Matrices and Determinants

Given matrix $A = \left[ {\matrix{ { - 1} & 2 \cr 1 & { - 1} \cr } } \right]$

For option A :

${R_1} \to {R_1} + {R_2}$

$A = \left[ {\matrix{ 0 & 1 \cr 1 & { - 1} \cr } } \right]$

$\therefore$ Option A can be obtained.

For option B :

${R_1} \leftrightarrow {R_2}$

$A = \left[ {\matrix{ 1 & { - 1} \cr { - 1} & 2 \cr } } \right]$

$\therefore$ Option B can be obtained.

Option C :

Not possible by a single elementary row operation.

Option D :

${R_2} \to {R_2} + 2{R_1}$

$A = \left[ {\matrix{ { - 1} & 2 \cr { - 1} & 3 \cr } } \right]$

$\therefore$ Option D can be obtained.

Given,

$x + y + z = 6$ ...... (1)

$2x + 5y + \alpha z = \beta $ ..... (2)

$x + 2y + 3z = 14$ ...... (3)

System of equation have infinite many solutions.

$\therefore$ ${\Delta _x} = {\Delta _y} = {\Delta _z} = 0$ and $\Delta = 0$

Now, $\Delta = \left| {\matrix{ 1 & 1 & 1 \cr 2 & 5 & \alpha \cr 1 & 2 & 3 \cr } } \right| = 0$

${C_1} \to {C_1} - {C_3}$

${C_2} \to {C_2} - {C_3}$

$ \Rightarrow \left| {\matrix{ 0 & 0 & 1 \cr {2 - \alpha } & {5 - \alpha } & \alpha \cr { - 2} & { - 1} & 3 \cr } } \right| = 0$

$ \Rightarrow - 2 + \alpha + 10 - 2\alpha = 0$

$ \Rightarrow 8 - \alpha = 0$

$ \Rightarrow \alpha = 8$

Now, $x + y + z = 6$

$2x + 5y + 8z = \beta $

$x + 2y + 3z = 14$

$\therefore$ ${\Delta _x} = \left| {\matrix{ 6 & 1 & 1 \cr \beta & 5 & 8 \cr {14} & 2 & 3 \cr } } \right| = 0$

${C_1} \to {C_1} - 6{C_3}$

${C_2} \to {C_2} - {C_3}$

$ \Rightarrow \left| {\matrix{ 0 & 0 & 1 \cr {\beta - 48} & { - 3} & 8 \cr { - 4} & { - 1} & 3 \cr } } \right| = 0$

$ \Rightarrow - \beta + 48 - 12 = 0$

$ \Rightarrow \beta = 36$

$\therefore$ $\alpha + \beta = 8 + 36 = 44$

AB is zero matrix

$ \Rightarrow |A| = |B| = 0$

So neither A nor B is invertible

If $|A| = 0$

$ \Rightarrow |\mathrm{adj}\,A| = 0$ so $\mathrm{adj}\,A$

$AX = 0$ is homogeneous system and $|A| = 0$

So, it is having infinitely many solutions

(A) $M = {A^4} - {B^4}$

${M^T} = {({A^4} - {B^4})^T} = {({A^T})^4} - {({B^T})^4}$

$ = {A^4} - {( - B)^4} = {A^4} - {B^4} = M$

(B) $M = AB - BA$

${M^T} = {(AB - BA)^T} = {(AB)^T} - {(BA)^T}$

$ = {B^T}{A^T} - {A^T}{B^T}$

$ = - BA - A( - B)$

$ = AB - BA = M$

(C) $M = {B^5} - {A^5}$

${M^T} = {({B^T})^5} - {({A^T})^5} = - ({B^5} + {A^5}) \ne - M$

(D) $M = AB + BA$

${M^T} = {(AB)^T} + {(BA)^T}$

$ = {B^T}{A^T} + {A^T}{B^T} = - BA - AB = - M$

$A = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right]$

$ \Rightarrow {A^2} = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right] \times \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right] = \left[ {\matrix{ 0 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr } } \right]$

$ \Rightarrow {A^3} = \left[ {\matrix{ 0 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr } } \right]\left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] = l$

Now ${B_0} = {A^{49}} + 2{A^{98}} = {({A^3})^{16}}\,.\,A + 2{({A^3})^{32}}\,.\,{A^2}$

${B_0} = A + 2{A^2} = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right] + \left[ {\matrix{ 0 & 0 & 2 \cr 2 & 0 & 0 \cr 0 & 2 & 0 \cr } } \right] = \left[ {\matrix{ 0 & 1 & 2 \cr 2 & 0 & 1 \cr 1 & 2 & 0 \cr } } \right]$

$|{B_0}| = 9$

Since, ${B_n} = Adj\,|{B_{n - 1}}| \Rightarrow |{B_n}| = |{B_{n - 1}}{|^2}$

Hence $|{B_4}| = |{B_3}{|^2} = |{B_2}{|^4} = |{B_1}{|^8} = |{B_0}{|^{16}}$

$ = |{3^2}{|^{16}} = {3^{32}}$

Characteristic equation of A is given by

$\left| {A - \lambda I} \right| = 0$

$\left| {\matrix{ {4 - \lambda } & { - 2} \cr \alpha & {\beta - \lambda } \cr } } \right| = 0$

$ \Rightarrow {\lambda ^2} - (4 + \beta )\lambda + (4\beta + 2\alpha ) = 0$

So, ${A^2} - (4 + \beta )A + (4\beta + 2\alpha )I = 0$

$|A| = 4\beta + 2\alpha = 18$

${A^2} = \left[ {\matrix{ 1 & 2 \cr { - 2} & { - 5} \cr } } \right]\left[ {\matrix{ 1 & 2 \cr { - 2} & { - 5} \cr } } \right] = \left[ {\matrix{ { - 3} & { - 8} \cr 8 & {21} \cr } } \right]$

