Matrices and Determinants

$\det A = \left| {\matrix{ { - 2} & {4 + d} & {\sin \theta - 2} \cr 1 & {\sin \theta + 2} & d \cr 5 & {2\sin \theta - d} & { - \sin \theta + 2 + 2d} \cr } } \right|$

(R₁ $ \to $ R₁ + R₃ $-$ 2R₂)

$ = \left| {\matrix{ 1 & 0 & 0 \cr 1 & {\sin \theta + 2} & d \cr 5 & {2\sin \theta - d} & {2 + 2d - \sin \theta } \cr } } \right|$

= (2 + sin $\theta $) ( 2 + 2d $-$ sin$\theta $) $-$ d(2sin$\theta $ $-$ d)

= 4 + 4d $-$ 2sin$\theta $ + 2sin$\theta $ + 2dsin$\theta $ $-$ sin²$\theta $ $-$ 2dsin$\theta $ + d²

d² + 4d + 4 $-$ sin²$\theta $

= (d + 2)² $-$ sin²$\theta $

For a given d, minimum value of

det(A) = (d + 2)² $-$ 1 = 8

$ \Rightarrow $  d = 1 or $-$ 5

x $-$ $4y + 7z = g$
      $3y$ $-$ $5z = h$
$-$$2x + 5y$ $-$ $9z = k$

$D = \left| {\matrix{ 1 & { - 4} & 7 \cr 0 & 3 & { - 5} \cr { - 2} & 5 & { - 9} \cr } } \right|$

$D = 1\left( { - 27 + 25} \right) - 2\left( {20 - 21} \right)$

$D = - 2 + 2 = 0$

If system is consistent then ${D_1} = {D_2} = {D_3} = 0$

$\left| {\matrix{ 1 & { - 4} & g \cr 0 & 3 & h \cr { - 2} & 5 & k \cr } } \right| = 0$

$1\left( {3k - 5h} \right) - 2\left( { - 4h - 3g} \right) = 0$

$3k - 5h + 8h + 6g = 0$

$6g + 3h + 3k = 0$

$2g + h + k = 0$

$A = \left[ {\matrix{ {{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \cr {{e^t}} & { - {e^{ - t}}\cos t - {e^{ - t}}\sin t} & { - {e^{ - t}}\sin t + {e^{ - t}}co{\mathop{\rm s}\nolimits} t} \cr {{e^t}} & {2{e^{ - t}}\sin t} & { - 2{e^{ - t}}\cos t} \cr } } \right]$

$\left| A \right| = {e^t}.\,{e^{ - t}}.{e^{ - t}}\left| {\matrix{ 1 & {\cos t} & {\sin t} \cr 1 & { - \cos t - \sin t} & { - \sin t + \cos t} \cr 1 & {2\sin t} & { - 2\cos t} \cr } } \right|$

Apply operations R₂ < R₂ $-$R₁, R₃ < R₃ $-$ R₁, R₁ < R₁

$\left| A \right| = {e^{ - t}}\left| {\matrix{ 1 & {\cos t} & {\sin t} \cr 0 & { - \sin t - 2\cos t} & { - 2\sin t + \cos t} \cr 0 & {2\sin t - \cos t} & { - 2\cos t - \sin t} \cr } } \right|$

Open the determinant by R₁

$\left| A \right| = 5{e^{ - t}}$

Invertible for all t $ \in $ R

$D = \left| {\matrix{ 1 & 1 & 1 \cr 2 & 3 & 2 \cr 2 & 3 & {{\alpha ^2} - 1} \cr } } \right|$

D = 3$a$² $-$ 3 $-$ 6 $-$ 2$a$² + 2 + 4 + 2$a$² $-$ 2 $-$ 4

D = ($a$² $-$ 3)

When D $ \ne $ 0 then system of equiation has unique solution.

$ \therefore $  3($a$² $-$ 3) $ \ne $ 0

$ \Rightarrow $  $\left| a \right|$ $ \ne \sqrt 3 $

When $3({a^2} - 3) = 0$

$ \Rightarrow $  $\left| a \right| = \sqrt 3 $ then D = 0

If D = 0 then two cases possible

(1)  System of equation has infinite many solution.

(2) System of equation has no solution and inconsistent.

Here D₁ = $\left| {\matrix{ 2 & 1 & 1 \cr 5 & 3 & 2 \cr {a + 1} & 3 & {{a^2} - 1} \cr } } \right| = {a^2} - a + 1$

D₂ = $\left| {\matrix{ 1 & 2 & 1 \cr 2 & 5 & 2 \cr 2 & {a + 1} & {{a^2} - 1} \cr } } \right| = {a^2} - 3$

D₃ = $\left| {\matrix{ 1 & 1 & 2 \cr 2 & 3 & 5 \cr 2 & 3 & {a + 1} \cr } } \right| = a - 4$

System of equation will have infinite solution if D₁ = D₂ = D₃ = 0.

And system of equation will have no solution if at last one of D₁, D₂, D₂ is non zero.

At $\left| a \right| = \sqrt 3 $ we get D = 0

But D₁ = 3 $ \pm $ $\sqrt 3 + 1$ $ \ne $ 0

and D₃ = $ \pm $ $\sqrt 3 - 4$ $ \ne $ 0

So, system of equations has no solution at $\left| a \right| = \sqrt 3 $ then system is in consistent

(A^$-$50) = (A^$-$1)⁵⁰

We know,

A^$-$1 = ${{adjA} \over {\left| A \right|}}$

$\left| A \right|$ = cos²$\theta $ + sin²$\theta $ = 1

cofactor of A = $\left[ {\matrix{ {\cos \theta } & { - \sin \theta } \cr {\sin \theta } & {\cos \theta } \cr } } \right]$

Adjoint of A = Transpose of cofactor matrix

$ \therefore $  Adj A = $\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$

$ \therefore $  A^$-$1 = $\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$

$ \therefore $  (A^$-$1)² = $\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$

= $\left[ {\matrix{ {\cos 2\theta } & {\sin 2\theta } \cr { - \sin 2\theta } & {\cos 2\theta } \cr } } \right]$

Similarly,

(A^$-$1)³ = $\left[ {\matrix{ {\cos 3\theta } & {\sin 3\theta } \cr { - \sin 3\theta } & {\cos 3\theta } \cr } } \right]$

:

:

:

(A^$-$1)⁵⁰ = $\left[ {\matrix{ {\cos 50\theta } & {\sin 50\theta } \cr { - \sin 50\theta } & {\cos 50\theta } \cr } } \right]$

when $\theta $ = ${\pi \over {12}}$ then

${A^{ - 50}}$ = $\left[ {\matrix{ {\cos {{25\pi } \over 6}} & {\sin {{25\pi } \over 6}} \cr { - \sin {{25\pi } \over 6}} & {\cos {{25\pi } \over 6}} \cr } } \right]$

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

Note:

$\cos {{25\pi } \over 6} = \cos \left( {4\pi + {\pi \over 6}} \right) = \cos {\pi \over 6} = {{\sqrt 3 } \over 2}$

It is given, that matrices

$P = \left[ {\matrix{ 1 & 1 & 1 \cr 0 & 2 & 2 \cr 0 & 0 & 3 \cr } } \right]$, $Q = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 6 \cr } } \right]$

$ \therefore $ ${P^{ - 1}} = {{adj(P)} \over {|P|}}$

as |P| = 6 and adj P = ${\left[ {\matrix{ 6 & 0 & 0 \cr { - 3} & 3 & 0 \cr 0 & { - 2} & 2 \cr } } \right]^T}$

$ \Rightarrow $ ${P^{ - 1}} = {1 \over 6}\left[ {\matrix{ 6 & { - 3} & 0 \cr 0 & 3 & { - 2} \cr 0 & 0 & 2 \cr } } \right]$

$ \therefore $ |R| = |PQP^$-$1| [$ \because $ R = PQP^$-$1 (given)]

$ \Rightarrow $ |R| = |P||Q||P^$-$1| = |Q| [$ \because $ |P| |P^$-$1| = |I| = 1]

$ = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 6 \cr } } \right] = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 5 \cr } } \right] + \left[ {\matrix{ 2 & x & 0 \cr 0 & 4 & 0 \cr x & x & 1 \cr } } \right]$

$ = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 5 \cr } } \right] + 2(4 - 0) - x(0 - 0) + 0(0 - 4x)$

$ = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 5 \cr } } \right] + 8$, for all x $ \in $ R

$ \because $ $PQ = \left[ {\matrix{ 1 & 1 & 1 \cr 0 & 2 & 2 \cr 0 & 0 & 3 \cr } } \right]\left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 6 \cr } } \right] = \left[ {\matrix{ {2 + x} & {4 + 2x} & {x + 6} \cr {2x} & {2x + 8} & {12} \cr {3x} & {3x} & {18} \cr } } \right]$

and $QP = \left[ {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 6 \cr } } \right]\left[ {\matrix{ 1 & 1 & 1 \cr 0 & 2 & 2 \cr 0 & 0 & 3 \cr } } \right] = \left[ {\matrix{ 2 & {2 + 2x} & {2 + 5x} \cr 0 & 8 & 8 \cr x & {3x} & {3x + 18} \cr } } \right]$

There is no common value of 'x', for which each corresponding element of matrices PQ and QP is equal.

