Matrices and Determinants
618 Questions
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 5th July Morning Shift
If $A=\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]$, then $A A^T$ is a
A.
symmetric matrix
B.
skew-symmetric matrix
C.
singular matrix
D.
inverse of $A$
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 5th July Morning Shift
If $A X=D$ represents the system of simultaneous linear equations $x+y+z=6, 5 x-y+2 z=3$ and $2 x+y-z=-5$, then (Adj $A$) $D=$
A.
$\left[\begin{array}{c}8 \\ -16 \\ 40\end{array}\right]$
B.
$\left[\begin{array}{c}32 \\ 64 \\ -160\end{array}\right]$
C.
$\left[\begin{array}{c}-16 \\ 32 \\ 80\end{array}\right]$
D.
$\left[\begin{array}{l}12 \\ 24 \\ 60\end{array}\right]$
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 5th July Morning Shift
If $A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right], B=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$, then $\operatorname{det}\left(A^6+B^6\right)=$
A.
$-68$
B.
$-106$
C.
$665$
D.
$720$
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 5th July Morning Shift
Let $G(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$. If $x+y=0$ then $G(x) G(y)=$
A.
null Matrix
B.
skew-symmetric Matrix
C.
identity Matrix
D.
symmetric Matrix
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 4th July Evening Shift
If $A=\left[\begin{array}{cc}2 & -3 \\ -4 & 1\end{array}\right]$, then $\left(A^T\right)^2+(12 A)^T=$
A.
$5\left[\begin{array}{cc}8 & 12 \\ -9 & 5\end{array}\right]$
B.
$5\left[\begin{array}{cc}8 & -9 \\ -12 & 5\end{array}\right]$
C.
$\left[\begin{array}{cc}40 & -45 \\ 60 & 25\end{array}\right]$
D.
$\left[\begin{array}{cc}40 & -60 \\ -45 & 25\end{array}\right]$
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 4th July Evening Shift
If $a, b, c$ are respectively the 5 th, 8 th, 13 th terms of an arithmetic progression, then $\left|\begin{array}{ccc}a & 5 & 1 \\ b & 8 & 1 \\ c & 13 & 1\end{array}\right|=$
A.
0
B.
1
C.
abc
D.
520
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 4th July Evening Shift
If $A=\left[\begin{array}{ccc}1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1\end{array}\right]$ is such that $A^2=I$, then
A.
$b=\frac{a c}{2}$
B.
$b=-\frac{a c}{2}$
C.
$b=\frac{a+c}{2}$
D.
$b=\sqrt{a c}$
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 4th July Evening Shift
Let $A=\left[\begin{array}{ccc}-2 & x & 1 \\ x & 1 & 1 \\ 2 & 3 & -1\end{array}\right]$. If the roots of the equation $\operatorname{det} A=0$ are $l, m$ then $l^3-m^3=$
A.
35
B.
$-$35
C.
19
D.
$-$19
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 4th July Morning Shift
For $i=1,2,3$ and $j=1,23$ If $a_i^2+b_i^2+c_i^2=1, a_i a_j+b_i b_j+c_i c_j=0, \forall i \neq j$ and $A=\left[\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right]$, then $\operatorname{det}\left(A A^T\right)=$
A.
0
B.
1
C.
$-$1
D.
3
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 4th July Morning Shift
If $A=\frac{1}{7}\left[\begin{array}{ccc}3 & -2 & 6 \\ -6 & -3 & 2 \\ -2 & 6 & 3\end{array}\right]$, then
A.
$A^{-1}=A$
B.
$A^{-1}=A^T$
C.
$A^{-1}$ does not exist
D.
$A^{-1}=-A$
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 4th July Morning Shift
If $A=\left[\begin{array}{cc}\alpha^2 & 5 \\ 5 & -\alpha\end{array}\right]$ and $\operatorname{det}\left(A^{10}\right)=1024$, then $\alpha=$
A.
$-$2
B.
$-$1
C.
$-$3
D.
0
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 4th July Morning Shift
Let $A=\left[\begin{array}{ccc}5 & \sin ^2 \theta & \cos ^2 \theta \\ -\sin ^2 \theta & -5 & 1 \\ \cos ^2 \theta & 1 & 5\end{array}\right]$. Then, maximum value of $\operatorname{det}(A)$ is
A.
$-125$
B.
200
C.
$-\frac{255}{2}$
D.
$145$
2022
AP-EAPCET
MCQ
AP EAPCET 2022 - 4th July Morning Shift
If $\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}=\frac{A x+B}{x^2+1}$ $+\frac{C x+D}{\left(x^2+1\right)^2}+\frac{E x+F}{\left(x^2+1\right)^3},$ then the value of $A+B+C+D+E+F=$
A.
21
B.
22
C.
28
D.
