Matrices and Determinants

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$

The given matrix $ P = [p_{ij}] $ is an $ n \times n $ matrix where $ p_{ij} = \omega^{i + j} $ and $ \omega $ is a complex cube root of unity with $\omega \neq 1$. We need to determine when $ P^2 \neq 0 $.

For $ n = 1 $:

$ P = [p_{ij}]_{1 \times 1} = [\omega^2] $

$ \Rightarrow P^2 = [\omega^4] \neq 0 $

For $ n = 2 $:

$ P = [p_{ij}]_{2 \times 2} = \begin{bmatrix} \omega^2 & \omega^3 \\ \omega^3 & \omega^4 \end{bmatrix} = \begin{bmatrix} \omega^2 & 1 \\ 1 & \omega \end{bmatrix} $

$ P^2 = \begin{bmatrix} \omega^2 & 1 \\ 1 & \omega \end{bmatrix} \begin{bmatrix} \omega^2 & 1 \\ 1 & \omega \end{bmatrix} = \begin{bmatrix} \omega^4 + 1 & \omega^2 + \omega \\ \omega^2 + \omega & 1 + \omega^2 \end{bmatrix} \neq 0 $

For $ n = 3 $:

$ P = [p_{ij}]_{3 \times 3} = \begin{bmatrix} \omega^2 & \omega^3 & \omega^4 \\ 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \end{bmatrix} = \begin{bmatrix} \omega^2 & 1 & \omega \\ 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \end{bmatrix} $

$ P^2 = \begin{bmatrix} \omega^2 & 1 & \omega \\ 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \end{bmatrix} \begin{bmatrix} \omega^2 & 1 & \omega \\ 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = 0 $

Thus, $ P^2 = 0 $ when $ n $ is a multiple of 3. Therefore, $ P^2 \neq 0 $ when $ n $ is not a multiple of 3.

The possible values of $ n $ where $ P^2 \neq 0 $ are:

$ \Rightarrow n = 55, 58, 56 $

In case of option (A),

${({N^T}MN)^T} = {N^T}{M^T}{({N^T})^T} = {N^T}{M^T}N$

Now, ${N^T}{M^T}N = \left\{ \matrix{ {N^T}MN,\,when\,{M^T} = M \hfill \cr - {N^T}MN,\,when\,{M^T} = - M \hfill \cr} \right.$

$\therefore$ N^TMN is symmetric or skew symmetric according as M is symmetric or skew symmetric.

In case of option (B),

${(MN - NM)^T} = {(MN)^T} - {(NM)^T}$

$ = {N^T}{M^T} - {M^T}{N^T}$

$ = NM - MN$ [$\because$ ${M^T} = M,\,{N^T} = N$]

$ = - (MN - NM)$

$\therefore$ $MN - NM$ is skew symmetric matrix.

In case of option (C),

${(MN)^T} = {N^T}{M^T} = NM$ [$\because$ ${M^T} = M,\,{N^T} = N$]

Now, MN cannot always be equal to NM

$\therefore$ MN is not symmetric matrix.

In case of option (D),

$(Adj\,M)(Adj\,N) = Adj(NM) \ne Adj(MN)$

Therefore, (C) and (D) are the correct options.

Given

${P^3} = {q^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

${P^2}Q = {Q^2}p\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

Subtracting $(1)$ and $(2)$, we get

${P^3} - {P^2}Q = {Q^3} - {Q^2}P$

$ \Rightarrow {P^2}\left( {P - Q} \right) + {Q^2}\left( {P - Q} \right) = 0$

$ \Rightarrow \left( {{P^2} + {Q^2}} \right)\left( {P - Q} \right) = 0$

$ \Rightarrow \left| {{p^2} + {Q^2}} \right| = 0$

as $P \ne Q$

Let $A{u_1} = \left( {\matrix{ 1 \cr 0 \cr 0 \cr } } \right)\,\,\,\,\,\,A{u_2} = \left( {\matrix{ 0 \cr 1 \cr 0 \cr } } \right)$