For x = 0, $Q = \left[ {\matrix{ 2 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 6 \cr } } \right]$

then, if $R\left[ {\matrix{ 1 \cr a \cr b \cr } } \right] = 6\left[ {\matrix{ 1 \cr a \cr b \cr } } \right]$

$ \Rightarrow $ $PQ{P^{ - 1}}\left[ {\matrix{ 1 \cr a \cr b \cr } } \right] = 6\left[ {\matrix{ 1 \cr a \cr b \cr } } \right]$ [$ \because $ R = PQP^$-$1]

$ \Rightarrow {1 \over 6}\left[ {\matrix{ 1 & 1 & 1 \cr 0 & 2 & 2 \cr 0 & 0 & 3 \cr } } \right]\left[ {\matrix{ 2 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 6 \cr } } \right]\left[ {\matrix{ 6 & { - 3} & 0 \cr 0 & 3 & { - 2} \cr 0 & 0 & 2 \cr } } \right]\left[ {\matrix{ 1 \cr a \cr b \cr } } \right] = 6\left[ {\matrix{ 1 \cr a \cr b \cr } } \right]$

$ \Rightarrow {1 \over 6}\left[ {\matrix{ 2 & 4 & 6 \cr 0 & 8 & {12} \cr 0 & 0 & {18} \cr } } \right]\left[ {\matrix{ 6 & { - 3} & 0 \cr 0 & 3 & { - 2} \cr 0 & 0 & 2 \cr } } \right]\left[ {\matrix{ 1 \cr a \cr b \cr } } \right] = 6\left[ {\matrix{ 1 \cr a \cr b \cr } } \right]$

$ \Rightarrow \left[ {\matrix{ {12} & 6 & 4 \cr 0 & {24} & 8 \cr 0 & 0 & {36} \cr } } \right]\left[ {\matrix{ 1 \cr a \cr b \cr } } \right] = 36\left[ {\matrix{ 1 \cr a \cr b \cr } } \right]$

$ \Rightarrow \left[ {\matrix{ {12} & + & {6a} & + & {4b} \cr 0 & + & {24a} & + & {8b} \cr 0 & + & 0 & + & {36b} \cr } } \right] = \left[ {\matrix{ {36} \cr {36a} \cr {36b} \cr } } \right]$

$ \Rightarrow 6a + 4b = 24$ and $12a = 8b$

$ \Rightarrow 3a + 2b = 12$ and $3a = 2b$

$ \Rightarrow a = 2$ and $b = 3$

So, a + b = 5.

Now, $R\left[ {\matrix{ \alpha \cr \beta \cr \gamma \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$

and $\alpha \widehat i + \beta \widehat j + \gamma \widehat k$ is a unit vector,

so det (R) = 0

$ \Rightarrow $ det (Q) = 0 [$ \because $ R = PQP^$-$1 So, |R| = |Q|]

$ \Rightarrow \left| {\matrix{ 2 & x & x \cr 0 & 4 & 0 \cr x & x & 6 \cr } } \right| = 0$

$ \Rightarrow 2(24 - 0) - x(0 - 0) + x(0 - 4x) = 0 \Rightarrow 48 - 4{x^2} = 0$

$ \Rightarrow {x^2} = 12 \Rightarrow x = \pm 2\sqrt 3 $

So, for x = 1, there does not exist a unit vector $\alpha \widehat i + \beta \widehat j + \gamma \widehat k$, for which $R\left[ {\matrix{ \alpha \cr \beta \cr \gamma \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$

Hence, options (b) and (d) are correct.

${P_1} = I = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right],\,{P_2} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right],\,{P_3} = \left[ {\matrix{ 0 & 1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 \cr } } \right],\,{P_4} = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right],\,{P_5} = \left[ {\matrix{ 0 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr } } \right],\,{P_6} = \left[ {\matrix{ 0 & 0 & 1 \cr 0 & 1 & 0 \cr 1 & 0 & 0 \cr } } \right]$ and $X = \sum\limits_{k = 1}^6 {{P_k}} \left[ {\matrix{ 2 & 1 & 3 \cr 1 & 0 & 2 \cr 3 & 2 & 1 \cr } } \right]P_k^T$

$ \because $ $P_1^T = {P_1},\,P_2^T = {P_2},\,P_3^T = {P_3},\,P_4^T = {P_5},\,P_5^T = {P_4}$ and $P_6^T = {P_6}$

and Let $Q = \left[ {\matrix{ 2 & 1 & 3 \cr 1 & 0 & 2 \cr 3 & 2 & 1 \cr } } \right]$

and $ \because $ Q^T = Q

Now,

$X = ({P_1}QP_1^T) + ({P_2}QP_2^T) + ({P_3}QP_3^T) + ({P_4}QP_4^T) + ({P_5}QP_5^T) + ({P_6}QP_6^T)$

So, ${X^T} = {({P_1}QP_1^T)^T} + {({P_2}QP_2^T)^T} + {({P_3}QP_3^T)^T} + {({P_4}QP_4^T)^T} + {({P_5}QP_5^T)^T} + {({P_6}QP_6^T)^T}$

= ${P_1}QP_1^T$ + ${P_2}QP_2^T$ + ${P_3}QP_3^T$ + ${P_4}QP_4^T$ + ${P_5}QP_5^T$ + ${P_6}QP_6^T$

[$ \because $ (ABC)^T = C^TB^TA^T and (A^T)^T = A and Q^T = Q]

$ \Rightarrow $ X^T = X

$ \Rightarrow $ X is a symmetric matrix.

The sum of diagonal entries of X = Tr(X)

$ = \sum\limits_{i = 1}^6 {{T_r}({P_i}QP_i^T) = \sum\limits_{i = 1}^6 {{T_r}(QP_i^T{P_i})} } $ [$ \because $ T_r(ABC) = T_r(BCA)]

$ = \sum\limits_{i = 1}^6 {{T_r}(QI)} $

[$ \because $ P_i's are orthogonal matrices]

$ = \sum\limits_{i = 1}^6 {{T_r}(Q) = 6{T_r}(Q) = 6 \times 3 = 18} $

Now, Let $R = \left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$, then

$XR = \sum\limits_{K = 1}^6 {({P_K}} QP_K^T)R = \sum\limits_{K = 1}^6 {({P_K}QP_K^TR)} $

$ = \sum\limits_{K = 1}^6 {({P_K}QR)} $ [$ \because $ $P_K^T$R = R]

$ = \sum\limits_{K = 1}^6 {{P_K}\left[ {\matrix{ 6 \cr 3 \cr 6 \cr } } \right]} $

$ = \sum\limits_{K = 1}^6 {{P_K}\left[ {\matrix{ 6 \cr 3 \cr 6 \cr } } \right]} = \left[ {\matrix{ 2 & 2 & 2 \cr 2 & 2 & 2 \cr 2 & 2 & 2 \cr } } \right]\left[ {\matrix{ 6 \cr 3 \cr 6 \cr } } \right]$

$ \Rightarrow XR = \left[ {\matrix{ {30} \cr {30} \cr {30} \cr } } \right] \Rightarrow XR = 30R$

$ \Rightarrow X\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right] = 30\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$

$ \Rightarrow $ (X $-$ 30I) R = 0 $ \Rightarrow $ |X $-$ 30I| = 0

So, (X $-$ 30I) is not invertible and value of $\alpha $ = 30.

Hence, options (a), (b) and (d) are correct.