29
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Evening Shift
The number of elements in the set $\left\{ {A = \left( {\matrix{
a & b \cr
0 & d \cr
} } \right):a,b,d \in \{ - 1,0,1\} \,and\,{{(I - A)}^3} = I - {A^3}} \right\}$, where I is 2 $\times$ 2 identity matrix, is :
Correct Answer: 8
Explanation:
${(I - A)^3} = {I^3} - {A^3} - 3A(I - A) = I - {A^3}$
$ \Rightarrow 3A(I - A) = 0$ or ${A^2} = A$
$ \Rightarrow \left[ {\matrix{ {{a^2}} & {ab + bd} \cr 0 & {{d^2}} \cr } } \right] = \left[ {\matrix{ a & b \cr 0 & d \cr } } \right]$
$ \Rightarrow {a^2} = a,b(a + d - 1) = 0,{d^2} = d$
If b $\ne$ 0, a + d = 1 $\Rightarrow$ 4 ways
If b = 0, a = 0, 1 & d = 0, 1 $\Rightarrow$ 4 ways
$\Rightarrow$ Total 8 matrices
$ \Rightarrow 3A(I - A) = 0$ or ${A^2} = A$
$ \Rightarrow \left[ {\matrix{ {{a^2}} & {ab + bd} \cr 0 & {{d^2}} \cr } } \right] = \left[ {\matrix{ a & b \cr 0 & d \cr } } \right]$
$ \Rightarrow {a^2} = a,b(a + d - 1) = 0,{d^2} = d$
If b $\ne$ 0, a + d = 1 $\Rightarrow$ 4 ways
If b = 0, a = 0, 1 & d = 0, 1 $\Rightarrow$ 4 ways
$\Rightarrow$ Total 8 matrices
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Morning Shift
If the system of linear equations
2x + y $-$ z = 3
x $-$ y $-$ z = $\alpha$
3x + 3y + $\beta$z = 3
has infinitely many solution, then $\alpha$ + $\beta$ $-$ $\alpha$$\beta$ is equal to _____________.
2x + y $-$ z = 3
x $-$ y $-$ z = $\alpha$
3x + 3y + $\beta$z = 3
has infinitely many solution, then $\alpha$ + $\beta$ $-$ $\alpha$$\beta$ is equal to _____________.
Correct Answer: 5
Explanation:
2 $\times$ (i) $-$ (ii) $-$ (iii) gives :
$-$ (1 + $\beta$)z = 3 $-$ $\alpha$
For infinitely many solution
$\beta$ + 1 = 0 = 3 $-$ $\alpha$ $\Rightarrow$ ($\alpha$, $\beta$) = (3, $-$1)
Hence, $\alpha$ + $\beta$ $-$ $\alpha$$\beta$ = 5
$-$ (1 + $\beta$)z = 3 $-$ $\alpha$
For infinitely many solution
$\beta$ + 1 = 0 = 3 $-$ $\alpha$ $\Rightarrow$ ($\alpha$, $\beta$) = (3, $-$1)
Hence, $\alpha$ + $\beta$ $-$ $\alpha$$\beta$ = 5
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Evening Shift
Let A be a 3 $\times$ 3 real matrix. If det(2Adj(2 Adj(Adj(2A)))) = 241, then the value of det(A2) equal __________.
Correct Answer: 4
Explanation:
adj (2A) = 22 adjA
$\Rightarrow$ adj(adj (2A)) = adj(4 adjA) = 16 adj (adj A)
= 16 | A | A
$\Rightarrow$ adj (32 | A | A) = (32 | A |)2 adj A
12(32| A |)2 |adj A | = 23 (32 | A |)6 | adj A |
23 . 230 | A |6 . | A |2 = 241
| A |8 = 28 $\Rightarrow$ | A | = $\pm$2
| A |2 = | A |2 = 4
$\Rightarrow$ adj(adj (2A)) = adj(4 adjA) = 16 adj (adj A)
= 16 | A | A
$\Rightarrow$ adj (32 | A | A) = (32 | A |)2 adj A
12(32| A |)2 |adj A | = 23 (32 | A |)6 | adj A |
23 . 230 | A |6 . | A |2 = 241
| A |8 = 28 $\Rightarrow$ | A | = $\pm$2
| A |2 = | A |2 = 4
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Evening Shift
If $A = \left[ {\matrix{
1 & 1 & 1 \cr
0 & 1 & 1 \cr
0 & 0 & 1 \cr
} } \right]$ and M = A + A2 + A3 + ....... + A20, then the sum of all the elements of the matrix M is equal to _____________.
Correct Answer: 2020
Explanation:
${A^n} = \left[ {\matrix{
1 & n & {{{{n^2} + n} \over 2}} \cr
0 & 1 & n \cr
0 & 0 & 1 \cr
} } \right]$
So, required sum
$ = 20 \times 3 + 2 \times \left( {{{20 \times 21} \over 2}} \right) + \sum\limits_{r = 1}^{20} {\left( {{{{r^2} + r} \over 2}} \right)} $
$ = 60 + 420 + 105 + 35 \times 41 = 2020$
So, required sum
$ = 20 \times 3 + 2 \times \left( {{{20 \times 21} \over 2}} \right) + \sum\limits_{r = 1}^{20} {\left( {{{{r^2} + r} \over 2}} \right)} $
$ = 60 + 420 + 105 + 35 \times 41 = 2020$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Morning Shift
For real numbers $\alpha$ and $\beta$, consider the following system of linear equations :
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 ______________.
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 ______________.
