Matrices and Determinants

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} $

Let $A=\left[\begin{array}{ccc}-1 & x & 3 \\ -4 & -5 & -6 \\ -7 & y & 9\end{array}\right]$

Cofactor of $-6=22$

$ \begin{array}{ll} \Rightarrow & (-1)[(-1)(y)-(-7)(x)]=22 \\ \Rightarrow & y-7 x=22 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Cofactor of $\quad-7=27$

$ \begin{array}{lrl} \Rightarrow &(x)(-6)-(-5)(3)=27 \\ \Rightarrow & -6 x+15 & =27 \\ \Rightarrow & -6 x & =12 \\ \Rightarrow & & x =-2 \end{array} $

$ \begin{aligned} &\text { On putting } x=-2 \text { in Eq. (i), }\\ &\begin{aligned} y & =7(-2)+22=-14+22=8 \\ \therefore 5 x+y & =-10+8=-2 \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \beta x+\alpha y-z=-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & 3 x-\beta y+\alpha z=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \alpha x+\beta y+z=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & \frac{\Delta_1}{\Delta}=x=-1 \\ & \Rightarrow \quad \frac{\Delta_2}{\Delta}=y=1 \\ & \text { and } \quad \frac{\Delta_3}{\Delta}=z=2 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { From Eqs. (i), (ii) and (iii), we have }\\ &\begin{array}{rlrl} & & -\beta+\alpha-2 & =-1 \\ & \Rightarrow & \alpha-\beta & =1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\\ & & -3-\beta+2 \alpha & =0 \\ & \Rightarrow& 2 \alpha-\beta & =3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)\\ & & -\alpha+\beta+2 & =1 \\\Rightarrow &&&-\alpha+\beta =-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi)\end{array} \end{aligned} $

Solving Eqs. (iv) and (v), we get

$ \begin{aligned} & & \alpha-\beta-2 \alpha+\beta & =1-3 \\ \Rightarrow & & -\alpha & =-2 \\ \Rightarrow & & \alpha & =2 \end{aligned} $

On putting $\alpha=2$ in Eq. (iv),

$ \begin{array}{rlrl} & & 2-\beta & =1 \\ & \Rightarrow & & \beta =1\\ & \therefore & & (\alpha, \beta) =(2,1) \end{array} $

$ \begin{array}{ll} \text { } & \left|\begin{array}{cc} 2+3 i & i \\ 1-2 i & -i \end{array}\right|=x+i y \\ \Rightarrow & -i(2+3 i)-i(1-2 i)=x+i y \\ \Rightarrow & -2 i-3 i^2-i+2 i^2=x+i y \\ \Rightarrow & -3 i-i^2=x+i y \\ \Rightarrow & -3 i+1=x+i y \\ \Rightarrow & 1-3 i=x+i y \end{array} $

On comparing real and imaginary parts, we get

$ \begin{array}{ll} & x=1 \text { and } y=-3 \\ \therefore & x+y=1+(-3)=-2 \end{array} $

             $ \begin{aligned} &\text { Given, } A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 4 & 3 & 2 \\ 3 & 4 & 5 \end{array}\right] \text { then, }\\ &A^T=\left[\begin{array}{lll} 1 & 4 & 3 \\ 2 & 3 & 4 \\ 3 & 2 & 5 \end{array}\right] \end{aligned} $

$ \begin{aligned} \left(A+A^T\right) & \left(A-A^T\right) \\ & =\left[\begin{array}{ccc} 2 & 6 & 6 \\ 6 & 6 & 6 \\ 6 & 6 & 10 \end{array}\right]\left[\begin{array}{ccc} 0 & -2 & 0 \\ 2 & 0 & -2 \\ 0 & 2 & 0 \end{array}\right] \\ & =\left[\begin{array}{ccc} 12 & 8 & -12 \\ 12 & 0 & -12 \\ 12 & 8 & -12 \end{array}\right] \\ & =4\left[\begin{array}{ccc} 3 & 2 & -3 \\ 3 & 0 & -3 \\ 3 & 2 & -3 \end{array}\right] \end{aligned} $

Given,

$ f(x)=\left|\begin{array}{ccc} x & x+1 & x+3 \\ x+2 & x+4 & x+7 \\ x+6 & x+9 & x+13 \end{array}\right| $