Given square matrix

$M = \left[ {\matrix{ 0 & 1 & a \cr 1 & 2 & 3 \cr 3 & b & 1 \cr } } \right]$

and adj $(M) = \left[ {\matrix{ { - 1} & 1 & { - 1} \cr 8 & { - 6} & 2 \cr { - 5} & 3 & { - 1} \cr } } \right]$

$ \because $ |adj (M)| = |M|² = $\left[ {\matrix{ { - 1} & 1 & { - 1} \cr 8 & { - 6} & 2 \cr { - 5} & 3 & { - 1} \cr } } \right]$

$ \Rightarrow $ $|M{|^2} = - 1(6 - 6) - 1( - 8 + 10) - 1(24 - 30) = - 2 + 6 = 4$

$ \Rightarrow $ $|M| = \pm 2$

$ \therefore $ det (adj M²) = $|{M^2}{|^2} = |M{|^4} = 16$

As we know A(adj A) = |A| I

$ \Rightarrow $ M = |M| (adj M)^-1 ....(i)

$ \because $ (adj M)^-1 = ${1 \over {|adj\,M|}}{\left[ {\matrix{ 0 & { - 2} & { - 6} \cr { - 2} & { - 4} & { - 2} \cr { - 4} & { - 6} & { - 2} \cr } } \right]^T}$

= ${1 \over 4}\left[ {\matrix{ 0 & { - 2} & { - 4} \cr { - 2} & { - 4} & { - 6} \cr { - 6} & { - 2} & { - 2} \cr } } \right]$

So$\left[ {\matrix{ 0 & 1 & a \cr 1 & 2 & 3 \cr 3 & b & 1 \cr } } \right] = {{|M|} \over 4}\left[ {\matrix{ 0 & { - 2} & { - 4} \cr { - 2} & { - 4} & { - 6} \cr { - 6} & { - 2} & { - 2} \cr } } \right]$

$ \Rightarrow $ M = $\left[ {\matrix{ 0 & 1 & 2 \cr 1 & 2 & 3 \cr 3 & 1 & 1 \cr } } \right]$

Now, If M$\left[ {\matrix{ \alpha \cr \beta \cr \gamma \cr } } \right] = \left[ {\matrix{ 1 \cr 2 \cr 3 \cr } } \right]$

$ \Rightarrow $ $\left[ {\matrix{ 0 & 1 & 2 \cr 1 & 2 & 3 \cr 3 & 1 & 1 \cr } } \right]\left[ {\matrix{ \alpha \cr \beta \cr \gamma \cr } } \right] = \left[ {\matrix{ 1 \cr 2 \cr 3 \cr } } \right]$

$ \Rightarrow $ $\beta $ + 2$\gamma $ = 1, $\alpha $ + 2$\beta $ + 3$\gamma $ = 2 and 3$\alpha $ + $\beta $ + $\gamma $ = 3

$ \Rightarrow $ $\alpha $ = 1, $\beta $ = -1 and $\gamma $ = 1

$ \therefore $ $\alpha $ - $\beta $ + $\gamma $ = 3

And ${(adj\,M)^{ - 1}} + adj\,({M^{ - 1}}) = 2{(adj\,M)^{ - 1}}$

[$ \because $ adj (M^-1) = (adj M)^-1]

= $2\left( { - {M \over 2}} \right) = - M$

[$ \because $ (adj M)-¹ = ${M \over {|M|}}$ from Eq. (i) ]

and $ \because $ a = 2 and b = 1, so a + b = 3

Hence, options (b), (c) and (d) are correct.

It is given that matrix

$M = \left[ {\matrix{ {{{\sin }^4}\theta } \cr {1 + {{\cos }^2}\theta } \cr } \matrix{ { - 1 - {{\sin }^2}\theta } \cr {{{\cos }^4}\theta } \cr } } \right] = \alpha I + \beta {M^{ - 1}}$, where


$\alpha $ = $\alpha $($\theta $) and $\beta $ = $\beta $($\theta $) are real numbers and I is the 2 $ \times $ 2 identity matrix.

Now,

$(M) = \left| M \right| = {\sin ^4}\theta {\cos ^4}\theta + 1 + {\sin ^2}\theta + {\cos ^2}\theta + {\sin ^2}\theta {\cos ^2}\theta$

= ${\sin ^4}\theta {\cos ^4}\theta + {\sin ^2}\theta {\cos ^2}\theta + 2$

and$\left[ {\matrix{ {{{\sin }^4}\theta } \cr {1 + {{\cos }^2}\theta } \cr } \matrix{ { - 1 - {{\sin }^2}\theta } \cr {{{\cos }^4}\theta } \cr } } \right] = \left[ {\matrix{ \alpha \cr 0 \cr } \,\matrix{ 0 \cr \alpha \cr } } \right]$$ + {\beta \over {\left| M \right|}}(adj\,M)$

$ \because $ $\left[ {{M^{ - 1}} = {{adj\,M} \over {\left| M \right|}}} \right]$

$ \Rightarrow $ $\left[ {\matrix{ {{{\sin }^4}\theta } \cr {1 + {{\cos }^2}\theta } \cr } \matrix{ { - 1 - {{\sin }^2}\theta } \cr {{{\cos }^4}\theta } \cr } } \right] = \left[ {\matrix{ \alpha \cr 0 \cr } \,\matrix{ 0 \cr \alpha \cr } } \right] + {\beta \over {\left| M \right|}}\left[ {\matrix{ {{{\cos }^4}\theta } \cr { - 1 - {{\cos }^2}\theta } \cr } \matrix{ {1 + {{\sin }^2}\theta } \cr {{{\sin }^4}\theta } \cr } } \right]$

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

$ \Rightarrow \beta = - \left| M \right|$ and $\alpha = {\sin ^4}\theta + {\cos ^4}\theta $

$ \Rightarrow \alpha = \alpha (\theta ) = 1 - {1 \over 2}{\sin ^2}(2\theta )$, and $\beta = \beta (\theta ) = - \left\{ {{{\left( {{{\sin }^2}\theta {{\cos }^2}\theta + {1 \over 2}} \right)}^2} + {7 \over 4}} \right\} = - \left\{ {{{\left( {{{{{\sin }^2}(2\theta )} \over 4} + {1 \over 2}} \right)}^2} + {7 \over 4}} \right\}$

Now, ${\alpha ^*} = {\alpha _{\min }} = {1 \over 2}$

and ${\beta ^*} = {\beta _{\min }} = - {{37} \over {16}}$

$ \because $ $\alpha $ is minimum at sin²(2$\theta $) = 1 and $\beta $ is minimum at sin²(2$\theta $) = 1

So, ${\alpha ^*} + {\beta ^*} = {1 \over 2} - {{37} \over {16}} = - {{29} \over {16}}$

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

A² = A.A = $\left[ {\matrix{ 1 & 0 & 0 \cr 1 & 1 & 0 \cr 1 & 1 & 1 \cr } } \right] \times \left[ {\matrix{ 1 & 0 & 0 \cr 1 & 1 & 0 \cr 1 & 1 & 1 \cr } } \right]$

=   $\left[ {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 0 \cr 3 & 2 & 1 \cr } } \right]$

A³ = A².A =  $\left[ {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 0 \cr 3 & 2 & 1 \cr } } \right] \times \left[ {\matrix{ 1 & 0 & 0 \cr 1 & 1 & 0 \cr 1 & 1 & 1 \cr } } \right]$

=   $\left[ {\matrix{ 1 & 0 & 0 \cr 3 & 1 & 0 \cr 6 & 3 & 1 \cr } } \right]$

Similarly

A⁴ =   $\left[ {\matrix{ 1 & 0 & 0 \cr 4 & 1 & 0 \cr {10} & 4 & 1 \cr } } \right]$

From this we can say,

Aⁿ =   $\left[ {\matrix{ 1 & 0 & 0 \cr n & 1 & 0 \cr {{{n\left( {n + 1} \right)} \over 2}} & n & 1 \cr } } \right]$

$\therefore\,\,\,$ A²⁰ =   $\left[ {\matrix{ 1 & 0 & 0 \cr {20} & 1 & 0 \cr {210} & {20} & 1 \cr } } \right]$

$\therefore\,\,\,$ Sum of the first column

= 1 + 20 + 210

= 231

System of linear equation have no solution,

$\therefore\,\,\,$ determinant of coefficient = 0

$\left| {\matrix{ {k + 2} & {10} \cr k & {k + 3} \cr } } \right| = 0$

$ \Rightarrow $ $\,\,\,\,$ (k + 2) (k + 3) $-$ 10 K = 0

$ \Rightarrow $ $\,\,\,\,$ k² $-$ 5k + 6 = 0

$\therefore\,\,\,\,$ k = 2, 3

When, k = 2 then equations become,

4x + 10y = 2

and 2x + 5y = 1

It has in finite number of solutions.

