to.........
x - 2y + 3z = 9
2x + y + z = b
x - 7y + az = 24,
has infinitely many solutions, then a - b is equal to.........
Correct Answer: 5
Explanation:
D = 0
$\left| {\matrix{ 1 & { - 2} & 3 \cr 2 & 1 & 1 \cr 1 & { - 7} & a \cr } } \right| = 0$
$1(a + 7) + 2(2a - 1) + 3( - 14 - 1) = 0$
$a + 7 + 4a - 2 - 45 = 0$
$5a = 40$
$a = 8$
${D_1} = \left| {\matrix{ 9 & { - 2} & 3 \cr b & 1 & 1 \cr {24} & { - 7} & 8 \cr } } \right| = 0$
$ \Rightarrow 9(8 + 7) + 2(8b - 24) + 3( - 7b - 24) = 0$
$ \Rightarrow 135 + 16b - 48 - 21b - 72 = 0$
$ \Rightarrow $ $15 = 5b$
$ \Rightarrow b = 3$
$a - b = 5$
$\left| {\matrix{ 1 & { - 2} & 3 \cr 2 & 1 & 1 \cr 1 & { - 7} & a \cr } } \right| = 0$
$1(a + 7) + 2(2a - 1) + 3( - 14 - 1) = 0$
$a + 7 + 4a - 2 - 45 = 0$
$5a = 40$
$a = 8$
${D_1} = \left| {\matrix{ 9 & { - 2} & 3 \cr b & 1 & 1 \cr {24} & { - 7} & 8 \cr } } \right| = 0$
$ \Rightarrow 9(8 + 7) + 2(8b - 24) + 3( - 7b - 24) = 0$
$ \Rightarrow 135 + 16b - 48 - 21b - 72 = 0$
$ \Rightarrow $ $15 = 5b$
$ \Rightarrow b = 3$
$a - b = 5$
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Evening Slot
Let S be the set of all integer solutions, (x, y, z),
of the system of equations
x – 2y + 5z = 0
–2x + 4y + z = 0
–7x + 14y + 9z = 0
such that 15 $ \le $ x2 + y2 + z2 $ \le $ 150. Then, the number of elements in the set S is equal to ______ .
x – 2y + 5z = 0
–2x + 4y + z = 0
–7x + 14y + 9z = 0
such that 15 $ \le $ x2 + y2 + z2 $ \le $ 150. Then, the number of elements in the set S is equal to ______ .
Correct Answer: 8
Explanation:
$x - 2y + 5z = 0$ ....(1)
$ - 2x + 4y + z = 0$ .....(2)
$ - 7x + 14y + 9z = 0$ ....(3)
2.(1) + (2) we get z = 0, x = 2y
15 $ \le $ 4y2 + y2 $ \le $ 150
$ \Rightarrow $ 3 $ \le $ y2 $ \le $ 30
$y \in \left[ { - \sqrt {30} , - \sqrt 3 } \right] \cup \left[ {\sqrt 3 ,\sqrt {30} } \right]$
$y = \pm 2,\, \pm 3,\, \pm 4,\, \pm 5$
$ \therefore $ no. of integer's in S is 8
$ - 2x + 4y + z = 0$ .....(2)
$ - 7x + 14y + 9z = 0$ ....(3)
2.(1) + (2) we get z = 0, x = 2y
15 $ \le $ 4y2 + y2 $ \le $ 150
$ \Rightarrow $ 3 $ \le $ y2 $ \le $ 30
$y \in \left[ { - \sqrt {30} , - \sqrt 3 } \right] \cup \left[ {\sqrt 3 ,\sqrt {30} } \right]$
$y = \pm 2,\, \pm 3,\, \pm 4,\, \pm 5$
$ \therefore $ no. of integer's in S is 8
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Morning Slot
Let A = $\left[ {\matrix{
x & 1 \cr
1 & 0 \cr
} } \right]$, x $ \in $ R and A4 = [aij].
If a11 = 109, then a22 is equal to _______ .
If a11 = 109, then a22 is equal to _______ .
Correct Answer: 10
Explanation:
${A^2} = \left[ {\matrix{
x & 1 \cr
1 & 0 \cr
} } \right]\left[ {\matrix{
x & 1 \cr
1 & 0 \cr
} } \right] = \left[ {\matrix{
{{x^2} + 1} & x \cr
x & 1 \cr
} } \right]$
${A^4} = \left[ {\matrix{ {{x^2} + 1} & x \cr x & 1 \cr } } \right]\left[ {\matrix{ {{x^2} + 1} & x \cr x & 1 \cr } } \right]$
$ = \left[ {\matrix{ {{{({x^2} + 1)}^2} + {x^2}} & {x({x^2} + 1) + x} \cr {x({x^2} + 1) + x} & {{x^2} + 1} \cr } } \right]$
Given ${({x^2} + 1)^2} + {x^2} = 109$
Let ${x^2} + 1$ = t
${t^2} + t - 1 = 109$
$ \Rightarrow $ (t $ - $ 10) (t + 11) = 0
$ \therefore $ t = 10 = x2 + 1 = a22
${A^4} = \left[ {\matrix{ {{x^2} + 1} & x \cr x & 1 \cr } } \right]\left[ {\matrix{ {{x^2} + 1} & x \cr x & 1 \cr } } \right]$
$ = \left[ {\matrix{ {{{({x^2} + 1)}^2} + {x^2}} & {x({x^2} + 1) + x} \cr {x({x^2} + 1) + x} & {{x^2} + 1} \cr } } \right]$
Given ${({x^2} + 1)^2} + {x^2} = 109$
Let ${x^2} + 1$ = t
${t^2} + t - 1 = 109$
$ \Rightarrow $ (t $ - $ 10) (t + 11) = 0
$ \therefore $ t = 10 = x2 + 1 = a22
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 8th January Morning Slot
The number of all 3 × 3 matrices A, with
enteries from the set {–1, 0, 1} such that the sum
of the diagonal elements of AAT is 3, is
Correct Answer: 672
Explanation:
Let A = $\left[ {\matrix{
{{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr
{{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr
{{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr
} } \right]$
$ \therefore $ AT = $\left[ {\matrix{ {{a_{11}}} & {{a_{21}}} & {{a_{31}}} \cr {{a_{12}}} & {{a_{22}}} & {{a_{32}}} \cr {{a_{13}}} & {{a_{23}}} & {{a_{33}}} \cr } } \right]$
diagonal elements of AAT are $a_{11}^2 + a_{12}^2 + a_{13}^2$ ,
$a_{21}^2 + a_{22}^2 + a_{23}^2$ , $a_{31}^2 + a_{32}^2 + a_{33}^2$
Given Sum = ($a_{11}^2 + a_{12}^2 + a_{13}^2$) +
($a_{21}^2 + a_{22}^2 + a_{23}^2$) + ($a_{31}^2 + a_{32}^2 + a_{33}^2$) = 3
This is only possible when three enteries must be either 1 or – 1 and all other six enteries are 0.