Applying $R_3 \rightarrow R_3-R_1, R_2 \rightarrow R_2-R_1$,

$ f(x)=\left|\begin{array}{ccc} x & x+1 & x+3 \\ 2 & 3 & 4 \\ 6 & 8 & 10 \end{array}\right| $

$ \begin{aligned} & \mathrm{At} x=5, f(5)=\left|\begin{array}{llr} 5 & 6 & 8 \\ 2 & 3 & 4 \\ 6 & 8 & 10 \end{array}\right| \\ & =5(30-32)-6(20-24)+8(16-18) \\ & =-10+24-16=-2 \end{aligned} $

Given, $A=\left[\begin{array}{lll}2 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 2\end{array}\right]$

$ \because \quad A^{-1}=\alpha A^2+\beta A+\gamma I $

Multiply by $A$ both sides,

$ \begin{aligned} A^{-1} \cdot A & =\alpha A^3+\beta A^2+\gamma I \cdot A \\ I & =\alpha A^3+\beta A^2+\gamma A \end{aligned} $

For matrix $A$

$\operatorname{det} .(A-\lambda I)=0$

$ \left|\begin{array}{ccc} 2-\lambda & 1 & 1 \\ 0 & 1-\lambda & 0 \\ 1 & 1 & 2-\lambda \end{array}\right|=0 $

$ \text { Expanding along } R_1 $

$ \begin{aligned} & (2-\lambda)(1-\lambda)(2-\lambda)-1(0)+1(\lambda-1)=0 \\ & \Rightarrow \quad(1-\lambda)\left(\lambda^2-4 \lambda+4\right)+\lambda-1=0 \\ & \Rightarrow \lambda^2-4 \lambda+4-\lambda^3+4 \lambda^2-4 \lambda+\lambda-1=0 \end{aligned} $

$ \begin{array}{ll} \Rightarrow & -\lambda^3+5 \lambda^2-7 \lambda+3=0 \\ \Rightarrow & \frac{1}{3} \lambda^3-\frac{5}{3} \lambda^2+\frac{7}{3} \lambda=1 \end{array} $

$ \begin{aligned} &\text { Compare with } \alpha A^3+\beta A^2+\gamma A=I \text {, }\\ &\begin{aligned} & \text { So, } \alpha=\frac{1}{3}, \beta=\frac{-5}{3}, \gamma=\frac{7}{3} \\ & \therefore 17 \alpha+5 \beta+\gamma=\frac{17}{3}-\frac{25}{3}+\frac{7}{3}=-\frac{1}{3} \end{aligned} \end{aligned} $

Here,

$A X=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right], \operatorname{Adj} A=\left[\begin{array}{ccc}1 & -1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & 1\end{array}\right]$ and also $|A|>0$.

Now, using $|\operatorname{adj} A|=|A|^{n-1} ; n$ is order of square matrix $A$.

$ \begin{aligned} & \quad\left|\begin{array}{ccc} 1 & -1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & 1 \end{array}\right|=|A|^{3-1} \\ & \Rightarrow \quad|A|^2=1(2)+1(2)-1(0)=4 \\ & \Rightarrow \quad|A|=2 \quad[\because|A|>0] \end{aligned} $

$ \text { } \begin{aligned} Now, X & =A^{-1}\left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right]=\frac{\operatorname{Adj}(A)}{|A|}\left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right] \\ & =\frac{1}{2}\left[\begin{array}{ccc} 1 & -1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & 1 \end{array}\right]\left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right] \\ \Rightarrow \quad X & =\frac{1}{2}\left[\begin{array}{c} -2 \\ 0 \\ 4 \end{array}\right]=\left[\begin{array}{c} -1 \\ 0 \\ 2 \end{array}\right] \end{aligned} $

$ \text { } \begin{aligned}Given, A & =\left[\begin{array}{ll} 0 & 1 \\ 1 & k \end{array}\right] \\ A^2 & =\left[\begin{array}{ll} 0 & 1 \\ 1 & k \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & k \end{array}\right]=\left[\begin{array}{cc} 1 & k \\ k & 1+k^2 \end{array}\right] \end{aligned} $