$\alpha {A^2} + \beta A = \left[ {\matrix{ { - 3\alpha } & { - 8\alpha } \cr {8\alpha } & {21\alpha } \cr } } \right] + \left[ {\matrix{ \beta & {2\beta } \cr { - 2\beta } & { - 5\beta } \cr } } \right]$

$ = \left[ {\matrix{ { - 3\alpha + \beta } & { - 8\alpha + 2\beta } \cr {8\alpha - 2\beta } & {21\alpha - 5\beta } \cr } } \right] = \left[ {\matrix{ 2 & 0 \cr 0 & 2 \cr } } \right]$

On Comparing

$8\alpha = 2\beta ,\, - 3\alpha + \beta = 2,\,21\alpha - 5\beta = 2$

$ \Rightarrow \alpha = 2,\,\beta = 8$

So, $\alpha + \beta = 10$

$A'BA = \left[ {\matrix{ 1 & 1 & 1 \cr } } \right]\left[ {\matrix{ {{9^2}} & { - {{10}^2}} & {{{11}^2}} \cr {{{12}^2}} & {{{13}^2}} & { - {{14}^2}} \cr { - {{15}^2}} & {{{16}^2}} & {{{17}^2}} \cr } } \right]A$

$ = \left[ {\matrix{ {{9^2} + {{12}^2} - {{15}^2}} & { - {{10}^2} + {{13}^2} + {{16}^2}} & {{{11}^2} - {{14}^2} + {{17}^2}} \cr } } \right]\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$

$ = \left[ {{9^2} + {{12}^2} - {{15}^2} - {{10}^2} + {{13}^2} + {{16}^2} + {{11}^2} - {{14}^2} + {{17}^2}} \right]$

$ = [({9^2} - {10^2}) + ({11^2} + {12^2}) + ({13^2} - {14^2}) + ({16^2} - {15^2}) + {17^2}]$

$ = [ - 19 + 265 + ( - 27) + 31 + 289]$

$ = [585 - 46] = [539]$

$\Delta = \left| {\matrix{ 8 & 1 & 4 \cr 1 & 1 & 1 \cr \lambda & { - 3} & 0 \cr } } \right|$

$ = 8(3) - 1( - \lambda ) + 4( - 3 - \lambda )$

$ = 24 + \lambda - 12 - 4\lambda $

$ = 12 - 3\lambda $

So for $\lambda = 4$, it is having infinitely many solutions.

${\Delta _x} = \left| {\matrix{ { - 2} & 1 & 4 \cr 0 & 1 & 1 \cr \mu & { - 3} & 0 \cr } } \right|$

$ = - 2(3) - 1( - \mu )+4( - \mu )$

$ = - 6 - 3\mu = 0$

For $\mu = - 2$

Distance of $(4, - 2,{{ - 1} \over 2})$ from $8x + y + 4z + 2 = 0$

$ = {{32 - 2 - 2 + 2} \over {\sqrt {64 + 1 + 16} }} = {{10} \over 3}$ units

$|(A + I)(adj\,A + I)| = 4$

$ \Rightarrow |A\,adj\,A + A + adj\,A + I| = 4$

$ \Rightarrow |(A)I + A + adj\,A + I| = 4$

$|A| = - 1 \Rightarrow |A + adj\,A| = 4$

$A = \left[ {\matrix{ a & b \cr c & d \cr } } \right]\,adj\,A = \left[ {\matrix{ a & { - b} \cr { - c} & d \cr } } \right]$

$ \Rightarrow \left| {\matrix{ {(a + d)} & 0 \cr 0 & {(a + d)} \cr } } \right| = 4$

$ \Rightarrow a + d = \, \pm \,2$

$\Delta = \left| {\matrix{ 2 & { - 3} & 5 \cr 1 & 3 & { - 1} \cr 3 & { - 1} & {{\lambda ^2} - |\lambda |} \cr } } \right| = 2\left( {3{\lambda ^2} - 3|\lambda | - 1} \right) + 3\left( {{\lambda ^2} - |\lambda | + 3} \right) + 5( - 1 - 9)$

$ = 9{\lambda ^2} - 9|\lambda | - 43$

$ = 9|\lambda {|^2} - 9|\lambda | - 43$

$\Delta = 0$ for 2 values of $|\lambda |$ out of which one is $-\mathrm{ve}$ and other is $+\mathrm{ve}$

So, 2 values of $\lambda$ satisfy the system of equations to obtain no solution.

Given,

$3(\sin 3\theta )x - y + z = 2$

$3(\cos 2\theta )x + 4y + 3z = 3$

$6x + 7y + 7z = 9$

For no solutions determinant of coefficient will be = 0

$\therefore$ $D = \left| {\matrix{ {3\sin 3\theta } & { - 1} & 1 \cr {3\cos 2\theta } & 4 & 3 \cr 6 & 7 & 7 \cr } } \right| = 0$

$ \Rightarrow 3\sin 3\theta (28 - 21) + 1(21\cos 2\theta - 18) + 1(21\cos 2\theta - 24) = 0$

$ \Rightarrow 21\sin 3\theta + 42\cos 2\theta - 42 = 0$

$ \Rightarrow \sin 3\theta + 2\cos 2\theta - 2 = 0$

$ \Rightarrow 3\sin \theta - 4{\sin ^3}\theta + 2(1 - 2{\sin ^2}\theta ) - 2 = 0$

