Matrices and Determinants

$|A| = 2$

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

$ = |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 $\alpha = 1 + 1 = 2$

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$

Given $\alpha$ + $\beta$ + $\gamma$ = 2$\pi$

$\Delta = \left| {\matrix{ 1 & {\cos \gamma } & {\cos \beta } \cr {\cos \gamma } & 1 & {\cos \alpha } \cr {\cos \beta } & {\cos \alpha } & 1 \cr } } \right|$

$ = 1 - {\cos ^2}\alpha - \cos \gamma (\cos \gamma - \cos \alpha \cos \beta ) + \cos \beta (\cos \alpha \cos \gamma - \cos \beta )$

$ = 1 - {\cos ^2}\alpha - {\cos ^2}\beta - {\cos ^2}\gamma + 2\cos \alpha \cos \beta \cos \gamma $

$ = {\sin ^2}\alpha - {\cos ^2}\beta - \cos \gamma (\cos \gamma - 2\cos \alpha \cos \beta )$

$ = - \cos (\alpha + \beta )\cos (\alpha - \beta ) - \cos \gamma (\cos (2\pi - (\alpha - \beta )) - 2\cos \alpha \cos \beta )$

$ = - \cos (2\pi - \gamma )\cos (\alpha - \beta ) - \cos \gamma (\cos (\alpha + \beta ) - 2\cos \alpha \cos \beta )$

$ = - \cos \gamma \cos (\alpha - \beta ) + \cos \gamma (\cos \alpha \cos \beta + \sin \alpha \sin \beta )$

$ = - \cos \gamma \cos (\alpha - \beta ) + \cos \gamma \cos (\alpha - \beta )$

$ = 0$

So, the system of equation has infinitely many solutions.

Given, system of linear equations

4x + $\lambda$y + 2z = 0

2x $-$ y + z = 0

$\mu$x + 2y + 3z = 0

For non-trivial solution, $\Delta$ = 0

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

$ \Rightarrow 4( - 3 - 2) - \lambda (6 - \mu ) + 2(4 + \mu ) = 0$

$ \Rightarrow - \lambda (6 - \mu ) - 2(6 - \mu ) = 0$

$ \Rightarrow (6 - \mu )(\lambda + 2) = 0$

$ \Rightarrow \lambda = - 2$ and $\mu \in R$ or $\mu$ = 6 and $\lambda \in R$.