$ \begin{aligned} A^3=A^2 \times A & =\left[\begin{array}{cc} 1 & k \\ k & 1+k^2 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & k \end{array}\right] \\ & =\left[\begin{array}{cc} k & 1+k^2 \\ 1+k^2 & 2 k+k^3 \end{array}\right] \end{aligned} $

After comparing with given value of $A^3$, we get

$ \begin{aligned} & a=k, b=1+k^2, c=1+k^2 \text { and } \\ & d=2 k+k^3 \end{aligned} $

Given, $d=228$

$ \Rightarrow \quad 2 k+k^3=228 \Rightarrow k^3+2 k-228=0 $

By hit and trial method, $k=6$

$ \begin{aligned} b+c & =1+k^2+1+k^2 \\ & =2\left(1+k^2\right) \\ & =2(1+36)=74 \end{aligned} $

$A$ and $B$ are squae matrices of order 3.

$C$ is a unit matrix of order 3.

$ D=(A B-C)^{-1} $

$\Rightarrow D$ and $A B-C$ are inverse of each other

$ \begin{aligned} & & D \times(A B-C) & =I \\ \Rightarrow & & |D| \times|A B-C| & =1 \end{aligned} $

$ \begin{aligned} \Rightarrow|A B-C|=\frac{1}{|D|} \text { or }|D| & =\frac{1}{|A B-C|} \\ \text { Now, }|B C-C| \times|B D A| & =\frac{1}{|D|} \times|B D A| \\ & =\frac{|B| \times|D| \times|A|}{|D|} \\ & =|B| \times|A|=|B A| \end{aligned} $

If $D$ and $A B-C$ are inverse of each other,

$ \text { } \begin{aligned}then \,\,\, (A B-C) \times D & =D \times(A B-C)=I \\ A B D-D & =D A B-D=I \end{aligned} $

$ \begin{aligned} &\begin{array}{ll} \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,& A B D=D A B=D+I \\ \therefore \,\,\,\,\,\,\,& A B D=D A B \end{array}\\ &\text { Hence, both statements are true. } \end{aligned} $

$ \text { } \begin{aligned} \text { } A & =\left[\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{array}\right], B=\left[\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ |A| & =\left[\left.\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{array} \right\rvert\,=-1(0-1)=1\right. \\ \operatorname{adj} A & =\left[\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{array}\right]^T \\ A^{-1} & =\frac{1}{|A|} \operatorname{adj} A=A \\ |B| & =-1(1)=-1 \\ \operatorname{adj} B & =\left[\begin{array}{ccc} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \end{array}\right] \end{aligned} $

$ \begin{aligned} \therefore B^{-1}= & \frac{1}{|B|} \operatorname{adj} B=B \\ & \left(A^{-1} B\right)^{-1}+\left(A B^{-1}\right)^{-1} \\ & (A B)^{-1}+(A B)^{-1} \\ & {\left[\because A^{-1}=A \text { and } B^{-1}=B\right] } \\ & =2(A B)^{-1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

$ \begin{aligned} &\begin{aligned} A B & =\left[\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{array}\right]\left[\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & =\left[\begin{array}{ccc} 0 & 0 & -1 \\ -1 & 0 & 0 \\ 0 & -1 & 0 \end{array}\right] \\ |A B| & =-1(1)=-1 \\ \operatorname{adj} A B & =\left[\begin{array}{lll} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right]^T=\left[\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right] \end{aligned}\\ &\text { From Eq. (i), }\\ &\begin{aligned} 2 A B^{-1} & =2 \frac{1}{|A B|} \operatorname{adj}(A B) \\ & =-2\left[\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & -2 & 0 \\ 0 & 0 & -2 \\ -2 & 0 & 0 \end{array}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & {\left[\begin{array}{ccc} 7 & 5 & \alpha \\ \beta & 2 & 11 \\ 3 & \gamma & 1 \end{array}\right]\left[\begin{array}{l} 1 \\ 3 \\ 2 \end{array}\right]=\left[\begin{array}{c} \alpha+\beta \\ -2 \alpha+\beta-2 \gamma \\ \alpha+2 \beta+3 \gamma \end{array}\right]} \\ & \Rightarrow \quad\left[\begin{array}{c} 22+2 \alpha \\ \beta+28 \\ 5+3 \gamma \end{array}\right]=\left[\begin{array}{c} \alpha+\beta \\ -2 \alpha+\beta-2 \gamma \\ \alpha+2 \beta+3 \gamma \end{array}\right] \end{aligned} \end{aligned} $