Then, $A{u_1} + A{u_2} = \left( {\matrix{ 1 \cr 0 \cr 0 \cr } } \right) + \left( {\matrix{ 0 \cr 1 \cr 0 \cr } } \right)$

$ \Rightarrow A\left( {{u_1} + {u_2}} \right) = \left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

Also, $A = \left( {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 0 \cr 3 & 2 & 1 \cr } } \right)$

$ \Rightarrow \left| A \right| = 1\left( 1 \right) - 0\left( 2 \right) + 0\left( {4 - 3} \right) = 1$

We know,

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

( as $\left| A \right| = 1$ )

Now, from equation $(1)$, we have

${u_1} + {u_2} = {A^{ - 1}}\left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right)$

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

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

Concept Involved If $\left| {{A_{n \times n}}} \right| = \Delta $, then $\left| {adj\,A} \right| = {\Delta ^{n - 1}}$

Here, ${P_{3 \times 3}} = \left[ {\matrix{ 1 & 4 & 4 \cr 2 & 1 & 7 \cr 1 & 1 & 3 \cr } } \right]$

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

$\therefore$ $\left| {adj\,P} \right| = \left| {\matrix{ 1 & 4 & 4 \cr 2 & 1 & 7 \cr 1 & 1 & 3 \cr } } \right|$

$ = 1(3 - 7) - 4(6 - 7) + 4(2 - 1)$

$ = - 4 + 4 + 4 = 4$

$ \Rightarrow \left| P \right| = \pm \,2$

We have ${P^T} = 2P + I$

We get ${P^T} - 2P = I$ ..... (i)

Taking transpose, we have ${({P^T} - 2P)^T} = {I^T}$

$ \Rightarrow P - 2{P^T} = I$ ..... (ii)

From (i) and (ii) on eliminating P^T we have

$ - 4P + P = 3I \Rightarrow P = - I$ $\therefore$ $P + I = 0$

Thus, $(P + 1)X = 0 \Rightarrow PX = - X$

Here, $P = {[{a_{ij}}]_{3 \times 3}} = \left[ {\matrix{ {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr } } \right]$

$Q = {[{b_{ij}}]_{3 \times 3}} = \left[ {\matrix{ {{b_{11}}} & {{b_{12}}} & {{b_{13}}} \cr {{b_{21}}} & {{b_{22}}} & {{b_{23}}} \cr {{b_{31}}} & {{b_{32}}} & {{b_{33}}} \cr } } \right]$

where, ${b_{ij}} = {2^{i + j}}{a_{ij}}$

$\therefore$ $\left| Q \right| = \left| {\matrix{ {4{a_{11}}} & {8{a_{12}}} & {16{a_{13}}} \cr {8{a_{21}}} & {16{a_{22}}} & {32{a_{23}}} \cr {16{a_{31}}} & {32{a_{32}}} & {64{a_{33}}} \cr } } \right|$

$ = 4 \times 8 \times 16\left| {\matrix{ {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr {2{a_{21}}} & {2{a_{22}}} & {2{a_{23}}} \cr {4{a_{31}}} & {4{a_{32}}} & {4{a_{33}}} \cr } } \right|$

$ = {2^9} \times 2 \times 4\left| {\matrix{ {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr } } \right|$

$ = {2^{12}}.\left| P \right| = {2^{12}}.2 = {2^{13}}$

$\Delta = 0 \Rightarrow \left| {\matrix{ 4 & k & 2 \cr k & 4 & 1 \cr 2 & 2 & 1 \cr } } \right| = 0$

$ \Rightarrow 4\left( {4 - 2} \right) - k\left( {k - 2} \right) + $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\left( {2k - 8} \right) = 0$

$ \Rightarrow 8 - {k^2} + 2k + 4k - 16 = 0$

$ \Rightarrow {k^2} - 6k + 8 = 0$

$ \Rightarrow \left( {k - 4} \right)\left( {k - 2} \right) = 0,k = 4,2$

$\therefore$ $A' = A,B' = B$

Now $\,\,\,\left( {A\left( {BA} \right)} \right)' = \left( {BA} \right)'A'$

$ = \left( {A'B'} \right)A' = \left( {AB} \right)A = A\left( {BA} \right)$

Similarly $\left( {\left( {AB} \right)A} \right)' = \left( {AB} \right)A$

So, $A\left( {BA} \right)\,\,\,\,$ and $A\left( {BA} \right)\,\,\,\,$ are symmetric matrices.

