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
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
2026
JEE Mains
MCQ
JEE Main 2026 (Online) 28th January Morning Shift
Let $A, B$ and $C$ be three $2 \times 2$ matrices with real entries such that $B=(I+A)^{-1}$ and $\mathrm{A}+\mathrm{C}=\mathrm{I}$.
If $\mathrm{BC}=\left[\begin{array}{cc}1 & -5 \\ -1 & 2\end{array}\right]$ and $\mathrm{CB}\left[\begin{array}{l}x_1 \\ x_2\end{array}\right]=\left[\begin{array}{l}12 \\ -6\end{array}\right]$, then $x_1+x_2$ is
A.
4
B.
2
C.
0
D.
-2
2026
JEE Mains
MCQ
JEE Main 2026 (Online) 24th January Evening Shift
Let $P=\left[p_{i j}\right]$ and $Q=\left[q_{i j}\right]$ be two square matrices of order 3 such that $q_{\mathrm{ij}}=2^{(\mathrm{i}+\mathrm{j}-1)} \mathrm{p}_{\mathrm{ij}}$ and $\operatorname{det}(\mathrm{Q})=2^{10}$. Then the value of $\operatorname{det}(\operatorname{adj}(\operatorname{adj} \mathrm{P}))$ is:
A.
81
B.
16
C.
124
D.
32
2026
JEE Mains
MCQ
JEE Main 2026 (Online) 24th January Evening Shift
Let $f(x)=\int \frac{7 x^{10}+9 x^8}{\left(1+x^2+2 x^9\right)^2} d x, x>0, \lim\limits_{x \rightarrow 0} f(x)=0$ and $f(1)=\frac{1}{4}$.
If $\mathrm{A}=\left[\begin{array}{ccc}0 & 0 & 1 \\ \frac{1}{4} & f^{\prime}(1) & 1 \\ \alpha^2 & 4 & 1\end{array}\right]$ and $\mathrm{B}=\operatorname{adj}(\operatorname{adj} \mathrm{A})$ be such that $|\mathrm{B}|=81$, then $\alpha^2$ is equal to
A.
2
B.
4
C.
3
D.
1
2026
JEE Mains
MCQ
JEE Main 2026 (Online) 23rd January Evening Shift
The system of linear equations
$ \begin{aligned} & x+y+z=6 \\ & 2 x+5 y+a z=36 \\ & x+2 y+3 z=b \end{aligned} $
has :
A.
unique solution for $a=8$ and $b=16$
B.
infinitely many solutions for $a=8$ and $b=14$
C.
infinitely many solutions for $a=8$ and $b=16$
D.
unique solution for $a=8$ and $b=14$
2026
JEE Mains
MCQ
JEE Main 2026 (Online) 23rd January Morning Shift
Among the statements :
I: If $\left|\begin{array}{ccc}1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1\end{array}\right|=\left|\begin{array}{ccc}0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0\end{array}\right|$, then $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=\frac{3}{2}$, and
II: If $\left|\begin{array}{ccc}x^2+x & x+1 & x-2 \\ 2 x^2+3 x-1 & 3 x & 3 x-3 \\ x^2+2 x+3 & 2 x-1 & 2 x-1\end{array}\right|=\mathrm{p} x+\mathrm{q}$, then $\mathrm{p}^2=196 \mathrm{q}^2$,
A.
both are true
B.
both are false
C.
only I is true
D.
only II is true
2026
JEE Mains
MCQ
JEE Main 2026 (Online) 22nd January Evening Shift
Let n be the number obtained on rolling a fair die. If the probability that the system
$ \begin{aligned} & x-\mathrm{n} y+z=6 \\ & x+(\mathrm{n}-2) y+(\mathrm{n}+1) z=8 \\ & \quad(\mathrm{n}-1) y+z=1 \end{aligned} $
has a unique solution is $\frac{k}{6}$, then the sum of $k$ and all possible values of $n$ is :
A.
22
B.
20
C.
24
D.
21
2026
JEE Mains
MCQ
JEE Main 2026 (Online) 22nd January Evening Shift
If $X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ is a solution of the system of equations $A X=B$, where $\operatorname{adj} A=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$, then $|x+y+z|$ is equal to :
A.