$ \Rightarrow \quad \begin{aligned} 22+2 \alpha & =\alpha+\beta \\ -\alpha+\beta & =22 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Also, $-2 \alpha+\beta-2 \gamma=\beta+28$

$ \Rightarrow \quad-2(\alpha+\gamma)=28 $

or $\quad \alpha+\gamma=-14\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Also, $5+3 \gamma=\alpha+2 \beta+3 \gamma$

$ \alpha+2 \beta=5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

On adding Eqs. (i) and (iii),

$ 3 \beta=27 \Rightarrow \beta=9 $

From Eq. (i), $-\alpha+9=22$

$ \alpha=-13 $

From Eq. (ii), $-13+\gamma=-14$

$ \begin{aligned} & \gamma=-1 \\ & \begin{aligned} \therefore 100+\frac{2 \alpha+11 \beta}{\gamma} & =100+\left[\frac{(-26)+99}{-1}\right] \\ & =100-73=27 \end{aligned} \end{aligned} $

Let $A=\left[\begin{array}{ccc}0 & 2 & a \\ b & 0 & 4 \\ -3 & c & 0\end{array}\right]$

If $A$ is a skew-symmetric, then $A^T=-A$

$ \begin{aligned} {\left[\begin{array}{ccc} 0 & b & -3 \\ 2 & 0 & c \\ a & 4 & 0 \end{array}\right] } & =\left[\begin{array}{ccc} 0 & -2 & -a \\ -b & 0 & -4 \\ 3 & -c & 0 \end{array}\right] \\ \Rightarrow \quad b=-2, a & =3 \text { and } c=-4 \\ {\left[\begin{array}{ll} a & b \\ b & a \end{array}\right]\left[\begin{array}{ll} b & c \\ c & b \end{array}\right] } & =\left[\begin{array}{cc} 3 & -2 \\ -2 & 3 \end{array}\right]\left[\begin{array}{ll} -2 & -4 \\ -4 & -2 \end{array}\right] \\ & =\left[\begin{array}{cc} 2 & -8 \\ -8 & 2 \end{array}\right] \end{aligned} $

$ \text {Let } A=\left[\begin{array}{ccc} -1 & 2 & b \\ a & 5 & 6 \\ 3 & c & 7 \end{array}\right] $

If $A$ is a symmetric matrix, then $A^T=A$

$ \left[\begin{array}{ccc} -1 & a & 3 \\ 2 & 5 & c \\ b & 6 & 7 \end{array}\right]=\left[\begin{array}{ccc} -1 & 2 & b \\ a & 5 & 6 \\ 3 & c & 7 \end{array}\right] $

On comparing both the matrices, we get $a=2, b=3$ and $c=6$

$ \begin{aligned} \therefore & \left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|=\left|\begin{array}{lll} 2 & 3 & 6 \\ 3 & 6 & 2 \\ 6 & 2 & 3 \end{array}\right| \\ & =2(18-4)-3(9-12)+6(6-36) \\ & =28+9-180=-143 \end{aligned} $

$A^2-4 A-5 I=0$

Post multiplying by $A^{-1}$ on both the sides, we get

$ \begin{aligned} & A-4 I-5 A^{-1}=0 \\ \Rightarrow & 5 A^{-1}=A-4 I \\ \Rightarrow & A^{-1}=\frac{1}{5}\{A-4 I\} \\ \therefore & A^{-1}=\frac{1}{5}\left\{\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]-\left[\begin{array}{ccc} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right]\right\} \\ \because & A=\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]=\frac{1}{5}\left[\begin{array}{ccc} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{array}\right] \end{aligned} $

$\because B$ is the unique solution of $A Y=Q$ and $A$ is an invertible matrix

$ \begin{aligned} \Rightarrow & & A B & =Q \\ \Rightarrow & & B & =A^{-1} Q \end{aligned} $

Now, $\quad A X=B$

$ \begin{aligned} \Rightarrow \quad X & =A^{-1} B \\ & =A^{-1}\left(A^{-1} Q\right)=\left(A^{-1}\right)^2 Q \end{aligned} $