When k = 3, equations becomes

5x + 10y = 3

3x + 6y = 2

Those equation has no solutions.

$\therefore\,\,\,\,$ When k = 3, then system of equations have no solutions.

$\left| {\matrix{ {x - 4} & {2x} & {2x} \cr {2x} & {x - 4} & {2x} \cr {2x} & {2x} & {x - 4} \cr } } \right|$

Applying c₁ $ \to $ c₁ + c₂ + c₃

$ = \,\,\,\,\left| {\matrix{ {5x - 4} & {2x} & {2x} \cr {5x - 4} & {x - 4} & {2x} \cr {5x - 4} & {2x} & {x - 4} \cr } } \right|$

Taking common (5x $-$ 4) from c₁

$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{ 1 & {2x} & {2x} \cr 1 & {x - 4} & {2x} \cr 1 & {2x} & {x - 4} \cr } } \right|$

Apply R₂ $ \to $R₂ $-$ R₁ and R₃ $ \to $R₃ $-$ R₁

$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{ 1 & {2x} & {2x} \cr 0 & { - \left( {x + 4} \right)} & 0 \cr 0 & 0 & { - \left( {x + 4} \right)} \cr } } \right|$

$ = \,\,\,\,\left( {5x - 4} \right){\left( {x + 4} \right)^2}$

So, (A + Bx) (x $-$ A)² = (5x $-$ 4) (x + 4)²

By comparing both sides we get, A = $-$ 4 and B = 5

System of equations has non-zero solution when determinant of coefficient = 0.

So, in this questions,

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

$ \Rightarrow \,\,\,\,$ ($-$ 3K + 8) $-$ K ($-$9 + 4) + 3(12 $-$ 2K) = 0

$ \Rightarrow \,\,\,\,$ $-$ 3K + 8 + 9K $-$ 4K + 36 $-$ 6K = 0

$ \Rightarrow \,\,\,\,$ $-$ 4K + 44 = 0

$ \Rightarrow \,\,\,\,$ K = 11

Now the equations become

x + 11y + 3z = 0 . . . (1)

3x + 11y $-$ 2z = 0 . . . (2)

2x + 4y $-$ 3z = 0 . . . (3)

By adding equation (1) and (3) we get,

3x + 15y = 0

$ \Rightarrow \,\,\,\,$ x = $-$ 5y

Putting x = $-$ 5y in equation (1) we get

$-$ 5y + 11y + 3z = 0

$ \Rightarrow \,\,\,\,$ 6y + 3z = 0

$ \Rightarrow \,\,\,\,$ z = $-$ 2y

$\therefore\,\,\,\,$ ${{xz} \over {{y^2}}}$

$ = {{\left( { - 5y} \right)\left( { - 2y} \right)} \over {{y^2}}}$

$ = {{10{y^2}} \over {{y^2}}}$

$ = 10$

As the given system of equations has no solution then

$\Delta $ = 0 and at least one of $\Delta $₁, $\Delta $₂ and $\Delta $₂ should not be zero.

$ \therefore $ $\Delta $ = $\left| {\matrix{ 1 & a & 1 \cr 1 & 2 & 2 \cr 1 & 5 & 3 \cr } } \right| = 0$

$ \Rightarrow $ - $a$ - 1 = 0

$ \Rightarrow $ a = - 1

$\Delta $₂ = $\left| {\matrix{ 1 & 3 & 1 \cr 1 & 6 & 2 \cr 1 & b & 3 \cr } } \right| \ne 0$

$ \Rightarrow $ b $ \ne $ 0

Given,

( A $-$ 3I) (A $-$ 5I) = O

$ \Rightarrow $ A² - 8A + 15I = O

Multiplying both sides by A^{- 1}, we get,

A^{- 1}A.A - 8A^{- 1}A + 15A^{- 1}I = A^{- 1}O

$ \Rightarrow $ A - 8I + 15A^{- 1} = O

$ \Rightarrow $ A + 15A^{- 1} = 8I

$ \Rightarrow $${A \over 2} + {{15{A^{ - 1}}} \over 2} = 4I$

Comparing with the equation $\alpha $A + $\beta $A^-1 = 4I, we get

$\alpha $ = ${1 \over 2}$ and $\beta $ = ${15 \over 2}$

$ \therefore $ $\alpha $ + $\beta $ = ${1 \over 2}$ + ${15 \over 2}$ = ${16 \over 2}$ = 8

According to questions,

A. $\left[ {\matrix{ 1 & 2 \cr 0 & 3 \cr } } \right]$ = $\left[ {\matrix{ \lambda & 0 \cr 0 & \lambda \cr } } \right]$

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

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

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

$ \Rightarrow $ A = $\left[ {\matrix{ \lambda & { - {2 \over 3}\lambda } \cr 0 & {{\lambda \over 3}} \cr } } \right]$

As $\left| {3A} \right|$ = 108

$ \Rightarrow $ 108 = $\left| {\matrix{ {3\lambda } & { - 2\lambda } \cr 0 & \lambda \cr } } \right|$

$ \Rightarrow $ 3$\lambda $² = 108

$ \Rightarrow $ $\lambda $² = 36

$ \Rightarrow $ $\lambda $ = $ \pm $6

When $\lambda $ = +6

then A = $\left[ {\matrix{ 6 & { - 4} \cr 0 & 2 \cr } } \right]$

$ \Rightarrow $ A² = $\left[ {\matrix{ {36} & { - 32} \cr 0 & 4 \cr } } \right]$

For $\lambda $ = -6

A = $\left[ {\matrix{ { - 6} & 4 \cr 0 & { - 2} \cr } } \right]$

$ \Rightarrow $ A² = $\left[ {\matrix{ {36} & { - 32} \cr 0 & 4 \cr } } \right]$

As system of linear equations have unique solutions so, determinant of coefficient $ \ne $ 0

$ \therefore $ $\left| {\matrix{ 1 & 1 & 1 \cr 2 & 1 & { - 1} \cr 3 & 2 & k \cr } } \right|$ $ \ne $ 0

$ \Rightarrow $ k + 2 - (2k + 3) + 1 $ \ne $ 0

$ \Rightarrow $ k $ \ne $ 0

$ \therefore $ k $ \in $ R - {0}

We have,

$\eqalign{ & - x + 2y + 5z = {b_1} \cr & 2x - 4y + 3z = {b_2} \cr & x - 2y + 2z = {b_3} \cr} $

has at least one solution.

$ \therefore $ $D = \left| {\matrix{ { - 1} & 2 & 5 \cr 2 & { - 4} & 3 \cr 1 & { - 2} & 2 \cr } } \right|$

and ${D_1} = {D_2} = {D_3} = 0$

$ \Rightarrow {D_1} = \left| {\matrix{ {{b_1}} & 2 & 5 \cr {{b_2}} & { - 4} & 3 \cr {{b_3}} & { - 2} & 2 \cr } } \right|$

$ = - 2{b_1} - 14{b_2} + 26{b_3} = 0$

$ \Rightarrow {b_1} + 7{b_2} = 13{b_3}$ ....(i)

(a) $ \Rightarrow D = \left| {\matrix{ 1 & 2 & 3 \cr 0 & 4 & 5 \cr 1 & 2 & 6 \cr } } \right| = 1(24 - 10) + 1(10 - 12)$

$ = 14 - 2 = 12 \ne 0$

Here, D $ \ne $ 0 $ \Rightarrow $ unique solution for any b₁, b₂, b₃.

(b) $D = \left| {\matrix{ 1 & 1 & 3 \cr 5 & 2 & 6 \cr { - 2} & { - 1} & { - 3} \cr } } \right|$

$ = 1( - 6 + 6) - 1( - 15 + 12) + 3( - 5 + 4) = 0$

For at least one solution

${D_1} = {D_2} = {D_3} = 0$

Now, ${D_1} = \left| {\matrix{ {{b_1}} & 1 & 3 \cr {{b_2}} & 2 & 6 \cr {{b_3}} & { - 1} & { - 3} \cr } } \right|$

$ = {b_1}( - 6 + 6) - {b_2}( - 3 + 3) + {b_3}(6 - 6) = 0$

${D_2} = \left| {\matrix{ 1 & {{b_1}} & 3 \cr 5 & {{b_2}} & 6 \cr { - 2} & {{b_3}} & { - 3} \cr } } \right|$

$ = {b_1}( - 15 + 12) + {b_2}( - 3 + 6) - {b_3}(6 - 15)$

$ = 3{b_1} + 3{b_2} + 9{b_3} = 0 \Rightarrow {b_1} + {b_2} + 3{b_3} = 0$

not satisfies the Eq. (i)

$ \therefore $ It has no solution.