Again $\left( {AB} \right)' = B'A' = BA$

Now if $BA=AB$, then $AB$ is symmetric matrix.

Given, ${M^T} = - M$, ${N^T} = - N$

and $MN = NM$ ..... (i)

$\therefore$ ${M^2}{N^2}{({M^T}N)^{ - 1}}{(M{N^{ - 1}})^T}$

$ = {M^2}{N^2}{N^{ - 1}}{({M^T})^{ - 1}}{({N^{ - 1}})^T}.{M^T}$

$ = {M^2}N(N{M^{ - 1}}){( - M)^{ - 1}}{({N^T})^{ - 1}}( - M)$

$ = {M^2}NI( - {M^{ - 1}}){( - N)^{ - 1}}( - M)$

$ = - {M^2}N{M^{ - 1}}{N^{ - 1}}M$

$ = - M.(MN){M^{ - 1}}{N^{ - 1}}M$

$ = - M(NM){M^{ - 1}}{N^{ - 1}}M$

$ = - MN(N{M^{ - 1}}){N^{ - 1}}M$

$ = - M(N{N^{ - 1}})M = - {M^2}$

Note : This question is wrong, as given. An odd order skew symmetric matrix can't be invertible. Had the matrix be of even order, it could have been correct.

Given, ${[\matrix{ a & b & c \cr } ]_{1 \times 3}}{\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right]_{3 \times 3}} = [\matrix{ 0 & 0 & 0 \cr } ]$

$ \Rightarrow \left[ {\matrix{ {a + 8b + 7c} \cr {9a + 2b + 3c} \cr {7a + 7b + 7c} \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$

$ \Rightarrow a + 8b + 7c = 0$ ..... (i)

$ \Rightarrow 9a + 2b + 3c = 0$ .... (ii)

$ \Rightarrow a + b + c = 0$ ....... (iii)

On multiplying Eq. (iii) by 2, then subtract from Eq. (ii), we get

$7a + c = 0$ ..... (iv)

Again multiplying Eq. (iii) by 3, then subtract from Eq. (ii), we get

$6a - b = 0$ ..... (v)

$\therefore$ $b = 6a$ and $c = - 7a$

As (a, b, c) lies on $2x + y + z = 1$

$ \Rightarrow 2a + b + c = 1$

$ \Rightarrow 2a + 6a - 7a = 1$

$\Rightarrow$ a = 1, b = 6 and c = $-$7

$\therefore$ $7a + b + c = 7 + 6 - 7 = 6$

Given, ${[\matrix{ a & b & c \cr } ]_{1 \times 3}}{\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right]_{3 \times 3}} = [\matrix{ 0 & 0 & 0 \cr } ]$

$ \Rightarrow \left[ {\matrix{ {a + 8b + 7c} \cr {9a + 2b + 3c} \cr {7a + 7b + 7c} \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$

$ \Rightarrow a + 8b + 7c = 0$ ..... (i)

$ \Rightarrow 9a + 2b + 3c = 0$ .... (ii)

$ \Rightarrow a + b + c = 0$ .... (iii)

On multiplying Eq. (iii) by 2, then subtract from Eq. (ii), we get

$7a + c = 0$ ..... (iv)

Again multiplying Eq. (iii) by 3, then subtract from Eq. (ii), we get

$6a - b = 0$ ..... (v)

$\therefore$ b = 6a and c = $-$ 7a

If a = 2, b = 12 and c = $-$14

$\therefore$ ${3 \over {{\omega ^a}}} + {1 \over {{\omega ^b}}} + {3 \over {{\omega ^c}}}$