$ \Rightarrow 3\sin \theta - 4{\sin ^3}\theta - 4{\sin ^2}\theta = 0$

$ \Rightarrow 4{\sin ^3}\theta + 4{\sin ^2}\theta - 3\sin \theta = 0$

$ \Rightarrow \sin \theta (4{\sin ^2}\theta + 4\sin \theta - 3) = 0$

$\therefore$ $\sin \theta = 0$

$ \Rightarrow \theta = \pi ,2\pi ,3\pi $ when $\theta \in (0,4\pi )$

or,

$4{\sin ^2}\theta + 4\sin \theta - 3 = 0$

$ \Rightarrow 4{\sin ^2}\theta + 6\sin \theta - 2\sin \theta - 3 = 0$

$ \Rightarrow 2\sin \theta (2\sin \theta + 3) - 1(2\sin \theta + 3) = 0$

$ \Rightarrow (2\sin \theta - 1)(2\sin \theta + 3) = 0$

$\therefore$ $\sin \theta = {1 \over 2}$

or,

$\sin \theta = - {3 \over 2}$ [not possible as $\sin \in [ - 1,1]$]

$\therefore$ $\sin \theta = {1 \over 2}$

$ \Rightarrow \theta = {\pi \over 6},{{5\pi } \over 6},{{13\pi } \over 6},{{17\pi } \over 6}$

$\therefore$ Possible values of $\theta = \pi ,2\pi ,3\pi ,{\pi \over 6},{{5\pi } \over 6},{{13\pi } \over 6},{{17\pi } \over 6}$

$\therefore$ Total 7 values of $\theta$ possible.

Given,

$A = \left[ {\matrix{ 1 & { - 2} & \alpha \cr \alpha & 2 & { - 1} \cr } } \right]$

and $B = \left[ {\matrix{ 2 & \alpha \cr { - 1} & 2 \cr 4 & { - 5} \cr } } \right]$

$AB = \left[ {\matrix{ 1 & { - 2} & \alpha \cr \alpha & 2 & { - 1} \cr } } \right]\left[ {\matrix{ 2 & \alpha \cr { - 1} & 2 \cr 4 & { - 5} \cr } } \right]$

$ = \left[ {\matrix{ {4 + 4\alpha } & { - 4\alpha - 4} \cr {2\alpha - 6} & {{\alpha ^2} + 9} \cr } } \right]$

Given,

$|AB| = 0$

$\therefore$ $\left| {\matrix{ {4 + 4\alpha } & { - 4\alpha - 4} \cr {2\alpha - 6} & {{\alpha ^2} + 9} \cr } } \right| = 0$

$ \Rightarrow (4\alpha + 4)\left| {\matrix{ 1 & { - 1} \cr {2\alpha - 6} & {{\alpha ^2} + 9} \cr } } \right| = 0$

$ \Rightarrow (4\alpha + 4)({\alpha ^2} + 9 + 2\alpha - 6) = 0$

$ \Rightarrow (4\alpha + 4)({\alpha ^2} + 2\alpha + 3) = 0$

$\therefore$ $\alpha - = - 1$

or ${\alpha ^2} + 2\alpha + 3 = 0$

${\alpha _1} + {\alpha _2} = - 2$

$\therefore$ Sum of all values of $\alpha = - 1 - 2 = - 3$

$\therefore$ Absolute value of $\alpha = | - 3| = 3$

Given,

$\det (A) = 2$,

$\det (B) = 3$

and $\det \left( {\left( {\det \left( {5\left( {\det A} \right)B} \right)} \right){A^2}} \right) = {2^a}{3^b}{5^c}$

$ \Rightarrow \left| {\det \left( {5\left( {\det A} \right)B} \right){A^2}} \right| = {2^a}{3^b}{5^c}$

$ \Rightarrow \left| {\left| {5\left( {\det A} \right)\left. B \right|{A^2}} \right.} \right| = {2^a}{3^b}{5^c}$

$ \Rightarrow \left| {\left| {5\left| {A\left| B \right.\left| {{A^2}} \right.} \right.} \right.} \right| = {2^a}{3^b}{5^c}$

$ \Rightarrow \left| {\left| {5\,.\,2\,.\,B} \right|{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}$

$\Rightarrow \left| {\left| {10B} \right|{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}$

$ \Rightarrow \left| {{{10}^2}\,.\,\left| B \right|{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}$

As $\left| {k\,.\,A} \right| = {k^n}|A|$

$ \Rightarrow \left| {100 \times 3{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}$

$ \Rightarrow {(300)^2}\,.\,|{A^2}| = {2^a}\,.\,{3^b}\,.\,{5^c}$

$ \Rightarrow {(300)^2}\,.\,|A{|^2} = {2^a}\,.\,{3^b}\,.\,{5^c}$

$ \Rightarrow {(300)^2}\,.\,{2^2} = {2^a}\,.\,{3^b}\,.\,{5^c}$

$ \Rightarrow 9 \times 100 \times 100 \times {2^2} = {2^a}\,.\,{3^b}\,.\,{5^c}$

$ \Rightarrow {3^2} \times {2^2} \times {5^2} \times {2^2} \times {5^2} \times {2^2} = {2^a}\,.\,{3^b}\,.\,{5^c}$

$ \Rightarrow {2^6}\,.\,{3^2}\,.\,{5^4} = {2^a}\,.\,{3^b}\,.\,{5^c}$

Comparing both sides, we get

$a = 6$, $b = 2$, $c = 4$

$\therefore$ $a + b + c = 6 + 2 + 4 = 12$

Given $A = \left[ {\matrix{ 2 & { - 1} \cr 0 & 2 \cr } } \right]$

and

$B = I - {5_{{C_1}}}(adj\,A) + {5_{{C_2}}}{(adj\,A)^2} - {5_{{C_3}}}{(adj\,A)^3} + {5_{{C_4}}}{(adj\,A)^4} - {5_{{C_5}}}{(adj\,A)^5}$