3
B.
2
C.
$\frac{3}{2}$
D.
1
2026
JEE Mains
MCQ
JEE Main 2026 (Online) 22nd January Morning Shift
If $\mathrm{A}=\left[\begin{array}{ll}2 & 3 \\ 3 & 5\end{array}\right]$, then the determinant of the matrix $\left(\mathrm{A}^{2025}-3 \mathrm{~A}^{2024}+\mathrm{A}^{2023}\right)$ is
A.
12
B.
24
C.
28
D.
16
2026
JEE Mains
MCQ
JEE Main 2026 (Online) 21st January Evening Shift
If the system of equations
$ 3x + y + 4z = 3 $
$ 2x + \alpha y - z = -3 $
$ x + 2y + z = 4 $
has no solution, then the value of $ \alpha $ is equal to:
A.
13
B.
4
C.
19
D.
23
2026
JEE Mains
MCQ
JEE Main 2026 (Online) 21st January Evening Shift
For the matrices $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix}$, if $(A^{15} + B) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$, then among the following which one is true?
A.
$x = 16$, $y = 3$
B.
$x = 5$, $y = 7$
C.
$x = 11$, $y = 2$
D.
$x = 18$, $y = 11$
2025
JEE Mains
MCQ
JEE Main 2025 (Online) 8th April Evening Shift
Let α be a solution of $x^2 + x + 1 = 0$, and for some a and b in
$R, \begin{bmatrix} 4 & a & b \end{bmatrix} \begin{bmatrix} 1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$. If $\frac{4}{\alpha^4} + \frac{m}{\alpha^a} + \frac{n}{\alpha^b} = 3$, then m + n is equal to _______
A.
11
B.
3
C.
8
D.
7
2025
JEE Mains
MCQ
JEE Main 2025 (Online) 8th April Evening Shift
Let $ A = \begin{bmatrix} 2 & 2+p & 2+p+q \\ 4 & 6+2p & 8+3p+2q \\ 6 & 12+3p & 20+6p+3q \end{bmatrix} $.
If $ \det(\text{adj}(\text{adj}(3A))) = 2^m \cdot 3^n $, $ m, n \in \mathbb{N} $, then $ m + n $ is equal to
A.
22
B.
20
C.
24
D.
26
2025
JEE Mains
MCQ
JEE Main 2025 (Online) 7th April Evening Shift
Let the system of equations
x + 5y - z = 1
4x + 3y - 3z = 7
24x + y + λz = μ
λ, μ ∈ ℝ, have infinitely many solutions. Then the number of the solutions of this system,
if x, y, z are integers and satisfy 7 ≤ x + y + z ≤ 77, is :
A.
4
B.
5
C.
3
D.
6
2025
JEE Mains
MCQ
JEE Main 2025 (Online) 7th April Morning Shift
Let $A$ be a $3 \times 3$ matrix such that $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} \mathrm{A}))|=81$.
If $S=\left\{n \in \mathbb{Z}:(|\operatorname{adj}(\operatorname{adj} A)|)^{\frac{(n-1)^2}{2}}=|A|^{\left(3 n^2-5 n-4\right)}\right\}$, then $\sum_\limits{n \in S}\left|A^{\left(n^2+n\right)}\right|$ is equal to :
A.
820
B.
866
C.
750
D.
732
2025
JEE Mains
MCQ
JEE Main 2025 (Online) 7th April Morning Shift
Let the system of equations :
$ \begin{aligned} & 2 x+3 y+5 z=9 \\ & 7 x+3 y-2 z=8 \\ & 12 x+3 y-(4+\lambda) z=16-\mu \end{aligned}$
have infinitely many solutions. Then the radius of the circle centred at $(\lambda, \mu)$ and touching the line $4 x=3 y$ is :
A.
$\frac{7}{5}$
B.
$\frac{21}{5}$
C.
7
D.
$\frac{17}{5}$
2025
JEE Mains
MCQ
JEE Main 2025 (Online) 4th April Evening Shift
Let the matrix $A=\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]$ satisfy $A^n=A^{n-2}+A^2-I$ for $n \geqslant 3$. Then the sum of all the elements of $\mathrm{A}^{50}$ is :
A.