$ \Rightarrow {3 \over {{\omega ^2}}} + {1 \over {{\omega ^{12}}}} + {3 \over {{\omega ^{ - 14}}}} = {3 \over {{\omega ^2}}} + 1 + 3{\omega ^2}$

$ = 3\omega + 1 + 3{\omega ^2}$

$ = 1 + 3(\omega + {\omega ^2})$

$ = 1 - 3 = - 2$

Given, ${[\matrix{ a & b & c \cr } ]_{1 \times 3}}{\left[ {\matrix{ 1 & 9 & 7 \cr 8 & 2 & 7 \cr 7 & 3 & 7 \cr } } \right]_{3 \times 3}} = [\matrix{ 0 & 0 & 0 \cr } ]$

$ \Rightarrow \left[ {\matrix{ {a + 8b + 7c} \cr {9a + 2b + 3c} \cr {7a + 7b + 7c} \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$

$ \Rightarrow a + 8b + 7c = 0$ ..... (i)

$ \Rightarrow 9a + 2b + 3c = 0$ .... (ii)

$ \Rightarrow a + b + c = 0$ .... (iii)

On multiplying Eq. (iii) by 2, then subtract from Eq. (ii), we get

$7a + c = 0$ ..... (iv)

Again multiplying Eq. (iii) by 3, then subtract from Eq. (ii), we get

$6a - b = 0$ ..... (v)

$\therefore$ b = 6a and c = $-$ 7a

If b = 6, a = 1 and c = $-$7

$\therefore$ $a{x^2} + bx + c = 0$

$ \Rightarrow {x^2} + 6x - 7 = 0$

$ \Rightarrow (x + 7)(x - 1) = 0$

$\therefore$ $x = 1,7$

$ \Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {{1 \over 1} - {1 \over 7}} \right)}^n} \Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {{6 \over 7}} \right)}^n}} } $

$ \Rightarrow 1 + {6 \over 7} + {\left( {{6 \over 7}} \right)^n} + ....\infty $

$ = {1 \over {1 - {6 \over 7}}} = {1 \over {1/7}} = 7$

$\left| A \right| \ne 0$, as non-singular.

$\therefore$ $\left| {\matrix{ 1 & a & b \cr \omega & 1 & c \cr {{\omega ^2}} & \omega & 1 \cr } } \right| \ne 0$

$ \Rightarrow 1(1 - c\omega ) - a(\omega - c{\omega ^2}) + b({\omega ^2} - {\omega ^2}) \ne 0$

$ \Rightarrow 1 - c\omega - a\omega + ac{\omega ^2} \ne 0$

$ \Rightarrow (1 - c\omega )(1 - a\omega ) \ne 0$

$ \Rightarrow a \ne {1 \over \omega },c \ne {1 \over \omega } \Rightarrow a = \omega ,c = \omega $

and $b \in \{ \omega ,{\omega ^2}\} \Rightarrow 2$ solutions

$\left[ {\matrix{ 1 & {...} & {...} \cr {...} & 1 & {...} \cr {...} & {...} & 1 \cr } } \right]\,\,$ are $6$ non-singular matrices because $6$

blanks will be filled by $5$ zeros and $1$ one.

Similarly, $\left[ {\matrix{ {...} & {...} & 1 \cr {...} & 1 & {...} \cr 1 & {...} & {...} \cr } } \right]\,\,$ are $6$ non-singular matrices.

So, required cases are more than $7,$ non-singular $3 \times 3$ matrices.

Let $A = \left( {\matrix{ a & b \cr c & d \cr } } \right)$ where $a,b,c,d$ $ \ne 0$

${A^2} = \left( {\matrix{ a & b \cr c & d \cr } } \right)\left( {\matrix{ a & b \cr c & d \cr } } \right)$

$ \Rightarrow {A^2} = \left( {\matrix{ {{a^2} + bc} & {ab + bd} \cr {ac + cd} & {bc + {d^2}} \cr } } \right)$

$ \Rightarrow {a^2} + bc = 1,\,bc + {d^2} = 1$

$ab + bd = ac + cd = 0$

$c \ne 0\,\,\,\,\,b \ne 0$

$ \Rightarrow a + d = 0 \Rightarrow Tr\left( A \right) = 0$

$\left| A \right| = ad - bc = - {a^2} - bc = - 1$

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

${D_1}\left| {\matrix{ 3 & 2 & 1 \cr 3 & 3 & 1 \cr 1 & 5 & 2 \cr } } \right| \ne 0$

$ \Rightarrow $ Given system, does not have any solution.