Correct Answer: 5
Explanation:
For infinite solutions
$\Delta$ = $\Delta$1 = $\Delta$2 = $\Delta$3 = 0
$\Delta$ = $\left| {\matrix{ 1 & 1 & { - 1} \cr 1 & 2 & \alpha \cr 2 & { - 1} & 1 \cr } } \right| = 0$
$\Delta = \left| {\matrix{ 3 & 0 & 0 \cr 1 & 2 & \alpha \cr 2 & { - 1} & 1 \cr } } \right| = 0$
$\Delta$ = 3(2 + $\alpha$) = 0
$\Rightarrow$ $\alpha$ = $-$2
${\Delta _2} = \left| {\matrix{ 1 & 2 & { - 1} \cr 1 & 1 & { - 2} \cr 2 & \beta & 1 \cr } } \right| = 0$
1(1 + 2$\beta$) $-$2(1 + 4) $-$ ($\beta$ $-$ 2) = 0
$\beta$ $-$ 7 = 0
$\beta$ = 7
$\therefore$ $\alpha$ + $\beta$ = 5
$\Delta$ = $\Delta$1 = $\Delta$2 = $\Delta$3 = 0
$\Delta$ = $\left| {\matrix{ 1 & 1 & { - 1} \cr 1 & 2 & \alpha \cr 2 & { - 1} & 1 \cr } } \right| = 0$
$\Delta = \left| {\matrix{ 3 & 0 & 0 \cr 1 & 2 & \alpha \cr 2 & { - 1} & 1 \cr } } \right| = 0$
$\Delta$ = 3(2 + $\alpha$) = 0
$\Rightarrow$ $\alpha$ = $-$2
${\Delta _2} = \left| {\matrix{ 1 & 2 & { - 1} \cr 1 & 1 & { - 2} \cr 2 & \beta & 1 \cr } } \right| = 0$
1(1 + 2$\beta$) $-$2(1 + 4) $-$ ($\beta$ $-$ 2) = 0
$\beta$ $-$ 7 = 0
$\beta$ = 7
$\therefore$ $\alpha$ + $\beta$ = 5
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Morning Shift
Let $f(x) = \left| {\matrix{
{{{\sin }^2}x} & { - 2 + {{\cos }^2}x} & {\cos 2x} \cr
{2 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr
{{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \cos 2x} \cr
} } \right|,x \in [0,\pi ]$. Then the maximum value of f(x) is equal to ______________.
Correct Answer: 6
Explanation:
$\left| {\matrix{
{ - 2} & { - 2} & 0 \cr
2 & 0 & { - 1} \cr
{{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \cos 2x} \cr
} } \right|\left( \matrix{
{R_1} \to {R_1} - {R_2} \hfill \cr
\& \,{R_2} \to {R_2} - {R_3} \hfill \cr} \right)$
= $ - 2({\cos ^2}x) + 2(2 + 2\cos 2x + {\sin ^2}x)$
= $4 + 4\cos 2x - 2({\cos ^2}x - {\sin ^2}x)$
$ \therefore $ $f(x) = 4 + \underbrace {2\cos 2x}_{\max = 1}$
$ \Rightarrow $ $f{(x)_{\max }} = 4 + 2 = 6$
= $ - 2({\cos ^2}x) + 2(2 + 2\cos 2x + {\sin ^2}x)$
= $4 + 4\cos 2x - 2({\cos ^2}x - {\sin ^2}x)$
$ \therefore $ $f(x) = 4 + \underbrace {2\cos 2x}_{\max = 1}$
$ \Rightarrow $ $f{(x)_{\max }} = 4 + 2 = 6$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Morning Shift
Let $M = \left\{ {A = \left( {\matrix{
a & b \cr
c & d \cr
} } \right):a,b,c,d \in \{ \pm 3, \pm 2, \pm 1,0\} } \right\}$. Define f : M $\to$ Z, as f(A) = det(A), for all A$\in$M, where z is set of all integers. Then the number of A$\in$M such that f(A) = 15 is equal to _____________.
Correct Answer: 16
Explanation:
| A | = ad $-$ bc = 15
where ${a,b,c,d \in \{ \pm 3, \pm 2, \pm 1,0\} }$
Case I ad = 9 & bc = $-$6
For ad possible pairs are (3, 3), ($-$3, $-$3)
For bc possible pairs are (3, $-$2), ($-$3, 2), ($-$2, 3), (2, $-$3)
So total matrix = 2 $\times$ 4 = 8
Case II ad = 6 & bc = $-$9
Similarly total matrix = 2 $\times$ 4 = 8
$\Rightarrow$ Total such matrices are = 16
where ${a,b,c,d \in \{ \pm 3, \pm 2, \pm 1,0\} }$
Case I ad = 9 & bc = $-$6
For ad possible pairs are (3, 3), ($-$3, $-$3)
For bc possible pairs are (3, $-$2), ($-$3, 2), ($-$2, 3), (2, $-$3)
So total matrix = 2 $\times$ 4 = 8
Case II ad = 6 & bc = $-$9
Similarly total matrix = 2 $\times$ 4 = 8
$\Rightarrow$ Total such matrices are = 16
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 22th July Evening Shift
Let $A = \left[ {\matrix{
0 & 1 & 0 \cr
1 & 0 & 0 \cr
0 & 0 & 1 \cr
} } \right]$. Then the number of 3 $\times$ 3 matrices B with entries from the set {1, 2, 3, 4, 5} and satisfying AB = BA is ____________.
Correct Answer: 3125
Explanation:
Let matrix $B = \left[ {\matrix{
a & b & c \cr
d & e & f \cr
g & h & i \cr
} } \right]$
$\because$ $AB = BA$
$\left[ {\matrix{ 0 & 1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 \cr } } \right]\left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right] = \left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right]\left[ {\matrix{ 0 & 1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 \cr } } \right]$
$\left[ {\matrix{ d & e & f \cr a & b & c \cr g & h & i \cr } } \right] = \left[ {\matrix{ b & a & c \cr e & d & f \cr h & g & i \cr } } \right]$
$ \Rightarrow d = b,e = a,f = c,g = h$
$\therefore$ Matrix $B = \left[ {\matrix{ a & b & c \cr b & a & c \cr g & g & i \cr } } \right]$
No. of ways of selecting a, b, c, g, i
$ = 5 \times 5 \times 5 \times 5 \times 5$
$ = {5^5} = 3125$
$\therefore$ No. of matrices B = 3125
$\because$ $AB = BA$
$\left[ {\matrix{ 0 & 1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 \cr } } \right]\left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right] = \left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right]\left[ {\matrix{ 0 & 1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 \cr } } \right]$
$\left[ {\matrix{ d & e & f \cr a & b & c \cr g & h & i \cr } } \right] = \left[ {\matrix{ b & a & c \cr e & d & f \cr h & g & i \cr } } \right]$
$ \Rightarrow d = b,e = a,f = c,g = h$
$\therefore$ Matrix $B = \left[ {\matrix{ a & b & c \cr b & a & c \cr g & g & i \cr } } \right]$
No. of ways of selecting a, b, c, g, i
$ = 5 \times 5 \times 5 \times 5 \times 5$
$ = {5^5} = 3125$
$\therefore$ No. of matrices B = 3125
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Evening Shift
Let $A = \{ {a_{ij}}\} $ be a 3 $\times$ 3 matrix,
where ${a_{ij}} = \left\{ {\matrix{ {{{( - 1)}^{j - i}}} & {if} & {i < j,} \cr 2 & {if} & {i = j,} \cr {{{( - 1)}^{i + j}}} & {if} & {i > j} \cr } } \right.$
then $\det (3Adj(2{A^{ - 1}}))$ is equal to _____________.