$ \begin{aligned} &\text { } f(x)=\left|\begin{array}{ccc} -\sin x & 2 \sin 2 x & 4 \cos ^2 x \\ \cos x & 4 \sin ^2 x & 2 \sin 2 x \\ 0 & -\cos x & \sin x \end{array}\right|\\ &\begin{gathered} =-\sin x\left|\begin{array}{cc} 4 \sin ^2 x & 2 \sin 2 x \\ -\cos x & \sin x \end{array}\right| \\ -\cos x\left|\begin{array}{cc} 2 \sin 2 x & 4 \cos ^2 x \\ -\cos x & \sin x \end{array}\right| \\ =-\sin x\left[4 \sin ^3 x+4 \cos ^2 x \sin x\right] \\ -\cos x\left[4 \sin ^2 x \cos x+4 \cos ^3 x\right] \\ {[\because \sin 2 x=2 \sin x \cos x]} \\ =4\left[-\sin ^4 x-\cos ^2 x \sin ^2 x\right. \\ \left.-\sin ^2 x \cos ^2 x-\cos ^4 x\right] \\ =-4\left[\sin ^4 x+\cos ^4 x+2 \sin ^2 x \cos ^2 x\right] \\ =-4\left(\sin ^2 x+\cos ^2 x\right)^2=-4 \\ \therefore\,\,\,\,\,\,\,\,\,\, f^{\prime}(x)=0 \\ f\left(\frac{5 \pi}{4}\right)+f^{\prime}\left(\frac{5 \pi}{4}\right)=-4+0=-4 \end{gathered} \end{aligned} $

$ \begin{aligned} & \text { } A+B=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2 \end{array}\right] . A B=\left[\begin{array}{lll} 1 & 2 & 2 \\ 1 & 1 & 0 \\ 1 & 2 & 1 \end{array}\right] \\ & (A+B)^2=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2 \end{array}\right]=\left[\begin{array}{lll} 5 & 8 & 8 \\ 4 & 5 & 2 \\ 2 & 8 & 4 \end{array}\right] \\ & (A+B)^2=A^2+B^2+A B+B A \\ & \therefore \quad A^2+B^2+B A=(A+B)^2-A B \\ & \text { or } A^2+B(A+B)=\left[\begin{array}{lll} 5 & 8 & 8 \\ 4 & 5 & 2 \\ 2 & 8 & 4 \end{array}\right] -\left[\begin{array}{lll} 1 & 2 & 2 \\ 1 & 1 & 0 \\ 1 & 2 & 1 \end{array}\right] =\left[\begin{array}{lll} 4 & 6 & 6 \\ 3 & 4 & 2 \\ 1 & 6 & 3 \end{array}\right]\end{aligned} $

$ \begin{aligned} & \text { } \quad|-B|=5 \\ & \Rightarrow \quad(-1)^3|B|=5 \Rightarrow|B|=-5 \\ & \left|B A^T\right|=15 \Rightarrow|B|\left|A^T\right|=15 \\ & \Rightarrow \quad(-5)|A|=15 \quad\left[\because\left|A^T\right|=|A|\right] \\ & \Rightarrow \quad|A|=-3 \end{aligned} $

$ \begin{array}{ll} \text { Again, } & \left|P^T A P\right|=-27 \\ \Rightarrow & \left|P^T\|A\| P\right|=-27 \\ \Rightarrow & |P||A||P|=-27\,\,\,\,\,\, \left[\because\left|P^T\right|=|P|\right]\\ \Rightarrow & |P|^2(-3)=-27 \Rightarrow|P|^2=9 \end{array} $

$ \therefore \,\,\,\,\,\,\,\,\,\,\,\quad|P|= \pm 3 $

$ \begin{aligned} & \text { Given, }|A|=\frac{1}{2} \\ & \qquad \begin{array}{l} \left|A^{-1}(\operatorname{Adj}(\operatorname{Adj} A))\right|^{-1}=\frac{1}{\left|A^{-1}(\operatorname{Adj}(\operatorname{Adj} A))\right|} \\ =\frac{1}{\left|A^{-1}\right| \cdot|\operatorname{Adj}(\operatorname{Adj} A)|}=\frac{1}{\left|A^{-1}\right| \cdot|\operatorname{Adj} A|^{3-1}} \\ =\frac{1}{\left|A^{-1}\right| \cdot\left(|A|^{3-1}\right)^2}=\frac{1}{\frac{1}{|A|} \cdot|A|^4} \\ =\frac{1}{|A|^3}=\frac{1}{\left(\frac{1}{2}\right)^3}=8 \end{array} \end{aligned} $