$ \Rightarrow $ No solution

Given,

A $\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$ has two distinct solution we know that three Planes cannot intersect at two distinct point.

Hence, number of $3 \times 3$ matrix $A$ is zero.

We must have $a^2-b^2=1 < p$

$ (a+b)(a-b)=1 < p $

Either $a-b=0$ or $a+b$ is a multiple of $p$ when $a=b$ number of matrices is $p$ and when $a+b$ $=$ multiple of $p$.

$ \Rightarrow a, b \text { has } p-1 $

$ \begin{aligned} \text { Total number of matrices } & =p+p-1 \\\\ & =2 p-1 \end{aligned} $

We have an odd prime $p$. Consider the set

$ T_p \;=\; \Bigl\{ A \;=\;\begin{pmatrix} a & b \\[6pt] c & a \end{pmatrix} \;:\; a,b,c \in \{0,1,2,\dots,p-1\} \Bigr\}. $

We want to count the number of such matrices $A$ that satisfy:

$\text{trace}(A) \not\equiv 0 \pmod{p}$.

Since $\text{trace}(A) = a + a = 2a$, this means

$ 2a \;\not\equiv\; 0 \pmod{p}. $

Because $p$ is an odd prime, $2$ is invertible $\bmod\,p$.

Hence $2a \equiv 0 \pmod{p}$ if and only if $a \equiv 0 \pmod{p}$.

Thus the condition

$\text{trace}(A) \not\equiv 0 \pmod{p}$

is equivalent to

$ a \;\neq\; 0 \pmod{p}. $

In other words, $a$ must be a nonzero element modulo $p$; so

$a\in\{1,2,\dots,p-1\}.$

$\det(A) \equiv 0 \pmod{p}$.

For the matrix

$\begin{pmatrix} a & b \\ c & a \end{pmatrix},$ we have

$ \det(A) \;=\; a^2 - bc. $

The condition $\det(A)\equiv 0\pmod{p}$ means

$ a^2 \;-\; bc \;\equiv\; 0 \pmod{p} \quad\Longleftrightarrow\quad bc \;\equiv\; a^2 \pmod{p}. $

Putting these two pieces together:

$a$ is nonzero ($a \in \{1,\dots,p-1\}$).

$bc \equiv a^2 \pmod{p}.$

---

Counting the solutions

We must count the number of triples $(a,b,c)$ with $a\neq 0$ and $bc \equiv a^2$.

Range for $a$

Since $a\in \{1,2,\dots,p-1\}$, there are $p-1$ possible values of $a$.

Condition $bc \equiv a^2$ for given $a$

Since $a \neq 0$, $a^2$ is also nonzero $\bmod\,p$.

If $b = 0$, then $bc \equiv 0$, which would only be equal to $a^2$ if $a^2 \equiv 0$. But $a^2\neq 0$ since $a\neq 0$. So $b=0$ gives no solutions.

Similarly, if $c=0$, then $bc=0$, which again cannot equal the nonzero $a^2$. So $c=0$ gives no solutions either.

Therefore $b$ and $c$ must both be nonzero modulo $p$.

Once $b$ is chosen to be any nonzero element in $\{1,\dots,p-1\}$, there is a unique $c \equiv a^2 \,b^{-1} \pmod{p}$. Hence:

For each nonzero $b$ (that is, $b\in\{1,\dots,p-1\}$), there is exactly one $c\in\{1,\dots,p-1\}$ satisfying $bc \equiv a^2$.

Consequently, for each fixed nonzero $a$, the number of $(b,c)$-pairs is $p-1$.

Putting it all together

We have $p-1$ choices for $a\neq 0$.

For each such $a$, there are $p-1$ pairs $(b,c)$ satisfying $bc\equiv a^2$.