(c) $D = \left| {\matrix{ { - 1} & 2 & { - 5} \cr 2 & { - 4} & {10} \cr 1 & { - 2} & 5 \cr } } \right|$

= $ - 1( - 20 + 20) - 2(10 - 10) - 5( - 4 + 4) = 0$

Here, b₂ = $-$2b₁ and b₃ = $-$b₁ satisfies the Eq. (i)

Planes are parallel.

(d) $D = \left| {\matrix{ 1 & 2 & 5 \cr 2 & 0 & 3 \cr 1 & 4 & { - 5} \cr } } \right|$

$ = 1(0 - 12) - 2( - 10 - 3) + 5(8 - 0) = 54$

$D \ne 0$

$ \therefore $ It has unique solutino for any b₁, b₂, b₃.

Given, A + B = 2B^T .......(1)

$ \Rightarrow $ (A + B)^T = (2B^T)^T

$ \Rightarrow $ A^T + B^T = 2B

$ \Rightarrow $ B = ${{{A^T} + {B^T}} \over 2}$

Now put this in equation (1)

So, A + ${{{A^T} + {B^T}} \over 2}$ = 2B^T

$ \Rightarrow $2A + A^T = 3B^T

$ \Rightarrow $ A = ${{3{B^T} - {A^T}} \over 2}$

Also, 3A + 2B = I₃ .......(2)

$ \Rightarrow $ $3\left( {{{3{B^T} - {A^T}} \over 2}} \right) + 2\left( {{{{A^T} + {B^T}} \over 2}} \right)$ = I₃

$ \Rightarrow $ 11B^T - A^T = 2I₃

$ \Rightarrow $ (11B^T - A^T)^T = (2I₃)^T

$ \Rightarrow $ 11B - A = 2I₃ ........(3)

Multiply (3) by 3 and then adding (2) and (3) we get,

35B = 7I₃

$ \Rightarrow $ B = ${{{I_3}} \over 5}$

From (3), 11${{{I_3}} \over 5}$ - A = 2I₃

$ \Rightarrow $ A = ${{{I_3}} \over 5}$

$ \therefore $ 5A = 5B = I₃

$ \Rightarrow $ 10A + 5B = 3I₃

The system of equations can be written in the matrix form as

$\left[ {\matrix{ 2 & 4 & { - \lambda } \cr 4 & \lambda & 2 \cr \lambda & 2 & 2 \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$

The system has infinite solutions; thus, we get

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

$ \Rightarrow 0 = 2(2\lambda - 4) - 4(8 - 2\lambda ) - \lambda (8 - {\lambda ^2})$

$ \Rightarrow 4\lambda - 8 - 32 + 8\lambda - 8\lambda + {\lambda ^3} = 0$

$ \Rightarrow {\lambda ^3} + 4\lambda - 40 = 0$

We can solve this by graphical method:

$y = {x^3} + 4x - 40$

For x = 0, y = $-$40: If we take y = $-$40, then we have

$ - 40 = {x^3} + 4x - 40$

$ \Rightarrow {x^3} + 4x = 0$

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

$ \Rightarrow x = 0,{x^2} + 4 = 0$

$ \Rightarrow x = \pm \,2i$

The given equation of line intersects x only at one point; therefore, the real value of $\lambda$ is only one.

We know, the formula

A^-1 = ${{adj\left( A \right)} \over {\left| A \right|}}$

$ \therefore $ adj (A) = $\left| \right.$A$\left| \right.$.A^$-$1

So, Option (A) is true.

We know, the formula

adj (adj (A)) = ${\left| A \right|^{n - 2}}.A$

Now if we put n = 3 as given that A is a 3 $ \times $ 3 matrix, we get

adj (adj (A)) = ${\left| A \right|^{3 - 2}}.A$ = $\left| A \right|.A$

So, Option (B) is also true.

We know, the formula

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

Now if we put n = 3 as given that A is a 3 $ \times $ 3 matrix, we get

adj (adj (A)) = ${\left| A \right|^{3 - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$ = ${\left| A \right|^{2}}{\left( {adj\left( A \right)} \right)^{ - 1}}$

So, Option (C) is also true.

Now in this formula

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

if we put n = 2, we get

adj (adj (A)) = ${\left| A \right|^{2 - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$ = ${\left| A \right|}{\left( {adj\left( A \right)} \right)^{ - 1}}$

But as A is a 3 $ \times $ 3 matrix so we can not take n = 2, so we can say for a 3 $ \times $ 3 matrix option (D) is not true.

So, Option (D) is false.

Given,

        $\left| {\matrix{ 0 & {\cos x} & { - \sin x} \cr {\sin x} & 0 & {\cos x} \cr {\cos x} & {\sin x} & 0 \cr } } \right|$ = 0

$ \Rightarrow $$\,\,\,$ 0 (0 $-$ cosx sinx) $-$ cosx (0 $-$ cos²x) $-$ sinx(sin²x) = 0

$ \Rightarrow $$\,\,\,$ cos³x $-$ sin³ x = 0

$ \Rightarrow $$\,\,\,$ tan³x = 1

$ \Rightarrow $$\,\,\,$ tanx = 1

$ \therefore $ $\,\,\,$ $\sum\limits_{x\, \in \,\,S} {\,\tan \left( {{\pi \over 3} + x} \right)} $

= ${{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}$

= ${{\sqrt 3 + 1} \over {1 - \sqrt 3 }}$

= ${{\left( {\sqrt 3 + 1} \right)} \over {\left( {1 - \sqrt 3 } \right)}} \times {{1 + \sqrt 3 } \over {1 + \sqrt 3 }}$

= ${{1 + 3 + 2\sqrt 3 } \over { - 2}}$

= $-$ 2 $-$ $\sqrt 3 $

$\left| {\matrix{ 1 & 1 & 1 \cr 1 & a & 1 \cr a & b & 1 \cr } } \right| = 0$

$ \Rightarrow $ 1 [a – b] – 1 [1 – a] + 1 [b – a²] = 0

$ \Rightarrow $ (a - 1)² = 0

$ \Rightarrow $ a = 1

For a = 1, the equations become

x + y + z = 1

x + y + z = 1

x + by + z = 0

These equations give no solution for b = 1

$ \Rightarrow $ S is singleton set.

We have, $A = \left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]$

$ \therefore $ A² = A.A = $\left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]\left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]$

= $\left[ {\matrix{ {4 + 12} & { - 6 - 3} \cr { - 8 - 4} & {12 + 1} \cr } } \right]$

= $\left[ {\matrix{ {16} & { - 9} \cr { - 12} & {13} \cr } } \right]$

Now, 3A² + 12A

= $3\left[ {\matrix{ {16} & { - 9} \cr { - 12} & {13} \cr } } \right] + 12\left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]$

= $\left[ {\matrix{ {48} & { - 27} \cr { - 36} & {39} \cr } } \right] + \left[ {\matrix{ {24} & { - 36} \cr { - 48} & {12} \cr } } \right]$

= $\left[ {\matrix{ {72} & { - 63} \cr { - 84} & {51} \cr } } \right]$

$ \therefore $ adj(3A² + 12A) = $\left[ {\matrix{ {51} & {63} \cr {84} & {72} \cr } } \right]$

For a matrix to be square of matrix with real entries, its determinant should be positive.

Option (A) : $\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & { - 1} \cr } } \right]$, determinant is NOT possible:

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

Option (B) : $\left[ {\matrix{ 1 & 0 & 0 \cr 0 & { - 1} & 0 \cr 0 & 0 & { - 1} \cr } } \right]$, determinant is possible:

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

Option (C) : $\left[ {\matrix{ { - 1} & 0 & 0 \cr 0 & { - 1} & 0 \cr 0 & 0 & { - 1} \cr } } \right]$, determinant is NOT possible :

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

Option (D) : $\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$, determinant is possible:

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

Thus, options (A) and (C) are NOT the square of a 3 $\times$ 3 matrix with real entries.