$ \begin{aligned} & \text { } \begin{array}{c} \text { } 2 x+3 y-2 z=-4 \\ 3 x-4 y+3 z=-5 \\ \Rightarrow \quad k x-2 y+z=-3 \\ D=\left|\begin{array}{ccc} 2 & 3 & -2 \\ 3 & -4 & 3 \\ k & -2 & 1 \end{array}\right|=k+7 \\ D_x=\left|\begin{array}{ccc} -4 & 3 & -2 \\ -5 & -4 & 3 \\ -3 & -2 & 1 \end{array}\right| \\ =-4(2)-3(4)-2(-2)=-16 \\ \therefore \quad \alpha=D_x / D=-2 \\ \Rightarrow \quad \frac{-16}{k+7}=-2 \Rightarrow k+7=8 \\ \Rightarrow \quad k=1 \text { or }\left|\begin{array}{ll} 3 & 5 \\ 1 & 2 \end{array}\right| \end{array} 1 \end{aligned} $

$ \begin{aligned} & \text { } \frac{x^2+7}{\left(x^2+1\right)(x-2)}=\frac{A}{x-2}+\frac{B x+C}{x^2+1} \\ & \Rightarrow x^2+7=A\left(x^2+1\right)+(B x+C)(x-2) \\ & \text { On putting } x=2, A=11 / 5 \\ & \text { On putting } x=0, C=\frac{-12}{5} \\ & \text { On putting } x=1, B=\frac{-6}{5} \end{aligned} $

$ \begin{aligned} &\text { ∴ The value of the required determinant }\\ &\begin{aligned} & =\left|\begin{array}{cc} \frac{11}{5} & \frac{-6}{5} \\ \frac{-12}{5} & \frac{2}{5} \end{array}\right| \\ & =\frac{22}{25}-\frac{72}{25}=-2 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, trace of } A=-4\\ &\begin{aligned} & \therefore \quad a+(-1)+(-4)=-4 \\ & \Rightarrow \quad a=1 \\ & \text { Now, } A B=\left[\begin{array}{ccc} -1 & 0 & 17 \\ -3 & 10 & 25 \\ 28 & -8 & 3 \end{array}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} {\left[\begin{array}{ccc} 1 & 3 & 5 \\ 5 & -1 & 3 \\ 2 & 3 & -4 \end{array}\right]\left[\begin{array}{ccc} b & 1 & 4 \\ 4 & c & 1 \\ -3 & 1 & d \end{array}\right] } & =\left[\begin{array}{ccc} -1 & 0 & 17 \\ -3 & 10 & 25 \\ 28 & -8 & 3 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ccc} b-3 & 3 c+6 & 7+5 d \\ 5 b-13 & 8-c & 19+3 d \\ 2 b+24 & 3 c-2 & 11-4 d \end{array}\right] & =\left[\begin{array}{ccc} -1 & 0 & 17 \\ -3 & 10 & 25 \\ 28 & -8 & 3 \end{array}\right] \end{aligned}\\ &\text { On comparing, }\\ &\begin{aligned} b-3=-1 \Rightarrow b & =2 \\ 3 c+6=0 \Rightarrow c & =-2 \\ 7+5 d=17 \Rightarrow d & =2 \\ \therefore a+b+c+d=1+2-2+2 & =3 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } \text { Given, }\left|\begin{array}{ccc} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{array}\right| \\ & \begin{array}{r} =\left(b^2 c^3-b^3 c^2\right)-\left(a^2 c^3-a^3 c^2\right) +\left(a^2 b^3-a^3 b^2\right) \\ =\left(a^2 b^3-a^2 c^3\right)+\left(b^2 c^3-a^3 b^2\right) +\left(a^3 c^2-b^3 c^2\right) \\ =a^2\left(b^3-c^3\right)+b^2\left(c^3-a^3\right)+c^2\left(a^3-b^3\right) \end{array} \end{aligned} $