Therefore, in total, the number of matrices is

$ (p-1)\,\times\,(p-1) \;=\; (p-1)^2. $

Hence the correct answer matches option C, which is

$ \boxed{(p-1)^2.} $

Let $p$ be an odd prime number, and consider the set

$ T_{p} \;=\; \Bigl\{ A \;=\;\begin{pmatrix} a & b\\[6pt] c & a \end{pmatrix} \;:\; a,b,c \in \{0,1,2,\dots,p-1\} \Bigr\}. $

We want to count the number of matrices $A \in T_p$ whose determinant is not divisible by $p$.

Recall that for

$ A \;=\; \begin{pmatrix} a & b \\ c & a \end{pmatrix}, $

the determinant is

$ \det(A) \;=\; a^2 \;-\; b\,c. $

Hence $\det(A)$ is not divisible by $p$ precisely when

$ a^2 \;-\; b\,c \;\not\equiv\; 0 \pmod{p}, \quad\text{i.e.,}\quad a^2 \;\neq\; b\,c \pmod{p}. $

---

Total number of matrices

Since $a,b,c$ each range over $\{0,1,\dots,p-1\}$, there are

$ p^3 $

total possible matrices in $T_p$.

---

Step 1: Count how many $(a,b,c)$ do satisfy $a^2 \equiv b\,c \pmod{p}$

It will be easier to first count the number of triples $(a,b,c)$ for which

$ a^2 \;\equiv\; b\,c \pmod{p}, $

and then subtract that from $p^3$.

---

Case A: $a = 0$

Then $a^2 = 0$. We need

$ b\,c \;\equiv\; 0 \pmod{p}. $

Over the field $\mathbf{F}_p$, the product $b\,c\equiv 0$ if and only if at least one of $b$ or $c$ is $0$ (since $p$ is prime).

If $b=0$, then $c$ can be anything in $\{0,\dots,p-1\}$.

This gives $p$ possibilities.

If $c=0$, then $b$ can be anything in $\{0,\dots,p-1\}$.

This also gives $p$ possibilities.

However, the pair $(b=0, c=0)$ is counted in both cases, so we must subtract it out once.

Hence, for $a=0$, the number of $(b,c)$ such that $b\,c \equiv 0$ is

$ p \;+\; p \;-\; 1 \;=\; 2p \;-\; 1. $

---

Case B: $a \neq 0$

Then $a$ can be any nonzero element in $\{1,2,\dots,p-1\}$; there are $p-1$ such choices. In that scenario, $a^2 \neq 0\pmod{p}$. We want

$ b\,c \;\equiv\; a^2 \pmod{p}. $

Over $\mathbf{F}_p$, a nonzero right-hand side $a^2$ can be written as $b\,c$ in the following way:

If $b=0$, then $b\,c = 0$, which cannot equal $a^2\neq 0$.

No solutions in that subcase.

If $b \neq 0$, then for each nonzero $b$ there is exactly one $c$ satisfying $b\,c \equiv a^2\pmod{p}$, namely $c \equiv a^2 b^{-1}\pmod{p}$.

Since $b$ can be any of the $p-1$ nonzero elements, we get $p-1$ solutions $(b,c)$ for each nonzero $a$.

Thus, for each $a\neq 0$, there are $p-1$ pairs $(b,c)$. Hence, for $p-1$ values of $a$, the total number of solutions in this case is

$ (p-1)\,\times\,(p-1) \;=\; (p-1)^2. $

---

Total number of solutions to $a^2 = b\,c$

Summing these two cases:

Case A $(a=0)$: $2p - 1$ solutions.

Case B $(a\neq 0)$: $(p-1)^2$ solutions.

Hence the total count of $(a,b,c)$ satisfying $a^2 \equiv b\,c$ is

$ (2p -1) \;+\; (p-1)^2 \;=\; (2p -1) \;+\; \bigl(p^2 \;-\;2p \;+\;1\bigr) \;=\; p^2. $

(A classic result: there are exactly $p^2$ solutions to $b\,c = a^2$ over $\mathbf{F}_p^3$.)