Sum of diagonal entries of M^TM is $\sum\limits_{}^{} {a_i^2} $

$\sum\limits_{i = 1}^9 {a_i^2} = 5$

Possibilities

I. 2, 1, 0, 0, 0, 0, 0, 0, 0, which gives ${{9!} \over {7!}}$ matrices

II. 1, 1, 1, 1, 1, 0, 0, 0, 0, which gives ${{9!} \over {4!\, \times 5!}}$ matrices

Total matrices = 9 $ \times $ 8 + 9 $ \times $ 7 $ \times $ 2 = 198

Given,

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

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

$ = \left[ {\matrix{ {13} & 3 \cr { - 9} & { - 2} \cr } } \right]$

A² $-$ 2A $-$ I

$ = \left[ {\matrix{ {13} & 3 \cr { - 9} & { - 2} \cr } } \right] - \left[ {\matrix{ { - 8} & { - 2} \cr 6 & 2 \cr } } \right] - \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right]$

$ = \left[ {\matrix{ {20} & 5 \cr { - 15} & { - 5} \cr } } \right]$

$\left| A \right| = \left| {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right|$ $=$ $-$ 4 + 3 $=$ $-$ 1

Now,

$\left| {{A^{2016}} - 2{A^{2015}} - {A^{2014}}} \right|$

$=$ ${\left| A \right|^{2014}}\left| {{A^2} - 2A - {\rm I}} \right|$

$ = {\left( { - 1} \right)^{2014}}\left| {\matrix{ {20} & 5 \cr { - 15} & { - 5} \cr } } \right|$

$=$ 1 $ \times $ ($-$ 100 + 75)

$=$ $-$ 25

Given,

A² $-$ 5A + 7I = 0

$ \Rightarrow $   A² $-$ 5A = $-$ 7I

$ \Rightarrow $   AAA^$-$1 $-$ 5AA^$-$1 = $-$ 7IA^$-$1

$ \Rightarrow $   AI $-$ 5I = $-$ 7A^$-$1

$ \Rightarrow $   A $-$ 5I = $-$ 7A^$-$1

$ \Rightarrow $   A^$-$1 = ${1 \over 7}$(5I $-$ A)

Hence, statement 1 is true.

Now A³ $-$ 2A² $-$ 3A + I

=   A(A²) $-$ 2A² $-$ 3A + I

=   A(5A $-$ 7I) $-$ 2A² $-$ 3A + I

=   5A² $-$ 7A $-$ 2A² $-$ 3A + I

=   3A² $-$ 10A + I

=   3(5A $-$ 7I) $-$ 10A + I

=   15A $-$ 21A $-$ 10A + I

=   5A $-$ 20I

=   5(A $-$ 4I)

So, statement 2 is also correct.

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

$ \therefore $   P^T = $\left[ {\matrix{ {{{\sqrt 3 } \over 2}} & { - {1 \over 2}} \cr {{1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr  } } \right]$

As    PP^T = P^TP = I

given, Q = PAP^T

$ \therefore $   P^TQ = P^TP AP^T

$ \Rightarrow $   P^TQ = IAP^T = AP^T [ as    P^TP = I]

Now,

P^T  Q²⁰¹⁵  P

=  P^TQ   .   Q²⁰¹⁴  .  P

=  AP^T  Q²⁰¹⁴  P

=  AP^T  .  Q  .  Q²⁰¹³  .  P

=  A²P^T  .  Q²⁰¹³  .  P

.

.

.
=  A²⁰¹⁴  .  P^TQP

=  A²⁰¹⁴   .   AP^TP

=  A²⁰¹⁵

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

$ \therefore $  A² =   $\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]$   =  $\left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right]$

A³  =   $\left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right]$ $\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]$  =  $\left[ {\matrix{ 1 & 3 \cr 0 & 1 \cr } } \right]$

A²⁰¹⁵  =  $\left[ {\matrix{ 1 & {2015} \cr 0 & 1 \cr } } \right]$

Given,

$\left| {\matrix{ {\cos x} & {\sin x} & {\sin x} \cr {\sin x} & {\cos x} & {\sin x} \cr {\sin x} & {\sin x} & {\cos x} \cr } } \right| = 0$

R₁  $ \to $  R₁  $-$  R₃

R₁  $ \to $  R₂  $-$  R₃

$\left| {\matrix{ {\cos x - \sin x} & 0 & {\sin x - \cos x} \cr 0 & {\cos x - \sin x} & {\sin x - \cos x} \cr {\sin x} & {\sin x} & { \cos x} \cr } } \right| = 0$

C₃  $ \to $  C₃  +  C₂

$\left| {\matrix{ {\cos x - \sin x} & 0 & {\sin x - \cos x} \cr 0 & {\cos x - \sin x} & 0 \cr {\sin x} & {\sin x} & {\sin x + \cos x} \cr } } \right| = 0$

Expanding using first column,

(cosx $-$ sinx)(cos $-$ sinx) (sinx + cos x)

      + sinx (cosx $-$ sinx) (sinx $-$ cosx) = 0

$ \Rightarrow $   (cosx $-$ sinx)² (sinx + cosx)

     $+$ sinx (cosx $-$ sinx)² = 0

$ \Rightarrow $    (cosx $-$ sinx)² (sinx + cosx $+$ sinx) = 0

$ \Rightarrow $    (2sinx + cosx )(cosx $-$ sinx)² = 0

$ \therefore $   cosx = -2sinx   or  cosx   =  sinx

$ \Rightarrow $ tanx = $ - {1 \over 2}$ or tanx = 1

$ \therefore $ x = $ - {\tan ^{ - 1}}\left( {{1 \over 2}} \right)$, ${\pi \over 4}$

$ \therefore $   Number of solutions   =  2

For non-trivial solution, we have

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

$ \Rightarrow 1(\lambda + 1) - \lambda ( - {\lambda ^2} + 1) - 1(\lambda + 1) = 0$

$ \Rightarrow \lambda ({\lambda ^2} - 1) = 0$

$ \Rightarrow \lambda = - 1,0,1$

$A\left( {Adj\,\,A} \right) = A\,{A^T}$

$ \Rightarrow {A^{ - 1}}A\left( {adj\,\,A} \right) = {A^{ - 1}}A\,{A^T}$

$Adj\,\,A = {A^T}$

$ \Rightarrow \left[ {\matrix{ 2 & b \cr { - 3} & {5a} \cr } } \right] = \left[ {\matrix{ {5a} & 3 \cr { - b} & 2 \cr } } \right]$

$ \Rightarrow a = {2 \over 5}\,\,$ and $\,\,b = 3$

$ \Rightarrow 5a + b = 5$

The given system of linear equation is

ax + 2y = $\lambda$

3x $-$ 2y = $\mu$

By Cramer's rule, we have

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

${\Delta _1} = \left| {\matrix{ \lambda & 2 \cr \mu & { - 2} \cr } } \right| = - 2\lambda - 2\mu = - 2(\lambda + \mu )$

${\Delta _2} = \left| {\matrix{ a & \lambda \cr 3 & \mu \cr } } \right| = (a\mu - 3\lambda )$

For unique solution: $\Delta$ $\ne$ 0 $\Rightarrow$ a + 3 $\ne$ 0 $\Rightarrow$ a $\ne$ $-$3, where $\lambda$ and $\mu$ can take any values.

Hence, option (B) is correct.

For infinite solution: $\Delta$ = 0, $\Delta$₁ and $\Delta$₂ are zero.

That is, a = $-$3 and $\lambda$ + $\mu$ = 0 or a$\mu$ $-$ 3$\lambda$ = 0.

Hence, options (C) are correct.

For no solution: $\Delta$ = 0 and either $\Delta$₁ or $\Delta$₂ is non-zero.

That is, a = $-$3 and $\lambda$ + $\mu$ $\ne$ 0.

Hence, option (D) is correct.