$ = {\left( {I - (adj\,A)} \right)^5}$

Cofactor of $A = \left[ {\matrix{ {{{( - 1)}^{1 + 1}}\,.\,2} & {{{( - 1)}^{1 + 2}}\,.\,0} \cr {{{( - 1)}^{2 + 1}}\,.\,( - 1)} & {{{( - 1)}^{2 + 2}}\,.\,2} \cr } } \right]$

$ = \left[ {\matrix{ 2 & 0 \cr 1 & 2 \cr } } \right]$

Transpose of cofactor of $A = \left[ {\matrix{ 2 & 1 \cr 0 & 2 \cr } } \right]$

$\therefore$ $adj\,A = \left[ {\matrix{ 2 & 1 \cr 0 & 2 \cr } } \right]$

Now, $I - adj\,A$

$ = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right] - \left[ {\matrix{ 2 & 1 \cr 0 & 2 \cr } } \right]$

$ = \left[ {\matrix{ { - 1} & { - 1} \cr 0 & { - 1} \cr } } \right]$

Now let,

$P = I - adj\,A = \left[ {\matrix{ { - 1} & { - 1} \cr 0 & { - 1} \cr } } \right]$

$\therefore$ ${P^2} = \left[ {\matrix{ { - 1} & { - 1} \cr 0 & { - 1} \cr } } \right]\left[ {\matrix{ { - 1} & { - 1} \cr 0 & { - 1} \cr } } \right]$

$ = \left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right]$

${P^4} = {P^2}\,.\,{P^2} = \left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & 4 \cr 0 & 1 \cr } } \right]$

${P^5} = {P^4}\,.\,P = \left[ {\matrix{ 1 & 4 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ { - 1} & { - 1} \cr 0 & { - 1} \cr } } \right] = \left[ {\matrix{ { - 1} & { - 5} \cr 0 & { - 1} \cr } } \right]$

$\therefore$ $B = \left[ {\matrix{ { - 1} & { - 5} \cr 0 & { - 1} \cr } } \right]$

Now sum of elements $ = - 1 - 5 - 1 + 0 = - 7$

$2x + y - z = 7$

$x - 3y + 2z = 1$

$x + 4y + \delta z = k$

$\Delta = \left| {\matrix{ 2 & 1 & { - 1} \cr 1 & { - 3} & 2 \cr 1 & 4 & \delta \cr } } \right| = - 7\delta - 21 = 0$

$\delta = - 3$

${\Delta _1} = \left| {\matrix{ 7 & 1 & { - 1} \cr 1 & { - 3} & 2 \cr k & 4 & { - 3} \cr } } \right|$

$ \Rightarrow 6 - k = 0 \Rightarrow k = 6$

$\delta + k = - 3 + 6 = 3$

Given, ${a_{ij}} = {2^{j - i}}$

Now, $A = \left[ {\matrix{ {{2^0}} & {{2^1}} & {{2^2}} \cr {{2^{ - 1}}} & {{2^0}} & {{2^1}} \cr {{2^{ - 2}}} & {{2^{ - 1}}} & {{2^0}} \cr } } \right]$

$ = \left[ {\matrix{ 1 & 2 & 4 \cr {{1 \over 2}} & 1 & 2 \cr {{1 \over 4}} & {{1 \over 2}} & 1 \cr } } \right]$

${A^2} = \left[ {\matrix{ 1 & 2 & 4 \cr {{1 \over 2}} & 1 & 2 \cr {{1 \over 4}} & {{1 \over 2}} & 1 \cr } } \right]\left[ {\matrix{ 1 & 2 & 4 \cr {{1 \over 2}} & 1 & 2 \cr {{1 \over 4}} & {{1 \over 2}} & 1 \cr } } \right]$

$ = \left[ {\matrix{ {1 + 1 + 1} & {2 + 2 + 2} & {4 + 4 + 4} \cr {{1 \over 2} + {1 \over 2} + {1 \over 2}} & {1 + 1 + 1} & {2 + 2 + 2} \cr {{1 \over 4} + {1 \over 4} + {1 \over 4}} & {{1 \over 2} + {1 \over 2} + {1 \over 2}} & {1 + 1 + 1} \cr } } \right]$

$ = \left[ {\matrix{ 3 & 6 & {12} \cr {{3 \over 2}} & 3 & 6 \cr {{3 \over 4}} & {{3 \over 2}} & 3 \cr } } \right]$

$ = 3\left[ {\matrix{ 1 & 2 & 4 \cr {{1 \over 2}} & 1 & 2 \cr {{1 \over 4}} & {{1 \over 2}} & 1 \cr } } \right]$

$ = 3A$

Similarly, ${A^3} = {3^2}A$

${A^4} = {3^3}A$

$\therefore$ ${A^2} + {A^3} + \,\,......\,\, + \,\,{A^{10}}$

$ = 3A + {3^2}A + {3^3}A + \,\,......\,\, + \,\,{3^9}A$

$ = A(3 + {3^2} + {3^3} + \,\,......\,\, + \,\,{3^9})$

$ = A\left( {{{3({3^9} - 1)} \over {3 - 1}}} \right) = {{3({3^9} - 1)} \over 2}A$ = $\left( {{{{3^{10}} - 3} \over 2}} \right)A$

$\Delta = \left| {\matrix{ 2 & 3 & { - 1} \cr 1 & 1 & 1 \cr 1 & { - 1} & {|\lambda |} \cr } } \right| = 0 \Rightarrow |\lambda | = 7$

But at $\lambda = 7,\,{D_x} = {D_y} = {D_z} = 0$

${P_1}:2x + 3y - z = - 2$

${P_2}:x + y + z = 4$

${P_3}:x - y + |\lambda |z = 4\lambda - 4$

So clearly $5{P_2} - 2{P_1} = {P_3}$, so at $\lambda = 7$, system of equation is having infinite solutions.

So $\lambda = - 7$ is correct answer.