where ${a_{ij}} = \left\{ {\matrix{ {{{( - 1)}^{j - i}}} & {if} & {i < j,} \cr 2 & {if} & {i = j,} \cr {{{( - 1)}^{i + j}}} & {if} & {i > j} \cr } } \right.$
then $\det (3Adj(2{A^{ - 1}}))$ is equal to _____________.
Correct Answer: 108
Explanation:
$A = \left[ {\matrix{
2 & { - 1} & 1 \cr
{ - 1} & 2 & { - 1} \cr
1 & { - 1} & 2 \cr
} } \right]$
$|A| = 4$
$\det (3adj(2{A^{ - 1}}))$
$ = {3^3}\left| {adj(2{a^{ - 1}})} \right|$
$ = {3^2}{\left| {2{A^{ - 1}}} \right|^2}$
$ = {3^3}{.2^2}|{A^{ - 1}}{|^2} = {3^3}{.2^2}.{1 \over {|A{|^2}}} = {3^2}{.2^2}.{1 \over {{4^2}}} = 108$
$|A| = 4$
$\det (3adj(2{A^{ - 1}}))$
$ = {3^3}\left| {adj(2{a^{ - 1}})} \right|$
$ = {3^2}{\left| {2{A^{ - 1}}} \right|^2}$
$ = {3^3}{.2^2}|{A^{ - 1}}{|^2} = {3^3}{.2^2}.{1 \over {|A{|^2}}} = {3^2}{.2^2}.{1 \over {{4^2}}} = 108$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Morning Shift
Let $A = \left( {\matrix{
1 & { - 1} & 0 \cr
0 & 1 & { - 1} \cr
0 & 0 & 1 \cr
} } \right)$ and B = 7A20 $-$ 20A7 + 2I, where I is an identity matrix of order 3 $\times$ 3. If B = [bij], then b13is equal to _____________.
Correct Answer: 910
Explanation:
Let $A = \left( {\matrix{
1 & { - 1} & 0 \cr
0 & 1 & { - 1} \cr
0 & 0 & 1 \cr
} } \right) = I + C$
where, $I = \left( {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right),C = \left( {\matrix{ 0 & { - 1} & 0 \cr 0 & 0 & { - 1} \cr 0 & 0 & 0 \cr } } \right)$
${C^2} = \left( {\matrix{ 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right),$
${C^3} = \left( {\matrix{ 0 & 0 & 0 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right) = {C^4} = {C^5} = ........$
$B = 7{A^{20}} - 20{A^7} + 2I$
$ = 7{(I + C)^{20}} + 20{(I + C)^7} + 2I$
$ = 7(I + 20C + {}^{20}{C_2}{C^2}) - 20(I + 7C + {}^7{C_2}{C^2}) + 2I$
So
${b_{13}} = 7 \times {}^{20}{C_2}{C^2} - 20 \times {}^7{C_2} = 910$
where, $I = \left( {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right),C = \left( {\matrix{ 0 & { - 1} & 0 \cr 0 & 0 & { - 1} \cr 0 & 0 & 0 \cr } } \right)$
${C^2} = \left( {\matrix{ 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right),$
${C^3} = \left( {\matrix{ 0 & 0 & 0 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right) = {C^4} = {C^5} = ........$
$B = 7{A^{20}} - 20{A^7} + 2I$
$ = 7{(I + C)^{20}} + 20{(I + C)^7} + 2I$
$ = 7(I + 20C + {}^{20}{C_2}{C^2}) - 20(I + 7C + {}^7{C_2}{C^2}) + 2I$
So
${b_{13}} = 7 \times {}^{20}{C_2}{C^2} - 20 \times {}^7{C_2} = 910$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Morning Shift
Let a, b, c, d in arithmetic progression with common difference $\lambda$. If $\left| {\matrix{
{x + a - c} & {x + b} & {x + a} \cr
{x - 1} & {x + c} & {x + b} \cr
{x - b + d} & {x + d} & {x + c} \cr
} } \right| = 2$, then value of $\lambda$2 is equal to ________________.