$P = \left[ {\matrix{ 3 & { - 1} & { - 2} \cr 2 & 0 & \alpha \cr 3 & { - 5} & 0 \cr } } \right]$

Now, $\left| P \right| = 3(5\alpha ) + 1( - 3\alpha ) - 2( - 10)$

$ = 12\alpha + 20$ ....... (i)

$\therefore$ $(P) = {\left[ {\matrix{ {5\alpha } & {2\alpha } & { - 10} \cr { - 10} & 6 & {12} \cr { - \alpha } & { - (3\alpha + 4)} & 2 \cr } } \right]^T}$

$ = \left[ {\matrix{ {5\alpha } & { - 10} & { - \alpha } \cr {2\alpha } & 6 & { - 3\alpha - 4} \cr { - 10} & {12} & 2 \cr } } \right]$ ...... (ii)

As, $PQ = kI$

$ \Rightarrow \left| P \right|\left| Q \right| = \left| {kI} \right|$

$ \Rightarrow \left| P \right|\left| Q \right| = {k^3}$

$ \Rightarrow \left| P \right|\left( {{{{k^2}} \over 2}} \right) = {k^3}$ [given, $\left| Q \right| = {{{k^2}} \over 2}$]

$ \Rightarrow \left| P \right| = 2k$ ......(iii)

$\because$ $PQ = ki$

$\therefore$ $Q = k{p^{ - 1}}I$

$ = k\,.\,{{adj\,P} \over {\left| P \right|}} = {{k(adj\,P)} \over {2k}}$ [from Eq. (iii)]

$ = {{adj\,P} \over 2}$

$ = {1 \over 2}\left[ {\matrix{ {5\alpha } & { - 10} & { - \alpha } \cr {2\alpha } & 6 & { - 3\alpha - 4} \cr { - 10} & {12} & 2 \cr } } \right]$

$\therefore$ ${q_{23}} = {{ - 3\alpha - 4} \over 2}$ [given, ${q_{23}} = - {k \over 8}$]

$ \Rightarrow - {{(3\alpha + 4)} \over 2} = - {k \over 8}$

$ \Rightarrow (3\alpha + 4) \times 4 = k$

$ \Rightarrow 12\alpha + 16 = k$ ...... (iv)

From Eq. (iii), $\left| P \right| = 2k$

$ \Rightarrow 12\alpha + 20 = 2k$ [from Eq. (i)] ...... (v)

On solving Eqs. (iv) and (v), we get

$\alpha$ = $-$1 and k = 4 ....... (vi)

$\therefore$ $4\alpha - k + 8 = - 4 - 4 + 8 = 0$

$\therefore$ Option (b) is correct.

Now, $\left| {P\,adj(Q)} \right| = \left| P \right|\left| {adj\,Q} \right|$

$ = 2k{\left( {{{{k^2}} \over 2}} \right)^2} = {{{k^5}} \over 2} = {{{2^{10}}} \over 2} = {2^9}$

$\therefore$ Option (c) is correct.

Here, $P = \left[ {\matrix{ 1 & 0 & 0 \cr 4 & 1 & 0 \cr {16} & 4 & 1 \cr } } \right]$

$\therefore$ ${P^2} = \left[ {\matrix{ 1 & 0 & 0 \cr 4 & 1 & 0 \cr {16} & 4 & 1 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 4 & 1 & 0 \cr {16} & 4 & 1 \cr } } \right]$

$ = \left[ {\matrix{ 1 & 0 & 0 \cr {4 + 4} & 1 & 0 \cr {16 + 32} & {4 + 4} & 1 \cr } } \right]$

$ = \left[ {\matrix{ 1 & 0 & 0 \cr {4 \times 2} & 1 & 0 \cr {16(1 + 2)} & {4 \times 2} & 1 \cr } } \right]$ ...... (i)

and ${P^3} = \left[ {\matrix{ 1 & 0 & 0 \cr {4 \times 2} & 1 & 0 \cr {16(1 + 2)} & {4 \times 2} & 1 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 4 & 1 & 0 \cr {16} & 4 & 1 \cr } } \right]$

$ = \left[ {\matrix{ 1 & 0 & 0 \cr {4 \times 3} & 1 & 0 \cr {16(1 + 2 + 3)} & {4 \times 3} & 1 \cr } } \right]$ ..... (ii)

From symmetry,

${P^{50}} = \left[ {\matrix{ 1 & 0 & 0 \cr {4 \times 50} & 1 & 0 \cr {16(1 + 2 + 3 + .... + 50)} & {4 \times 50} & 1 \cr } } \right]$

$\because$ ${P^{50}} - Q = I$ [given]

$\therefore$ $\left[ {\matrix{ {1 - {q_{11}}} & { - {q_{12}}} & { - {q_{13}}} \cr {200 - {q_{21}}} & {1 - {q_{22}}} & { - {q_{23}}} \cr {16 \times {{50} \over 2}(51) - {q_{31}}} & {200 - {q_{32}}} & {1 - {q_{33}}} \cr } } \right]$

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

$ \Rightarrow 200 - {q_{21}} = 0$, ${{16 \times 50 \times 51} \over 2} - {q_{31}} = 0$, $200 - {q_{32}} = 0$

$\therefore$ ${q_{21}} = 200$, ${q_{32}} = 200$, ${q_{31}} = 20400$

Thus, ${{{q_{31}} + {q_{32}}} \over {{q_{21}}}} = {{20400 + 200} \over {200}}$

$ = {{20600} \over {200}} = 103$

$\left[ {\matrix{ 1 & 2 & 2 \cr 2 & 1 & { - 2} \cr a & 2 & b \cr } } \right]\left[ {\matrix{ 1 & 2 & a \cr 2 & 1 & 2 \cr 2 & { - 2} & b \cr } } \right] = \left[ {\matrix{ 9 & 0 & 0 \cr 0 & 9 & 0 \cr 0 & 0 & 9 \cr } } \right]$

$ \Rightarrow \left[ {\matrix{ {1 + 4 + 4} & {2 + 2 - 4} & {a + 4 + 2b} \cr {2 + 2 - 4} & {4 + 1 + 4} & {2a + 2 - 2b} \cr {a + 4 + 2b} & {2a + 2 - 2b} & {{a^2} + 4 + {b^2}} \cr } } \right]$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {\matrix{ 9 & 0 & 0 \cr 0 & 9 & 0 \cr 0 & 0 & 9 \cr } } \right]$

$ \Rightarrow a + 4 + 2b = 0$ $ \Rightarrow a + 2b = - 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

$2a + 2 - 2b = 0 \Rightarrow 2a - 2b = - 2$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a - b = - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

On solving $(i)$ and $(ii)$ we get

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1 + b + 2b = - 4\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$b=-1$ and $a=-2$

$\left( {a,b} \right) = \left( { - 2, - 1} \right)$

$\left. {\matrix{ {2{x_1} - 2{x_2} + {x^3} = \lambda {x_1}} \cr {2{x_1} - 3{x_2} + 2{x_3} = \lambda {x_2}} \cr {\,\,\,\,\,\,\,\,\,\, - {x_1} + 2{x_2} = \lambda {x_3}} \cr } } \right\}$

$\eqalign{ & \Rightarrow \,\,\,\,\,\,\,\left( {2 - \lambda } \right){x_1} - 2{x_2} + {x_3} = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,2{x_1} - \left( {3 + \lambda } \right){x_2} + 2{x_3} = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - {x_1} + 2{x_2} - \lambda {x_3} = 0 \cr} $

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

i.e. $\,\,\,\left| {\matrix{ {2 - \lambda } & { - 2} & 1 \cr 2 & { - \left( {3 + \lambda } \right)} & 2 \cr { - 1} & 2 & { - \lambda } \cr } } \right| = 0$

$ \Rightarrow \left( {2 - \lambda } \right)\left[ {\lambda \left( {3 + \lambda } \right) - 4} \right] + $

$\,\,\,\,\,\,\,\,\,2\left[ { - 2\lambda + 2} \right] + 1\left[ {4 - \left( {3 + \lambda } \right)} \right] = 0$

$ \Rightarrow {\lambda ^3} + {\lambda ^2} - 5\lambda + 3 = 0$

$ \Rightarrow \lambda = 1,1,3$

Hence, $\lambda $ has $2$ values.

Given,

${X^T} = - X,{Y^T} = - Y,{Z^T} = Z$

(a) Let $P = {Y^3}{Z^4} - {Z^4}{Y^3}$

Then, ${P^T} = {({Y^3}{Z^4})^T} - {({Z^4}{Y^3})^T}$

$ = {({Z^T})^4}{({Y^T})^3} - {({Y^T})^3}{({Z^T})^4}$

$ = - {Z^4}{Y^3} + {Y^3}{Z^4} = P$

$\therefore$ P is symmetric matrix.

(b) Let $P = {X^{44}} + {Y^{44}}$

Then, ${P^T} = {({X^T})^{44}} + {({Y^T})^{44}}$

$ = {X^{44}} + {Y^{44}} = P$

$\therefore$ P is symmetric matrix.

(c) Let $P = {X^4}{Z^3} - {Z^3}{X^4}$

Then, ${P^T} = {({X^4}{Z^3})^T} - {({Z^3}{X^4})^T}$

$ = {({Z^T})^3}{({X^T})^4} - {({X^T})^4}{({Z^T})^3}$

$ = {Z^3}{X^4} - {X^4}{Z^3}$

$ = - P$

$\therefore$ P is skew-symmetric matrix.