$|A| = 2$

$||A| = adj\,(5\,adj\,{A^3})|$

$ = |25|A|adj\,(adj\,{A^3})|$

$ = {25^3}|A{|^3}\,.\,|adj\,{A^3}{|^2}$

$ = {25^3}\,.\,{2^3}\,.\,|{A^3}{|^4}$

$ = {25^3}\,.\,{2^3}\,.\,{2^{12}} = {10^6}\,.\,512$

$f(x) = \left| {\matrix{ a & { - 1} & 0 \cr {ax} & a & { - 1} \cr {a{x^2}} & {ax} & a \cr } } \right|,\,a \in R$

$f(x) = a({a^2} + ax) + 1({a^2}x + a{x^2})$

$ = a{(x + a)^2}$

$f'(x) = 2a(x + a)$

Now, $2f'(10) - f'(5) + 100 = 0$

$ \Rightarrow 2.\,2a(10 + a) - 2a(5 + a) + 100 = 0$

$ \Rightarrow 2a(a + 15) + 100 = 0$

$ \Rightarrow {a^2} + 15a + 50 = 0$

$ \Rightarrow a = - 10,\, - 5$

$\therefore$ Sum of squares of values of a = 125.

A and B are two matrices of order 3 $\times$ 3.

and $AB = I$,

$|A| = {1 \over 8}$

Now, $|A||B| = 1$

$|B| = 8$

$\therefore$ $|adj(B(adj(2A))| = |B(adj(2A)){|^2}$

$ = |B{|^2}|adj(2A){|^2}$

$ = {2^6}|2A{|^{2 \times 2}}$

$ = {2^6}.\,{2^{12}}.\,{1 \over {{2^{12}}}} = 64$

$x + 2y + z = 2$

$\alpha x + 3y - z = \alpha $

$ - \alpha x + y + 2z = - \alpha $

$\Delta = \left| {\matrix{ 1 & 2 & 1 \cr \alpha & 3 & { - 1} \cr { - \alpha } & 1 & 2 \cr } } \right| = 1(6 + 1) - 2(2\alpha - \alpha ) + 1(\alpha + 3\alpha )$

$ = 7 + 2\alpha $

$\Delta = 0 \Rightarrow \alpha = - {7 \over 2}$

${\Delta _1} = \left| {\matrix{ 2 & 2 & 1 \cr \alpha & 3 & { - 1} \cr { - \alpha } & 1 & 2 \cr } } \right| = 14 + 2\alpha \ne 0$ for $\alpha = - {7 \over 2}$

$\therefore$ For no solution $\alpha = - {7 \over 2}$

Given system of equations

$\alpha x + y + z = 5$

$x + 2y + 3z = 4$, has infinite solution

$x + 3y + 5z = \beta $

$\therefore$ $\Delta = \left| {\matrix{ \alpha & 1 & 1 \cr 1 & 2 & 3 \cr 1 & 3 & 5 \cr } } \right| = 0 \Rightarrow \alpha (1) - 1(2) + 1(1) = 0$

$ \Rightarrow \alpha = 1$

and ${\Delta _1} = \left| {\matrix{ 5 & 1 & 1 \cr 4 & 2 & 3 \cr \beta & 3 & 5 \cr } } \right| = 0$

$ \Rightarrow 5(1) - 1(20 - 3\beta ) + 1(12 - 2\beta ) = 0$

$ \Rightarrow \beta = 3$

and ${\Delta _2} = \left| {\matrix{ 1 & 5 & 1 \cr 1 & 4 & 3 \cr 1 & \beta & 5 \cr } } \right| = 0 \Rightarrow (20 - 3\beta ) - 5(2) + 1(\beta - 4) = 0$

$ \Rightarrow - 2\beta + 6 = 0$

$ \Rightarrow \beta = 3$

Similarly,

$ \Rightarrow {\Delta _3} = \left| {\matrix{ 1 & 1 & 5 \cr 1 & 2 & 4 \cr 1 & 3 & \beta \cr } } \right| = 0 \Rightarrow \beta = 3$

$\therefore$ ($\alpha$, $\beta$) = (1, 3)

We know, $|adj\,A| = |A{|^{n - 1}}$

Now, $|adj\,24A| = |adj\,3(adj\,2A)|$

$ \Rightarrow |24A{|^{3 - 1}} = |3\,adj\,2A{|^{3 - 1}}$

$ \Rightarrow |24A{|^2} = |3\,adj\,2A{|^2}$

Also, we know, $|KA| = {K^n}|A|$

$ \Rightarrow {\left( {{{(24)}^2}} \right)^2}|A{|^2} = {\left( {{{(3)}^3}} \right)^2}|adj\,2A{|^2}$

$ \Rightarrow {(24)^6}|A{|^2} = {3^6}\,.\,{\left( {|2A{|^{3 - 1}}} \right)^2}$

$ \Rightarrow {(24)^6}|A{|^2} = {3^6}\,.\,|2A{|^4}$

$ \Rightarrow {(24)^6}|A{|^2} = {3^6}\,.\,{\left( {{2^3}} \right)^4}\,.\,|A{|^4}$

$ \Rightarrow {3^6}\,.\,{8^6}\,.\,|A{|^2} = {3^6}\,.\,{8^4}\,.\,|A{|^4}$

$ \Rightarrow {8^2} = |A{|^2}$

$ \Rightarrow |A{|^2} = 64$ = 2⁶

$\left| {\matrix{ 3 & { - 2} & 1 \cr 5 & { - 8} & 9 \cr 2 & 1 & a \cr } } \right| = 0 \Rightarrow - 14a - 42 = 0 \Rightarrow a = - 3$

Now 3 (equation (1)) $-$ (equation (2)) $-$ 2 (equation (3)) is

$3(3x - 2y + z - b) - (5x - 8y + 9z - 3) - 2(2x + y + az + 1) = 0$

$ \Rightarrow - 3b + 3 - 2 = 0 \Rightarrow b = {1 \over 3}$

So for no solution $a = - 3$ and $b \ne {1 \over 3}$

The system may be inconsistent if

$\left| {\matrix{ { - k} & 3 & { - 14} \cr { - 15} & 4 & { - k} \cr { - 4} & 1 & 3 \cr } } \right| = 0 \Rightarrow k = \, \pm \,11$

Hence if system is consistent then $k \in R - \{ 11, - 11\} $.