$ \begin{aligned} &\text { Let } \alpha, \beta, \gamma \text { be real numbers, then }\\ &\begin{aligned} & A=\left[\begin{array}{ccc} 7 & 3 & \alpha \\ \beta & 1 & -11 \\ -5 & \gamma & 19 \end{array}\right] \\ & \Rightarrow A\left[\begin{array}{c} 5 \\ -13 \\ 11 \end{array}\right]=\left[\begin{array}{c} -290 \\ -119 \\ 210 \end{array}\right] \\ & \Rightarrow {\left[\begin{array}{ccc} 7 & 3 & \alpha \\ \beta & 1 & -11 \\ -5 & \gamma & 19 \end{array}\right]\left[\begin{array}{c} 5 \\ -13 \\ 11 \end{array}\right]=\left[\begin{array}{c} -290 \\ -119 \\ 210 \end{array}\right] } \\ & 11 \alpha-4=-290 \Rightarrow 11 \alpha=-286 \\ & \Rightarrow \quad \alpha=-265 \beta-134=-119 \\ & \Rightarrow 5 \beta=15 \Rightarrow \beta=3-13 \gamma+184=210 \\ & \Rightarrow-13 \gamma=26 \Rightarrow \gamma=-2 \\ & \quad A=\left[\begin{array}{ccc} 7 & 3 & -26 \\ 3 & 1 & -11 \\ -5 & -2 & 19 \end{array}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} |A| & =7(-3)-3(2)-26(-1) \\ & =-27+26=-1 \\ A^{-1} & =\frac{\operatorname{adj}(A)}{|A|} \\ \Rightarrow \operatorname{adj}(A) & =A^{-1}|A| \end{aligned} $

$ \begin{aligned} (\operatorname{adj} A)^{-1}+\operatorname{adj} A^{-1} & =(\operatorname{adj} A)^{-1}+(\operatorname{adj} A)^{-1} \\ & =2(\operatorname{adj} A)^{-1} \\ =2\left(A^{-1}|A|\right)^{-1} & =2 A \cdot|A|^{-1} \\ =\frac{2 A}{|A|}=\frac{2 A}{-1} & =-2 A \end{aligned} $

Given, $[\alpha, \beta \gamma]\left[\begin{array}{ccc}1 & 2 & 3 \\ 2 & 3 & -5 \\ 1 & 2 & 5\end{array}\right]=\left[\begin{array}{lll}3 & 5 & 2\end{array}\right]$

$ \begin{aligned} & \text { Let } A=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 3 & -5 \\ 1 & 2 & 5 \end{array}\right] \\ & {\left[\begin{array}{lll} \alpha & \beta & \gamma \end{array}\right] \cdot A=\left[\begin{array}{lll} 3 & 5 & 2 \end{array}\right]} \\ & {\left[\begin{array}{lll} \alpha & \beta & \gamma \end{array}\right]=\left[\begin{array}{lll} 3 & 5 & 2 \end{array}\right] A^{-1}} \\ & |A|=1(15+10)-2(10+5)+3(4-3) \\ & |A|=25-30+3=-2 \\ & \operatorname{adj}(A)=\left[\begin{array}{ccc} 25 & -4 & -19 \\ -15 & 2 & 11 \\ 1 & 0 & -1 \end{array}\right] \end{aligned} $

Now, $\left[\begin{array}{lll}\alpha & \beta & \gamma\end{array}\right]=\left[\begin{array}{lll}3 & 5 & 2\end{array}\right] \frac{\operatorname{adj} A}{|A|}$

$ \left[\begin{array}{lll} \alpha & \beta & \gamma \end{array}\right]=\frac{1}{-2}\left[\begin{array}{lll} 3 & 5 & 2 \end{array}\right]\left[\begin{array}{ccc} 25 & -4 & -19 \\ -15 & 2 & 11 \\ 1 & 0 & -1 \end{array}\right] $

$\left[\begin{array}{lll}\alpha & \beta & \gamma\end{array}\right]=-\frac{1}{2}\left[\begin{array}{lll}2 & -2 & -4\end{array}\right]=\left[\begin{array}{lll}-1 & 1 & 2\end{array}\right]$

On comparing $\alpha=-1, \beta=1$ and $\gamma=2$

$ \therefore \quad \alpha^3+\beta^3+\gamma^3=-1+1+8=8 $