44
B.
39
C.
52
D.
53
2025
JEE Mains
MCQ
JEE Main 2025 (Online) 3rd April Morning Shift
Let $A$ be a matrix of order $3 \times 3$ and $|A|=5$. If $|2 \operatorname{adj}(3 A \operatorname{adj}(2 A))|=2^\alpha \cdot 3^\beta \cdot 5^\gamma, \alpha, \beta, \gamma \in N$, then $\alpha+\beta+\gamma$ is equal to
A.
26
B.
27
C.
25
D.
28
2025
JEE Mains
MCQ
JEE Main 2025 (Online) 2nd April Evening Shift
If the system of equations
$ \begin{aligned} & 2 x+\lambda y+3 z=5 \\ & 3 x+2 y-z=7 \\ & 4 x+5 y+\mu z=9 \end{aligned} $
has infinitely many solutions, then $\left(\lambda^2+\mu^2\right)$ is equal to :
A.
30
B.
26
C.
22
D.
18
2025
JEE Mains
MCQ
JEE Main 2025 (Online) 2nd April Evening Shift
Let $A$ be a $3 \times 3$ real matrix such that $A^2(A-2 I)-4(A-I)=O$, where $I$ and $O$ are the identity and null matrices, respectively. If $A^5=\alpha A^2+\beta A+\gamma I$, where $\alpha, \beta$, and $\gamma$ are real constants, then $\alpha+\beta+\gamma$ is equal to :
A.
76
B.
12
C.
4
D.
20
2025
JEE Mains
MCQ
JEE Main 2025 (Online) 2nd April Morning Shift
Let $\mathrm{A}=\left[\begin{array}{cc}\alpha & -1 \\ 6 & \beta\end{array}\right], \alpha>0$, such that $\operatorname{det}(\mathrm{A})=0$ and $\alpha+\beta=1$. If I denotes $2 \times 2$ identity matrix, then the matrix $(I+A)^8$ is :
A.
$\left[\begin{array}{cc}257 & -64 \\ 514 & -127\end{array}\right]$
B.
$\left[\begin{array}{cc}766 & -255 \\ 1530 & -509\end{array}\right]$
C.
$\left[\begin{array}{cc}1025 & -511 \\ 2024 & -1024\end{array}\right]$
D.
$\left[\begin{array}{ll}4 & -1 \\ 6 & -1\end{array}\right]$
2025
JEE Mains
MCQ
JEE Main 2025 (Online) 2nd April Morning Shift
Let $a \in R$ and $A$ be a matrix of order $3 \times 3$ such that $\operatorname{det}(A)=-4$ and $A+I=\left[\begin{array}{lll}1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2\end{array}\right]$, where $I$ is the identity matrix of order $3 \times 3$. If $\operatorname{det}((a+1) \operatorname{adj}((a-1) A))$ is $2^{\mathrm{m}} 3^{\mathrm{n}}, \mathrm{m}$, $\mathrm{n} \in\{0,1,2, \ldots, 20\}$, then $\mathrm{m}+\mathrm{n}$ is equal to :
A.
14
B.
17
C.
15
D.
16
2025
JEE Mains
MCQ
JEE Main 2025 (Online) 2nd April Morning Shift
If the system of linear equations
$ \begin{aligned} & 3 x+y+\beta z=3 \\ & 2 x+\alpha y-z=-3 \\ & x+2 y+z=4 \end{aligned} $
has infinitely many solutions, then the value of $22 \beta-9 \alpha$ is :
A.
31
B.
37
C.
43
D.
49
2025
JEE Mains
MCQ
JEE Main 2025 (Online) 29th January Evening Shift
Let $A = [a_{ij}]$ be a $2 \times 2$ matrix such that $a_{ij} \in \{0, 1\}$ for all $i$ and $j$. Let the random variable $X$ denote the possible values of the determinant of the matrix $A$. Then, the variance of $X$ is:
A.
$\frac{5}{8}$
B.
$\frac{1}{4}$
C.
$\frac{3}{4}$
D.
$\frac{3}{8}$