Let $A = \left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right]$

$A = \left[ {\matrix{ 1 \cr 1 \cr 0 \cr } } \right] = \left[ {\matrix{ 1 \cr 1 \cr 0 \cr } } \right] \Rightarrow \left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right]\left[ {\matrix{ 1 \cr 1 \cr 0 \cr } } \right] = \left[ {\matrix{ 1 \cr 1 \cr 0 \cr } } \right] \Rightarrow \matrix{ {a + b = 1} \cr {d + e = 1} \cr {g + h = 0} \cr } $

$A = \left[ {\matrix{ 1 \cr 0 \cr 1 \cr } } \right] = \left[ {\matrix{ { - 1} \cr 0 \cr 1 \cr } } \right] \Rightarrow \left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right]\left[ {\matrix{ 1 \cr 0 \cr 1 \cr } } \right] = \left[ {\matrix{ { - 1} \cr 0 \cr 1 \cr } } \right] \Rightarrow \matrix{ {a + c = - 1} \cr {d + f = 1} \cr {g + i = 0} \cr } $

$A = \left[ {\matrix{ 0 \cr 0 \cr 1 \cr } } \right] = \left[ {\matrix{ 1 \cr 1 \cr 2 \cr } } \right] \Rightarrow \left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right]\left[ {\matrix{ 0 \cr 0 \cr 1 \cr } } \right] = \left[ {\matrix{ 1 \cr 1 \cr 2 \cr } } \right] \Rightarrow \matrix{ {c = 1} \cr {f = 1} \cr {i = 2} \cr } $

Solving will get

$a = - 2,\,b = 3,\,c = 1,\,d = - 1,\,e = 2,\,f = 1,\,g = - 1,\,h = 1,\,i = 2$

$A = \left[ {\matrix{ { - 2} & 3 & 1 \cr { - 1} & 2 & 1 \cr { - 1} & 1 & 2 \cr } } \right] \Rightarrow A = 2I = \left[ {\matrix{ { - 4} & 3 & 1 \cr { - 1} & 0 & 1 \cr { - 1} & 1 & 0 \cr } } \right]$

$(A - 2I)x = \left[ {\matrix{ 4 \cr 1 \cr 1 \cr } } \right]$

$ \Rightarrow - 4{x_1} + 3{x_2} + {x_3} = 4$ ..... (i)

$ - {x_1} + {x_3} = 1$ ...... (ii)

$ - {x_1} + {x_2} = 1$ ...... (iii)

So 3(iii) + (ii) = (i)

$\therefore$ Infinite solution

$A = \left[ {\matrix{ 0 & { - 2} \cr 2 & 0 \cr } } \right]$

${A^2} = \left[ {\matrix{ 0 & { - 2} \cr 2 & 0 \cr } } \right]\left[ {\matrix{ 0 & { - 2} \cr 2 & 0 \cr } } \right] = \left[ {\matrix{ { - 4} & 0 \cr 0 & { - 4} \cr } } \right] = - 4I$

$M = {A^2} + {A^4} + {A^6} + \,\,.....\,\, + \,\,{A^{20}}$

$ = - 4I + 16I - 64I\,\, + $ ..... upto 10 terms

$ = - I$ [$4 - 16 + 64$ .... + upto 10 terms]

$ = - I\,.\,4\left[ {{{{{( - 4)}^{10}} - 1} \over { - 4 - 1}}} \right] = {4 \over 5}({2^{20}} - 1)I$

$N = {A^1} + {A^3} + {A^5} + \,\,....\,\, + \,\,{A^{19}}$

$ = A - 4A + 16A\,\, + $ ..... upto 10 terms

$ = A\left( {{{{{( - 4)}^{10}} - 1} \over { - 4 - 1}}} \right) = - \left( {{{{2^{20}} - 1} \over 5}} \right)A$

${N^2} = {{{{({2^{20}} - 1)}^2}} \over {{2^5}}}{A^2} = {{ - 4} \over {24}}{({2^{20}} - 1)^2}I$

$M{N^2} = {{ - 16} \over {125}}{({2^{20}} - 1)^3}I = KI\,\,\,\,\,(K \ne \pm \,1)$

${(M{N^2})^T} = {(KI)^T} = KI$

$\therefore$ A is correct

Given system of equations

$x + y + az = 2$ ..... (i)

$3x + y + z = 4$ ..... (ii)

$x + 2z = 1$ ..... (iii)

Solving (i), (ii) and (iii), we get

x = 1, y = 1, z = 0 (and for unique solution a $\ne$ $-$3)

Now, ($\alpha$, 1), (1, $\alpha$) and (1, $-$1) are collinear

$\therefore$ $\left| {\matrix{ \alpha & 1 & 1 \cr 1 & \alpha & 1 \cr 1 & { - 1} & 1 \cr } } \right| = 0$

$ \Rightarrow \alpha (\alpha + 1) - 1(0) + 1( - 1 - \alpha ) = 0$

$ \Rightarrow {\alpha ^2} - 1 = 0$

$\therefore$ $\alpha = \, \pm \,1$

$\therefore$ Sum of absolute values of $\alpha = 1 + 1 = 2$

$\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ \alpha & 2 \alpha & 3 \\ 1 & 3 \alpha & 5\end{array}\right|$