---

Step 2: Count how many $(a,b,c)$ satisfy $a^2 \neq b\,c$

Since there are $p^3$ total triples $(a,b,c)$, the number that satisfy

$ a^2 \;\neq\; b\,c \pmod{p} $

is simply

$ p^3 \;-\; p^2. $

This directly corresponds to matrices $A$ whose determinant $a^2 - b\,c$ is $\not\equiv 0\pmod{p}$.

Hence, the number of matrices in $T_p$ with $\det(A)$ not divisible by $p$ is

$ \boxed{p^3 - p^2}. $

---

The expression $p^3 - p^2$ matches Option D.

We know that $\left| {adj\left( {adj\,\,A} \right)} \right| = {\left| {Adj\,\,A} \right|^{2 - 1}}$b

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

$\therefore$ $\,\,\,\,\,\,$ Both the statements are true and statement $-2$ is a correct explanation for statement - $1.$

$\left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| + \left| {\matrix{ {a + 1} & {b + 1} & {c - 1} \cr {a - 1} & {b - 1} & {c + 1} \cr {{{\left( { - 1} \right)}^{n + 2}}a} & {{{\left( { - 1} \right)}^{n + 1}}b} & {{{\left( { - 1} \right)}^n}c} \cr } } \right| = 0$

$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| + \left| {\matrix{ {a + 1} & {a - 1} & {{{\left( { - 1} \right)}^{n + 2}}a} \cr {b + 1} & {b - 1} & {{{\left( { - 1} \right)}^{n + 1}}b} \cr {c - 1} & {c + 1} & {{{\left( { - 1} \right)}^n}c} \cr } } \right| = 0$

(Taking transpose of second determinant)

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_1} \Leftrightarrow {C_3}$

$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| - \left| {\matrix{ {{{\left( { - 1} \right)}^{n + 2}}a} & {a - 1} & {a + 1} \cr {{{\left( { - 1} \right)}^{n + 2}}\left( { - b} \right)} & {b - 1} & {b + 1} \cr {{{\left( { - 1} \right)}^{n + 2}}c} & {c + 1} & {c - 1} \cr } } \right| = 0$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2} \Leftrightarrow {C_3}$

$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| + {\left( 1 \right)^{n + 2}}\left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| = 0$

$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| = 0$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2} - {C_1},{C_3} - {C_1}$

$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ a & 1 & { - 1} \cr { - b} & {2b + 1} & {2b - 1} \cr c & { - 1} & 1 \cr } } \right| = 0$

${R_1} + {R_3}$

$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ {a + c} & 0 & 0 \cr { - b} & {2b + 1} & {2b - 1} \cr c & { - 1} & 1 \cr } } \right| = 0$

$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left( {a + c} \right)\left( {2b + 1 + 2b - 1} \right) = 0$

$ \Rightarrow 4b\left( {a + c} \right)\left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right] = 0$

$ \Rightarrow 1 + {\left( { - 1} \right)^{n + 2}} = 0$ $\,\,\,\,\,$ as $\,\,\,\,\,b\left( {a + c} \right) \ne 0$

$ \Rightarrow n$ should be an odd integer.

Here,

Matrix [A]_{3 × 3} is symmetric

Here, 5 be the number of 1’s and 4 be the 0’s entry in matrix A

Let x be the number of 1’s on main diagonal.

Since A is symmetric so, y be the number of 1’s below the main diagonal

So , x + 2y = 5

x = 1 and y = 2

or x = 3 and y = 1

Case I : x = 1 and y = 2

Here, Main diagonal entries in 0, 0 and 1. Hence, we can choose main diagonal in 3 ways, and element above the diagonal is 1, 1, 0 . Hence, we can choose it element above the main diagonal in 3 ways , element of below diagonal depends on above the main diagonal.

Hence, total way = 3 × 3 = 9

Case II : x = 3 and y = 1

Here, Main diagonal entries in 1, 1 and 1.
Hence, we can choose main diagonal 1 ways and element above the diagonal is 1, 0, 0. Hence, we can choose it element above the main diagonal in 3 ways, element of below diagonal depends on above the main diagonal.