Correct Answer: 1
Explanation:
$\left| {\matrix{
{x + a - c} & {x + b} & {x + a} \cr
{x - 1} & {x + c} & {x + b} \cr
{x - b + d} & {x + d} & {x + c} \cr
} } \right| = 2$
${C_2} \to {C_2} - {C_3}$
$ \Rightarrow \left| {\matrix{ {x - 2\lambda } & \lambda & {x + a} \cr {x - 1} & \lambda & {x + b} \cr {x + 2\lambda } & \lambda & {x + c} \cr } } \right| = 2$
${R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}$
$ \Rightarrow \left| {\matrix{ {x - 2\lambda } & 1 & {x + a} \cr {2\lambda - 1} & 0 & \lambda \cr {4\lambda } & 0 & {2\lambda } \cr } } \right| = 2$
$ \Rightarrow 1(4{\lambda ^2} - 4{\lambda ^2} + 2\lambda ) = 2$
$ \Rightarrow {\lambda ^2} = 1$
${C_2} \to {C_2} - {C_3}$
$ \Rightarrow \left| {\matrix{ {x - 2\lambda } & \lambda & {x + a} \cr {x - 1} & \lambda & {x + b} \cr {x + 2\lambda } & \lambda & {x + c} \cr } } \right| = 2$
${R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}$
$ \Rightarrow \left| {\matrix{ {x - 2\lambda } & 1 & {x + a} \cr {2\lambda - 1} & 0 & \lambda \cr {4\lambda } & 0 & {2\lambda } \cr } } \right| = 2$
$ \Rightarrow 1(4{\lambda ^2} - 4{\lambda ^2} + 2\lambda ) = 2$
$ \Rightarrow {\lambda ^2} = 1$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Evening Shift
Let I be an identity matrix of order 2 $\times$ 2 and P = $\left[ {\matrix{
2 & { - 1} \cr
5 & { - 3} \cr
} } \right]$. Then the value of n$\in$N for which Pn = 5I $-$ 8P is equal to ____________.
Correct Answer: 6
Explanation:
$P = \left[ {\matrix{
2 & { - 1} \cr
5 & { - 3} \cr
} } \right]$
$\left| {\matrix{ {2 - \lambda } & { - 1} \cr 5 & { - 3 - \lambda } \cr } } \right| = 0$
$ \Rightarrow $ $\lambda$2 + $\lambda$ $-$ 1 = 0
$ \Rightarrow $ P2 + P $-$ I = 0
$ \Rightarrow $ P2 = I $-$ P
$ \Rightarrow $ P4 = I + P2 $-$ 2P
$ \Rightarrow $ P4 = 2I $-$ 3P
Now, P4 . P2 = (2I $-$ 3P)(I $-$ P) = 2I $-$ 5P + 3P2
$ \Rightarrow $ P6 = 5I $-$ 8P
So n = 6
$\left| {\matrix{ {2 - \lambda } & { - 1} \cr 5 & { - 3 - \lambda } \cr } } \right| = 0$
$ \Rightarrow $ $\lambda$2 + $\lambda$ $-$ 1 = 0
$ \Rightarrow $ P2 + P $-$ I = 0
$ \Rightarrow $ P2 = I $-$ P
$ \Rightarrow $ P4 = I + P2 $-$ 2P
$ \Rightarrow $ P4 = 2I $-$ 3P
Now, P4 . P2 = (2I $-$ 3P)(I $-$ P) = 2I $-$ 5P + 3P2
$ \Rightarrow $ P6 = 5I $-$ 8P
So n = 6
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Evening Shift
Let $A = \left[ {\matrix{
a & b \cr
c & d \cr
} } \right]$ and $B = \left[ {\matrix{
\alpha \cr
\beta \cr
} } \right] \ne \left[ {\matrix{
0 \cr
0 \cr
} } \right]$ such that AB = B and a + d = 2021, then the value of ad $-$ bc is equal to ___________.
Correct Answer: 2020
Explanation:
$A = \left[ {\matrix{
a & b \cr
c & d \cr
} } \right],\,B = \left[ {\matrix{
\alpha \cr
\beta \cr
} } \right]$
$AB = B$
$\left[ {\matrix{ a & b \cr c & d \cr } } \right]\left[ {\matrix{ \alpha \cr \beta \cr } } \right] = \left[ {\matrix{ \alpha \cr \beta \cr } } \right]$
$\left[ {\matrix{ {a\alpha + b\beta } \cr {c\alpha + d\beta } \cr } } \right] = \left[ {\matrix{ \alpha \cr \beta \cr } } \right]$$ \Rightarrow $ $\eqalign{ & a\alpha + b\beta = \alpha \,......(1) \cr & c\alpha + d\beta = \beta \,......(2) \cr} $
$\alpha (a - 1) = - b\beta $ and $c\alpha = \beta (1 - d)$
${\alpha \over \beta } = {{ - b} \over {a - 1}}$ & ${\alpha \over \beta } = {{1 - d} \over c}$
$ \therefore $ ${{ - b} \over {a - 1}} = {{1 - d} \over c}$
$ - bc = (a - 1)(1 - d)$
$ - bc = a - ad - 1 + d$
$ad - bc = a + d - 1$
$ = 2021 - 1 = 2020$
$AB = B$
$\left[ {\matrix{ a & b \cr c & d \cr } } \right]\left[ {\matrix{ \alpha \cr \beta \cr } } \right] = \left[ {\matrix{ \alpha \cr \beta \cr } } \right]$
$\left[ {\matrix{ {a\alpha + b\beta } \cr {c\alpha + d\beta } \cr } } \right] = \left[ {\matrix{ \alpha \cr \beta \cr } } \right]$$ \Rightarrow $ $\eqalign{ & a\alpha + b\beta = \alpha \,......(1) \cr & c\alpha + d\beta = \beta \,......(2) \cr} $
$\alpha (a - 1) = - b\beta $ and $c\alpha = \beta (1 - d)$
${\alpha \over \beta } = {{ - b} \over {a - 1}}$ & ${\alpha \over \beta } = {{1 - d} \over c}$
$ \therefore $ ${{ - b} \over {a - 1}} = {{1 - d} \over c}$
$ - bc = (a - 1)(1 - d)$
$ - bc = a - ad - 1 + d$
$ad - bc = a + d - 1$
$ = 2021 - 1 = 2020$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Evening Shift
If 1, log10(4x $-$ 2) and log10$\left( {{4^x} + {{18} \over 5}} \right)$ are in arithmetic progression for a real number x, then the value of the determinant $\left| {\matrix{
{2\left( {x - {1 \over 2}} \right)} & {x - 1} & {{x^2}} \cr
1 & 0 & x \cr
x & 1 & 0 \cr
} } \right|$ is equal to :
Correct Answer: 2
Explanation:
1, $lo{g_{10}}({4^x} - 2),\,lo{g_{10}}\left( {{4^x} + {{18} \over 5}} \right)$ in AP.