$ \therefore $ Number of matrices = 9C3 $ \times $ 2 $ \times $ 2 $ \times $ 2
= 672
$ \therefore $ AT = $\left[ {\matrix{ {{a_{11}}} & {{a_{21}}} & {{a_{31}}} \cr {{a_{12}}} & {{a_{22}}} & {{a_{32}}} \cr {{a_{13}}} & {{a_{23}}} & {{a_{33}}} \cr } } \right]$
diagonal elements of AAT are $a_{11}^2 + a_{12}^2 + a_{13}^2$ ,
$a_{21}^2 + a_{22}^2 + a_{23}^2$ , $a_{31}^2 + a_{32}^2 + a_{33}^2$
Given Sum = ($a_{11}^2 + a_{12}^2 + a_{13}^2$) +
($a_{21}^2 + a_{22}^2 + a_{23}^2$) + ($a_{31}^2 + a_{32}^2 + a_{33}^2$) = 3
This is only possible when three enteries must be either 1 or – 1 and all other six enteries are 0.
$ \therefore $ Number of matrices = 9C3 $ \times $ 2 $ \times $ 2 $ \times $ 2
= 672
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 7th January Evening Slot
If the system of linear equations,
x + y + z = 6
x + 2y + 3z = 10
3x + 2y + $\lambda $z = $\mu $
has more than two solutions, then $\mu $ - $\lambda $2 is equal to ______.
x + y + z = 6
x + 2y + 3z = 10
3x + 2y + $\lambda $z = $\mu $
has more than two solutions, then $\mu $ - $\lambda $2 is equal to ______.
Correct Answer: 13
Explanation:
Given system of equation more than
2 solutions.
Hence system of equation has infinite many
solution.
$ \therefore $ $\Delta $ = $\Delta $1 = $\Delta $2 = $\Delta $3 = 0
$\Delta $ = $\left| {\matrix{ 1 & 1 & 1 \cr 1 & 2 & 3 \cr 3 & 2 & \lambda \cr } } \right|$ = 0
$ \Rightarrow $ 1(2λ – 6) – 1(λ – 9) + 1(– 4) = 0
$ \Rightarrow $ 2λ – 6 – λ + 9 – 4 = 0
$ \Rightarrow $ λ = 1
$\Delta $1 = $\left| {\matrix{ 6 & 1 & 1 \cr {10} & 2 & 3 \cr \mu & 2 & \lambda \cr } } \right|$ = 0
6(2λ – 6) – 1(10λ – 3μ) + 1(20 – 2μ) = 0
$ \Rightarrow $ 12λ – 36 – 10λ + 3μ + 20 – 2μ = 0
$ \Rightarrow $ 2λ + μ = 16
$ \Rightarrow $ 2 + μ = 16
$ \Rightarrow $ $\mu $ = 14
$ \therefore $ $\mu $ - $\lambda $2 = 14 - 1 = 13
$ \therefore $ $\Delta $ = $\Delta $1 = $\Delta $2 = $\Delta $3 = 0
$\Delta $ = $\left| {\matrix{ 1 & 1 & 1 \cr 1 & 2 & 3 \cr 3 & 2 & \lambda \cr } } \right|$ = 0
$ \Rightarrow $ 1(2λ – 6) – 1(λ – 9) + 1(– 4) = 0
$ \Rightarrow $ 2λ – 6 – λ + 9 – 4 = 0
$ \Rightarrow $ λ = 1
$\Delta $1 = $\left| {\matrix{ 6 & 1 & 1 \cr {10} & 2 & 3 \cr \mu & 2 & \lambda \cr } } \right|$ = 0
6(2λ – 6) – 1(10λ – 3μ) + 1(20 – 2μ) = 0
$ \Rightarrow $ 12λ – 36 – 10λ + 3μ + 20 – 2μ = 0
$ \Rightarrow $ 2λ + μ = 16
$ \Rightarrow $ 2 + μ = 16
$ \Rightarrow $ $\mu $ = 14
$ \therefore $ $\mu $ - $\lambda $2 = 14 - 1 = 13