$ \begin{aligned} &=1(10 \alpha-9 \alpha)-1(5 \alpha-3)+1\left(3 \alpha^{2}-2 \alpha\right) \\\\ &=\alpha-5 \alpha+3+3 \alpha^{2}-2 \alpha \\\\ &=3 \alpha^{2}-6 \alpha+3 \end{aligned} $

For inconsistency $\Delta=0$ i.e. $\alpha=1$

Now check for $\alpha=1$

$ x+y+z=1\quad\quad...(i) $

$x+2 y+3 z=-1\quad\quad...(ii)$

$x+3 y+5 z=4\quad\quad...(iii)$

By (ii) $\times 2-$ (i) $\times 1$

$ x+3 y+5 z=-3 $

so equations are inconsistent for $\alpha=1$

Given, $A = {\left[ {\matrix{ 1 & 0 & a \cr { - 1} & 1 & 0 \cr { - a} & 0 & 1 \cr } } \right]_{3 \times 3}}$

S = {$\sqrt{n}$ : 1 $\le$ n $\le$ 50 and n is odd}

$ \therefore $ S = $\left\{ {1,\sqrt 3 ,\sqrt 5 ,\sqrt 7 ,....,\sqrt {49} } \right\}$

We know,

$\left| {adj\,A} \right| = {\left| A \right|^{n - 1}}$

Here, n = order of matrix.

Here, n = 3

$\therefore$ $\left| {adj\,A} \right| = {\left| A \right|^{3 - 1}} = {\left| A \right|^2}$

Now, $\left| A \right| = \left| {\matrix{ 1 & 0 & a \cr { - 1} & 1 & 0 \cr { - a} & 0 & 1 \cr } } \right|$

$ = 1(1 - 0) - 0 + a(0 - ( - a))$

$ = {a^2} + 1$

$\therefore$ $\left| {adj\,A} \right| = {\left| A \right|^2} = {({a^2} + 1)^2}$

Now, $\sum\limits_{a\, \in \,S}^{} {\det (adj\,A)} $

$ = \sum\limits_{a\, \in \,S}^{} {{{({a^2} + 1)}^2}} $

= ${\left( {{1^2} + 1} \right)^2} + {\left( {{{\left( {\sqrt 3 } \right)}^2} + 1} \right)^2} + {\left( {{{\left( {\sqrt 5 } \right)}^2} + 1} \right)^2} + .... + {\left( {{{\left( {\sqrt {49} } \right)}^2} + 1} \right)^2}$

= ${\left( {{1^2} + 1} \right)^2} + {\left( {3 + 1} \right)^2} + {\left( {5 + 1} \right)^2} + .... + {\left( {49 + 1} \right)^2}$

= ${2^2} + {4^2} + {6^2} + .... + {50^2}$

= ${2^2}\left( {{1^2} + {2^2} + {3^2} + .... + {{25}^2}} \right)$

= $4.{{25.26.51} \over 6} = 100.221$

$\therefore$ $100K = 100.221$

$ \Rightarrow K = 221$

$A=\left(\begin{array}{cc}\frac{5}{2} & \frac{3}{2} \\ -\frac{3}{2} & -\frac{1}{2}\end{array}\right)$

$A^{2}=\left(\begin{array}{cc}4 & 3 \\ -3 & -2\end{array}\right)$

$A^{3}=\left(\begin{array}{cc}\frac{11}{2} & \frac{9}{2} \\ -\frac{9}{2} & -\frac{7}{2}\end{array}\right)$

$A^{4}=\left(\begin{array}{cc}7 & 6 \\ -6 & -5\end{array}\right)$

and so on

$A^n=\left(\begin{array}{cc}\frac{3 n}{2}+1 & \frac{3 n}{2} \\ -\frac{3 n}{2} & -\frac{3 n}{2}+1\end{array}\right)$

Now

$ \begin{aligned} A^{2022} & =\left(\begin{array}{ll} \frac{3 \times 2022}{2}+1 & \frac{3 \times 2022}{2} \\ \frac{-3 \times 2022}{2} & \frac{-3 \times 2022}{2}+1 \end{array}\right) \\\\ & =\left(\begin{array}{cc} 3034 & 3033 \\ -3033 & -3032 \end{array}\right) \end{aligned} $

$\therefore$ Option $(\mathrm{A})$ is correct

Given:

$ \begin{gathered} x+y+z=1 ..........(1)\\\\ 10 x+100 y+1000 z=0 .........(2)\\\\ q r x+p r y+p q z=0 ............(3) \end{gathered} $

Now equation (3) can be re-written as

$ \frac{x}{p}=\frac{y}{q}=\frac{z}{r}=0 \quad \because\{p, q, r \neq 0\} $

Now given $p, q$ and $r$ are $10^{\text {th }}, 100^{\text {th }}$ and $1000^{\text {th }}$ term of an. H.P.,

So let $p=\frac{y}{a+9 d}, q=\frac{1}{a+99 d}, r=\frac{1}{a+999 d}$

Now from equation (3)

$ (a+9 d) x+(a+99 d) y+(a+999 d) z=0 $

Now from and (1), (2) and (3) we get

$ \begin{aligned} \Delta & =\left|\begin{array}{ccc} 1 & 1 & 1 \\ 10 & 100 & 1000 \\ a+9 d & a+99 d & a+999 d \end{array}\right|=0 \\\\ \Delta x & =\left|\begin{array}{ccc} 1 & 1 & 1 \\ 0 & 100 & 1000 \\ 0 & a+99 d & a+999 d \end{array}\right|=900(d-a) \\\\ \Delta y & =\left|\begin{array}{ccc} 10 & 0 & 1000 \\ 1+9 d & 0 & a+999 d \end{array}\right|=990(a-d) \\\\ \Delta z & =\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 10 & 100 & 0 \\ a+9 d & a+99 d & 0 \end{array}\right|=90(d-a) \end{aligned} $

(I) If $\frac{q}{r}=10 \Rightarrow a=d$

$ \Delta=\Delta_x=\Delta_y=\Delta_z=0 $

And equation (1) and equation (2) represents non-parallel plane equation (2) and equation (3) represents same plane

$\Rightarrow$ Infinitely many solutions.