$ \therefore $ 2$ \times $$lo{g_{10}}({4^x} - 2) = 1 + \,lo{g_{10}}\left( {{4^x} + {{18} \over 5}} \right)$
$lo{g_{10}}{({4^x} - 2)^2} = \,lo{g_{10}}\left( {10.\left( {{4^x} + {{18} \over 5}} \right)} \right)$
${({4^x} - 2)^2} = 10.\left( {{4^x} + {{18} \over 5}} \right)$
${({4^x})^2} + 4 - {4.4^x} = {10.4^x} + 36$
${({4^x})^2} - {14.4^x} - 32 = 0$
${({4^x})^2} + {2.4^x} - {16.4^x} - 32 = 0$
${4^x}({4^x} + 2) - 16.({4^x} + 2) = 0$
$({4^x} + 2)({4^x} - 16) = 0$
4x = -2 (Not Possible)
Or 4x = 16
$ \Rightarrow $ x = 2
Therefore $\left| {\matrix{ {2(x - 1/2)} & {x - 1} & {{x^2}} \cr 1 & 0 & x \cr x & 1 & 0 \cr } } \right|$
$ = \left| {\matrix{ 3 & 1 & 4 \cr 1 & 0 & 2 \cr 2 & 1 & 0 \cr } } \right|$
$ = 3( - 2) - 1(0 - 4) + 4(1 - 0)$
$ = - 6 + 4 + 4 = 2$
$ \therefore $ 2$ \times $$lo{g_{10}}({4^x} - 2) = 1 + \,lo{g_{10}}\left( {{4^x} + {{18} \over 5}} \right)$
$lo{g_{10}}{({4^x} - 2)^2} = \,lo{g_{10}}\left( {10.\left( {{4^x} + {{18} \over 5}} \right)} \right)$
${({4^x} - 2)^2} = 10.\left( {{4^x} + {{18} \over 5}} \right)$
${({4^x})^2} + 4 - {4.4^x} = {10.4^x} + 36$
${({4^x})^2} - {14.4^x} - 32 = 0$
${({4^x})^2} + {2.4^x} - {16.4^x} - 32 = 0$
${4^x}({4^x} + 2) - 16.({4^x} + 2) = 0$
$({4^x} + 2)({4^x} - 16) = 0$
4x = -2 (Not Possible)
Or 4x = 16
$ \Rightarrow $ x = 2
Therefore $\left| {\matrix{ {2(x - 1/2)} & {x - 1} & {{x^2}} \cr 1 & 0 & x \cr x & 1 & 0 \cr } } \right|$
$ = \left| {\matrix{ 3 & 1 & 4 \cr 1 & 0 & 2 \cr 2 & 1 & 0 \cr } } \right|$
$ = 3( - 2) - 1(0 - 4) + 4(1 - 0)$
$ = - 6 + 4 + 4 = 2$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Morning Shift
If $A = \left[ {\matrix{
2 & 3 \cr
0 & { - 1} \cr
} } \right]$, then the value of det(A4) + det(A10 $-$ (Adj(2A))10) is equal to _____________.
Correct Answer: 16
Explanation:
$A = \left[ {\matrix{
2 & 3 \cr
0 & { - 1} \cr
} } \right]$
$|A|\, = - 2 \Rightarrow |A{|^4} = 16$
${A^2} = \left[ {\matrix{ 4 & 3 \cr 0 & 1 \cr } } \right]$
${A^3} = \left[ {\matrix{ 8 & 9 \cr 0 & { - 1} \cr } } \right]$
$ \therefore $ ${A^{10}} = \left[ {\matrix{ {{2^{10}}} & {{2^{10}} - 1} \cr 0 & 1 \cr } } \right] = \left[ {\matrix{ {1024} & {1023} \cr 0 & 1 \cr } } \right]$
$2A = \left[ {\matrix{ 4 & 6 \cr 0 & { - 2} \cr } } \right]$
$adj(2A) = \left[ {\matrix{ { - 2} & { - 6} \cr 0 & 4 \cr } } \right]$
$adj(2A) = - 2\left[ {\matrix{ 1 & 3 \cr 0 & { - 2} \cr } } \right]$
${(adj(2A))^{10}} = {2^{10}}{\left[ {\matrix{ 1 & 3 \cr 0 & { - 2} \cr } } \right]^{10}}$
$ = {2^{10}}\left[ {\matrix{ 1 & { - ({2^{10}} - 1)} \cr 0 & {{2^{10}}} \cr } } \right]$
$ = {2^{10}}\left[ {\matrix{ 1 & { - 1023} \cr 0 & {1024} \cr } } \right]$
${A^{10}} - {(adj(2A))^{10}} = \left[ {\matrix{ 0 & {{2^{11}} \times 1023} \cr 0 & {1 - {{(1024)}^2}} \cr } } \right]$
$|{A^{10}} - adj{(2A)^{10}}| = 0$
$ \therefore $ det(A4) + det(A10 $-$ (Adj(2A))10)
= 16 + 0 = 16
$|A|\, = - 2 \Rightarrow |A{|^4} = 16$
${A^2} = \left[ {\matrix{ 4 & 3 \cr 0 & 1 \cr } } \right]$
${A^3} = \left[ {\matrix{ 8 & 9 \cr 0 & { - 1} \cr } } \right]$
$ \therefore $ ${A^{10}} = \left[ {\matrix{ {{2^{10}}} & {{2^{10}} - 1} \cr 0 & 1 \cr } } \right] = \left[ {\matrix{ {1024} & {1023} \cr 0 & 1 \cr } } \right]$
$2A = \left[ {\matrix{ 4 & 6 \cr 0 & { - 2} \cr } } \right]$