(d) Let $P = {X^{23}} + {Y^{23}}$

Then, ${P^T} = {({X^T})^{23}} + {({Y^T})^{23}}$

$ = - {X^{23}} - {Y^{23}}$

$ = - P$

$\therefore$ P is skew-symmetric matrix.

$\left| {\matrix{ {{{(1 - \alpha )}^2}} & {{{(1 + 2\alpha )}^2}} & {{{(1 + 3\alpha )}^2}} \cr {{{(2 + \alpha )}^2}} & {{{(2 + 2\alpha )}^2}} & {{{(2 + 3\alpha )}^2}} \cr {{{(3 + \alpha )}^2}} & {{{(3 + 2\alpha )}^2}} & {{{(3 + 3\alpha )}^2}} \cr } } \right| = - 648\alpha $

Applying ${R_3} \to {R_3} - {R_2},{R_2} \to {R_2},{R_1}$, we get

$\left| {\matrix{ {{{(1 - \alpha )}^2}} & {{{(1 + 2\alpha )}^2}} & {{{(1 + 3\alpha )}^2}} \cr {3 + 2\alpha } & {3 + 4\alpha } & {3 + 6\alpha } \cr {5 + 2\alpha } & {5 + 4\alpha } & {5 + 6\alpha } \cr } } \right| = - 648\alpha $

Applying ${R_3} \to {R_3} - {R_2}$, we get

$\left| {\matrix{ {{{(1 - \alpha )}^2}} & {{{(1 + 2\alpha )}^2}} & {{{(1 + 3\alpha )}^2}} \cr {3 + 2\alpha } & {3 + 4\alpha } & {3 + 6\alpha } \cr 2 & 2 & 2 \cr } } \right| = - 648\alpha $

Applying ${C_3} \to {C_3} - {C_2},{C_2} \to {C_2} - {C_1}$, we get

$\left| {\matrix{ {{{(1 - \alpha )}^2}} & {{{(1 + 2\alpha )}^2}} & {{{(1 + 3\alpha )}^2}} \cr {3 + 2\alpha } & {2\alpha } & {2\alpha } \cr 2 & 0 & 0 \cr } } \right| = - 648\alpha $

Expanding, $2{\alpha ^2}(3\alpha + 2) - 2{\alpha ^2}(5\alpha + 2) = - 324\alpha $

$ \Rightarrow - 4{\alpha ^3} = - 324\alpha \Rightarrow \alpha ({\alpha ^2} - 81) = 0$

$\therefore$ $\alpha = 0, - 9,9$

$BB' = B\left( {{A^{ - 1}}A'} \right)' = B\left( {A'} \right)'\left( {{A^{ - 1}}} \right)' = BA\left( {{A^{ - 1}}} \right)'$

$ = \left( {{A^{ - 1}}A'} \right)\left( {A\left( {{A^{ - 1}}} \right)'} \right)$

$ = {A^{ - 1}}A.A'.\left( {{A^{ - 1}}} \right)'\,\,\,\,\,\,$ $\left\{ {} \right.$ as $\,\,\,\,\,\,$ $AA' = A'A$ $\left. \, \right\}$

$ = I\left( {{A^{ - 1}}A} \right)'$

$ = I.I = {I^2} = I$

Consider

$\left| {\matrix{ 3 & {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} \cr {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} \cr {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} & {1 + f\left( 4 \right)} \cr } } \right|$

$\,\,\,\,\,\,\,\,\,\,\, = \left| {\matrix{ {1 + 1 + 1} & {1 + \alpha + \beta } & {1 + {\alpha ^2} + {\beta ^2}} \cr {1 + \alpha + \beta } & {1 + {\alpha ^2} + {\beta ^2}} & {1 + {\alpha ^3} + {\beta ^3}} \cr {1 + {\alpha ^2} + {\beta ^2}} & {1 + {\alpha ^3} + {\beta ^3}} & {1 + {\alpha ^4} + {\beta ^4}} \cr } } \right|$

$\,\,\,\,\,\,\,\,\,\,\, = \left| {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & \beta \cr 1 & {{\alpha ^2}} & {{\beta ^2}} \cr } } \right| \times \left| {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & \beta \cr 1 & {{\alpha ^2}} & {{\beta ^2}} \cr } } \right|$

$\,\,\,\,\,\,\,\,\,\,\, = {\left| {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & \beta \cr 1 & {{\alpha ^2}} & {{\beta ^2}} \cr } } \right|^2}$

$\,\,\,\,\,\,\,\,\,\,\, = {\left[ {\left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {\alpha - \beta } \right)} \right]^2}$

So, $k=1$

Note : A square matrix M is invertible if det(M) or | M | $\ne$ 0.

Let $M = \left[ {\matrix{ a & b \cr b & c \cr } } \right]$

(a) Given that $\left[ {\matrix{ a \cr b \cr } } \right] = \left[ {\matrix{ b \cr c \cr } } \right]$

$ \Rightarrow a = b = c = \alpha $ (let)

$ \Rightarrow M = \left[ {\matrix{ \alpha & \alpha \cr \alpha & \alpha \cr } } \right]$

$\Rightarrow$ | M | = 0

$\Rightarrow$ M is non-invertible.

(b) Given that [bc] = [ab]

$ \Rightarrow a = b = c = \alpha $ (let)

Again | M | = 0

$\Rightarrow$ M is non-invertible.

(c) As given $M = \left[ {\matrix{ a & 0 \cr 0 & c \cr } } \right]$

$ \Rightarrow \left| M \right| = ac \ne 0$ ($\because$ a and c are non-zero)

$\Rightarrow$ M is invertible.

(d) $M = \left[ {\matrix{ a & b \cr b & c \cr } } \right] \Rightarrow \left| M \right| = ac - {b^2} \ne 0$

$\because$ ac is not equal to square of an integer.

$\therefore$ M is invertible.

Given MN = NM. Therefore, a² $-$ b² = (a + b) (a $-$ b) of algebra of numbers is applicable.

Now, M² = N⁴ $\Rightarrow$ M² $-$ N⁴ = 0 (Null matrix)

$\Rightarrow$ (M + N²) (M $-$ N²) = 0

Since M $\ne$ N² (given), the possibilities are

(M + N²) = 0 and M $-$ N² $\ne$ 0 (1)

or (M + N²) $\ne$ 0 and M + N² $\ne$ 0 (2)

Now, we know if A and B are non-null square matrix nd AB = 0, then A and B both are singular, that is, | A | = 0 and | B | = 0 and AB = 0.

Note : For example, let A be non-singular. Therefore, B = In(2) = A^$-$1 AB = 0 (since AB = 0 is assumed).

Hence, B is singular, which is a contradiction and A has to be singular. Similarly, B also has to be singular.

Therefore, from Eqs. (1) and (2), we conclude the only possibility is | M + N² | = 0.

Now checking options :

(A) | M² + MN² | = | M || M + N² | = 0

Hence, option (A) is correct.

(B) (M² + MN²) U = 0

Since, M² + MN² is singular. Therefore, U has infinitely many possible values (non-trivial solutions). Hence, option (B) is true.

(C) False since | M² + MN² | = 0.

(D) Since | M² + MN² | = 0, U is not a necessarily a zero matrix.

From the given system, we have

${{k + 1} \over k} = {8 \over {k + 3}} \ne {{4k} \over {3k - 1}}$

( as System has no solution)

$ \Rightarrow {k^2} + 4k + 3 = 8k$

$ \Rightarrow k = 1,3$

If $k = 1$ then ${8 \over {1 + 3}} \ne {{4.1} \over 2}$ which is false

And if $k = 3$

Then ${8 \over 6} \ne {{4.3} \over {9 - 1}}$ which is true, therefore $k=3$

Hence for only one value of $k.$ System has no solution.

$\left| P \right| = 1\left( {12 - 12} \right) - \alpha \left( {4 - 6} \right) + $

$\,\,\,\,\,\,\,\,\,\,\,3\left( {4 - 6} \right) = 2\alpha - 6$

Now, $adj\,\,A = P\,$ $\,\,\,\,\,\,\,\, \Rightarrow \left| {adj\,A} \right| = \left| P \right|$

$ \Rightarrow {\left| A \right|^2} = \left| P \right|$

$ \Rightarrow \left| P \right| = 16$

$ \Rightarrow 2\alpha - 6 = 16$

$ \Rightarrow \alpha = 11$