Now finding solution by taking $z=\lambda$ then

$ \begin{aligned} & x+y =1-\lambda \\\\ & x+10 y =-100 \lambda \\\\ & x =\frac{10}{9}+100 \lambda \\\\ & y =\frac{-1}{9}-11 \lambda \\\\ & \Rightarrow (x, y, z) \in\left(\frac{10}{9}+10 \lambda, \frac{-1}{9}-11 \lambda, \lambda\right) \end{aligned} $

So $\mathrm{P}$ is not valid for any value of $\lambda \rightarrow \mathrm{Q}$

(II) $\frac{p}{r} \neq 100 \Rightarrow a \neq d$

$ \Delta=0 \,\text{and}\, \Delta_{x^{\prime}} \Delta_{y^{\prime}} \Delta_z \neq 0 $

So no solution.

$ \text { II } \rightarrow \mathrm{S} $

(III) If $ \frac{p}{q} \neq 10 \Rightarrow a \neq d \text { then } \Delta_z \neq 0 $

So no solution.

$ \text { III } \rightarrow \mathrm{S} $

(IV) If $ \Rightarrow a=d \text { then } \Delta_z=0 \Rightarrow \Delta_x=\Delta_y=0 $

So infinitely many solutions.

$ \mathrm{IV} \rightarrow \mathrm{R} $

$|A|=-21$

$\operatorname{tr}\left(A^3\right)=2024$

$ \begin{aligned} A & =\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \\ A^2 & =\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \\ & =\left[\begin{array}{ll} a^2+b c & a b+b d \\ a c+d c & b c+d^2 \end{array}\right] \\ A^3 & =\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \cdot\left[\begin{array}{ll} a^2+b c & a b+b d \\ a c+d c & b c+d^2 \end{array}\right] \end{aligned} $

$ =\left[\begin{array}{c} a^3+2 a b c+b d c \\ a^2 c+a d c+b c^2+d^2 c \end{array}\right. $

$ \left.\begin{array}{c} a^2 b+a b d+b^2 c+b d^2 \\ a b c+2 b c d+d^3 \end{array}\right] $

Given, $|A|=a d-b c=-21$, and $\operatorname{tr}\left(A^3\right)=a^3+d^3+3 a b c+3 b c d=2024$

On putting $b c=a d+21$

$ \begin{aligned} a^3+d^3+3 a(a d+21)+3(a d+ & 21) d =2024 \end{aligned} $

$ \begin{aligned} \Rightarrow a^3+d^3+3 a^2 d+63 a+3 a d^2 & +63 d =2024 \\ \Rightarrow a^3+d^3+3 a^2 d+3 a d^2+63(a+ & d) =2024 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad(a+d)^3+63(a+d)=2024 \\ & \Rightarrow \quad(a+d)\left[(a+d)^2+63\right]=2024 \\ & \therefore \operatorname{tr}(A)=a+d \end{aligned} $

On putting $a+d=6$

$ \begin{aligned} \Rightarrow \mathrm{LHS} & =6[36+63]=6 \times 99 \\ & =594 \neq \mathrm{RHS} \end{aligned} $

On putting $a+d=11$

$ \begin{aligned} \Rightarrow \mathrm{LHS} & =11[121+63] \\ & =11 \times 184=2024=\mathrm{RHS} \end{aligned} $

Therefore, $\operatorname{tr}(A)$ i.e. $a+d$ will be 11.

Given $A=\left[\begin{array}{lll}a & b & c \\ d & e & f \\ g & h & i\end{array}\right]$

∴ Given that $A$ is a skew-symmetric matrix.

$ \begin{aligned} & A^T=-A \\ & \Rightarrow \quad\left[\begin{array}{lll} a & d & g \\ b & e & h \\ c & f & i \end{array}\right]=-\left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right] \\ & \Rightarrow \quad\left[\begin{array}{lll} a & d & g \\ b & e & h \\ c & f & i \end{array}\right]=\left[\begin{array}{lll} -a & -b & -c \\ -d & -e & -f \\ -g & -h & -i \end{array}\right] \end{aligned} $

On comparing corresponding elements, we get

$ \begin{aligned} a & =-a \\ \Rightarrow \quad 2 a & =0 \Rightarrow a=0 \\ e & =-e \Rightarrow 2 e=0 \Rightarrow e=0 \\ i & =-i \Rightarrow 2 i=0 \Rightarrow i=0 \end{aligned} $

$ \begin{aligned} &\text { Now, } d=-b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ &\begin{aligned} & g=-c \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & b=-d \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & h=-f \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\\ & c=-g \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)\\ & f=-h \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi)\end{aligned} \end{aligned} $

Eq. (iii) divide by Eq. (v),

$ \begin{aligned} \frac{b}{c} & =\frac{-d}{-g} \\ \Rightarrow \quad \frac{b}{c} & =\frac{-d}{g} \times(-1) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vii)\end{aligned} $

Eq. (iv) gives $\frac{h}{f}=-1$ and Eq. (vi) gives

$ \frac{f}{h}=-1 $

But for $h=0, \frac{f}{h}$ is not defined.

On putting $(-1)=\frac{h}{f}$ in Eq. (viii),

$ \frac{b}{c}=\frac{-d}{g} \times \frac{h}{f}=\frac{-d h}{g f} $