$adj(2A) = \left[ {\matrix{ { - 2} & { - 6} \cr 0 & 4 \cr } } \right]$
$adj(2A) = - 2\left[ {\matrix{ 1 & 3 \cr 0 & { - 2} \cr } } \right]$
${(adj(2A))^{10}} = {2^{10}}{\left[ {\matrix{ 1 & 3 \cr 0 & { - 2} \cr } } \right]^{10}}$
$ = {2^{10}}\left[ {\matrix{ 1 & { - ({2^{10}} - 1)} \cr 0 & {{2^{10}}} \cr } } \right]$
$ = {2^{10}}\left[ {\matrix{ 1 & { - 1023} \cr 0 & {1024} \cr } } \right]$
${A^{10}} - {(adj(2A))^{10}} = \left[ {\matrix{ 0 & {{2^{11}} \times 1023} \cr 0 & {1 - {{(1024)}^2}} \cr } } \right]$
$|{A^{10}} - adj{(2A)^{10}}| = 0$
$ \therefore $ det(A4) + det(A10 $-$ (Adj(2A))10)
= 16 + 0 = 16
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
Let $A = \left[ {\matrix{
{{a_1}} \cr
{{a_2}} \cr
} } \right]$ and $B = \left[ {\matrix{
{{b_1}} \cr
{{b_2}} \cr
} } \right]$ be two 2 $\times$ 1 matrices with real entries such that A = XB, where
$X = {1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & { - 1} \cr 1 & k \cr } } \right]$, and k$\in$R.
If $a_1^2$ + $a_2^2$ = ${2 \over 3}$(b$_1^2$ + b$_2^2$) and (k2 + 1) b$_2^2$ $\ne$ $-$2b1b2, then the value of k is __________.
$X = {1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & { - 1} \cr 1 & k \cr } } \right]$, and k$\in$R.
If $a_1^2$ + $a_2^2$ = ${2 \over 3}$(b$_1^2$ + b$_2^2$) and (k2 + 1) b$_2^2$ $\ne$ $-$2b1b2, then the value of k is __________.
Correct Answer: 1
Explanation:
$XB = A$
$ \Rightarrow $ ${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & { - 1} \cr 1 & k \cr } } \right]\left[ {\matrix{ {{b_1}} \cr {{b_2}} \cr } } \right] = \left[ {\matrix{ {{a_1}} \cr {{a_2}} \cr } } \right]$
$ \Rightarrow $ ${1 \over {\sqrt 3 }}\left[ {\matrix{ {{b_1} - {b_2}} \cr {{b_1} + k{b_2}} \cr } } \right] = \left[ {\matrix{ {{a_1}} \cr {{a_2}} \cr } } \right]$
${b_1} - {b_2} = \sqrt 3 {a_1} \Rightarrow 3a_1^2 = b_1^2 + b_2^2 - 2{b_1}{b_2}$
${b_1} + k{b_2} = \sqrt 3 {a_2} \Rightarrow 3a_2^2 = b_1^2 + {k^2}b_2^2 + 2k{b_1}{b_2}$
$3\left( {a_1^2 + a_2^2} \right) = 2b_1^2 + \left( {{k^2} + 1} \right)b_2^2 + 2{b_1}{b_2}(k - 1)$
$ \Rightarrow $ ${a_1^2 + a_2^2} $ = ${2 \over 3}b_1^2 + {{\left( {{k^2} + 1} \right)} \over 3}b_2^2 + {2 \over 3}{b_1}{b_2}\left( {k - 1} \right)$
Given $a_1^2$ + $a_2^2$ = ${2 \over 3}$(b$_1^2$ + b$_2^2$)
$ \therefore $ ${2 \over 3}$(b$_1^2$ + b$_2^2$) = ${2 \over 3}b_1^2 + {{\left( {{k^2} + 1} \right)} \over 3}b_2^2 + {2 \over 3}{b_1}{b_2}\left( {k - 1} \right)$
$ \Rightarrow $ ${2 \over 3}b_2^2 = {{\left( {{k^2} + 1} \right)} \over 3}b_2^2 + {2 \over 3}{b_1}{b_2}\left( {k - 1} \right)$
Comparing both sides, We get
${{\left( {{k^2} + 1} \right)} \over 3} = {2 \over 3}$
$ \Rightarrow $ k2 = 1
$ \Rightarrow $ k = $ \pm $ 1 ......(1)
and ${2 \over 3}\left( {k - 1} \right) = 0$ $ \Rightarrow $ k = 1 ....(2)
From (1) and (2),
k = 1
$ \Rightarrow $ ${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & { - 1} \cr 1 & k \cr } } \right]\left[ {\matrix{ {{b_1}} \cr {{b_2}} \cr } } \right] = \left[ {\matrix{ {{a_1}} \cr {{a_2}} \cr } } \right]$
$ \Rightarrow $ ${1 \over {\sqrt 3 }}\left[ {\matrix{ {{b_1} - {b_2}} \cr {{b_1} + k{b_2}} \cr } } \right] = \left[ {\matrix{ {{a_1}} \cr {{a_2}} \cr } } \right]$
${b_1} - {b_2} = \sqrt 3 {a_1} \Rightarrow 3a_1^2 = b_1^2 + b_2^2 - 2{b_1}{b_2}$
${b_1} + k{b_2} = \sqrt 3 {a_2} \Rightarrow 3a_2^2 = b_1^2 + {k^2}b_2^2 + 2k{b_1}{b_2}$
$3\left( {a_1^2 + a_2^2} \right) = 2b_1^2 + \left( {{k^2} + 1} \right)b_2^2 + 2{b_1}{b_2}(k - 1)$
$ \Rightarrow $ ${a_1^2 + a_2^2} $ = ${2 \over 3}b_1^2 + {{\left( {{k^2} + 1} \right)} \over 3}b_2^2 + {2 \over 3}{b_1}{b_2}\left( {k - 1} \right)$
Given $a_1^2$ + $a_2^2$ = ${2 \over 3}$(b$_1^2$ + b$_2^2$)
$ \therefore $ ${2 \over 3}$(b$_1^2$ + b$_2^2$) = ${2 \over 3}b_1^2 + {{\left( {{k^2} + 1} \right)} \over 3}b_2^2 + {2 \over 3}{b_1}{b_2}\left( {k - 1} \right)$
$ \Rightarrow $ ${2 \over 3}b_2^2 = {{\left( {{k^2} + 1} \right)} \over 3}b_2^2 + {2 \over 3}{b_1}{b_2}\left( {k - 1} \right)$
Comparing both sides, We get
${{\left( {{k^2} + 1} \right)} \over 3} = {2 \over 3}$
$ \Rightarrow $ k2 = 1
$ \Rightarrow $ k = $ \pm $ 1 ......(1)
and ${2 \over 3}\left( {k - 1} \right) = 0$ $ \Rightarrow $ k = 1 ....(2)
From (1) and (2),
k = 1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
Let $P = \left[ {\matrix{
{ - 30} & {20} & {56} \cr
{90} & {140} & {112} \cr
{120} & {60} & {14} \cr
} } \right]$ and
$A = \left[ {\matrix{ 2 & 7 & {{\omega ^2}} \cr { - 1} & { - \omega } & 1 \cr 0 & { - \omega } & { - \omega + 1} \cr } } \right]$ where
$\omega = {{ - 1 + i\sqrt 3 } \over 2}$, and I3 be the identity matrix of order 3. If the
determinant of the matrix (P$-$1AP$-$I3)2 is $\alpha$$\omega$2, then the value of $\alpha$ is equal to ______________.
$A = \left[ {\matrix{ 2 & 7 & {{\omega ^2}} \cr { - 1} & { - \omega } & 1 \cr 0 & { - \omega } & { - \omega + 1} \cr } } \right]$ where
$\omega = {{ - 1 + i\sqrt 3 } \over 2}$, and I3 be the identity matrix of order 3. If the
determinant of the matrix (P$-$1AP$-$I3)2 is $\alpha$$\omega$2, then the value of $\alpha$ is equal to ______________.
Correct Answer: 36
Explanation:
$|{P^{ - 1}}AP - I{|^2}$
$ = |({P^{ - 1}}AP - I){({P^{ - 1}}AP - 1)^2}|$
$ = |{P^{ - 1}}AP{P^{ - 1}}AP - 2{P^{ - 1}}AP + I|$
$ = |{P^{ - 1}}{A^2}P - 2{P^{ - 1}}AP + {P^{ - 1}}IP|$
$ = |{P^{ - 1}}({A^2} - 2A + I)P|$
$ = |{P^{ - 1}}{(A - I)^2}P|$
$ = |{P^{ - 1}}||A - I{|^2}|P|$
$ = |A - I{|^2}$
$ = \left| {\matrix{ 1 & 7 & {{\omega ^2}} \cr { - 1} & { - \omega - 1} & 1 \cr 0 & { - \omega } & { - \omega } \cr } } \right|$
$ = {(1(\omega (\omega + 1) + \omega ) - 7\omega + {\omega ^2}.\omega )^2}$
$ = {({\omega ^2} + 2\omega - 7\omega + 1)^2}$
$ = {({\omega ^2} - 5\omega + 1)^2}$
$ = {( - 6\omega )^2}$
$ = 36{\omega ^2} $
$ \therefore $ $\alpha$$\omega$2 = $36{\omega ^2} $
$\Rightarrow \alpha = 36$
$ = |({P^{ - 1}}AP - I){({P^{ - 1}}AP - 1)^2}|$
$ = |{P^{ - 1}}AP{P^{ - 1}}AP - 2{P^{ - 1}}AP + I|$
$ = |{P^{ - 1}}{A^2}P - 2{P^{ - 1}}AP + {P^{ - 1}}IP|$
$ = |{P^{ - 1}}({A^2} - 2A + I)P|$
$ = |{P^{ - 1}}{(A - I)^2}P|$
$ = |{P^{ - 1}}||A - I{|^2}|P|$
$ = |A - I{|^2}$
$ = \left| {\matrix{ 1 & 7 & {{\omega ^2}} \cr { - 1} & { - \omega - 1} & 1 \cr 0 & { - \omega } & { - \omega } \cr } } \right|$
$ = {(1(\omega (\omega + 1) + \omega ) - 7\omega + {\omega ^2}.\omega )^2}$
$ = {({\omega ^2} + 2\omega - 7\omega + 1)^2}$
$ = {({\omega ^2} - 5\omega + 1)^2}$
$ = {( - 6\omega )^2}$
$ = 36{\omega ^2} $
$ \therefore $ $\alpha$$\omega$2 = $36{\omega ^2} $
$\Rightarrow \alpha = 36$
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