Hence, total ways = 3 × 4 = 12

So, Total matrices = 9 + 3 = 12

We have

$\left[ {\matrix{ 0 & a & b \cr a & 0 & c \cr b & c & 1 \cr } } \right]$

We can see that either $b = 0$ or $c = 0 \Rightarrow |A| \ne 0$; therefore two matrices.

$\left[ {\matrix{ 0 & a & b \cr a & 1 & c \cr b & c & 0 \cr } } \right]$

Now, either $a = 0$ or $c = 0 \Rightarrow |A| \ne 0$; therefore, two matrices

$\left[ {\matrix{ 1 & a & b \cr a & 0 & c \cr b & c & 0 \cr } } \right]$

Now, either $a = 0$ or $b = 0 \Rightarrow |A| \ne 0$; therefore two matrices.

$\left[ {\matrix{ 1 & a & b \cr a & 1 & c \cr b & c & 1 \cr } } \right]$

When $a = b = 0 \Rightarrow |A| = 0$

When $a = c = 0 \Rightarrow |A| = 0$

When $b = c = 0 \Rightarrow |A| = 0$

Therefore, there are only six matrices.

The six matrices A for which $|A| = 0$ are as follows:

$\left[ {\matrix{ 0 & 0 & 1 \cr 0 & 0 & 1 \cr 1 & 1 & 1 \cr } } \right] \Rightarrow $ inconsistent.

$\left[ {\matrix{ 0 & 1 & 0 \cr 1 & 1 & 1 \cr 0 & 1 & 0 \cr } } \right] \Rightarrow $ inconsistent.

$\left[ {\matrix{ 1 & 1 & 1 \cr 1 & 0 & 0 \cr 1 & 0 & 0 \cr } } \right] \Rightarrow $ infinite solutions.

$\left[ {\matrix{ 1 & 1 & 0 \cr 1 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] \Rightarrow $ inconsistent.

$\left[ {\matrix{ 1 & 0 & 1 \cr 0 & 1 & 0 \cr 1 & 0 & 1 \cr } } \right] \Rightarrow $ inconsistent.

$\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 1 \cr 0 & 1 & 1 \cr } } \right] \Rightarrow $ infinite solutions.

The given equations are

$\matrix{ { - x + cy + bz = 0} \cr {cx - y + az = 0} \cr {bx + ay - z = 0} \cr } $

As $x,y,z$ are not all zero

$\therefore$ The above system should not have unique (zero) solution

$ \Rightarrow \Delta = 0 \Rightarrow \left| {\matrix{ { - 1} & c & b \cr c & { - 1} & a \cr b & a & { - 1} \cr } } \right| = 0$

$ \Rightarrow - 1\left( {1 - {a^2}} \right) - c\left( { - c - ab} \right) + b\left( {ac + b} \right) = 0$

$ \Rightarrow - 1 + {a^2} + {b^2} + {c^2} + 2abc = 0$

$ \Rightarrow {a^2} + {b^2} + {c^2} + 2abc = 1$

Let $A = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$ $\,\,\,$ then ${A^2} = 1$

$ \Rightarrow {a^2} + bc = 1\,\,\,\,ab + bd = 0$

$ac + cd = 0\,\,\,\,bc + {d^2} = 1$

From these four relations,

${a^2} + bc = bc + {d^2} \Rightarrow {a^2} = {d^2}$

and $\,\,b\left( {a + d} \right) = 0 = c\left( {a + d} \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a = - d$

We can take $a = 1,b = 0,c = 0,d = - 1$

as one possible set of values, then

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

Clearly $A \ne I\,\,\,$ and $\,\,\,\,A \ne - I\,\,$ and $\,\,\,A = - 1$

$\therefore$ $\,\,\,\,\,$ Statement $1$ is true.

Also if $A \ne I\,\,\,\,\,tr\left( A \right) = 0$

$\therefore$ $\,\,\,\,\,$ Statement $2$ is false.

As all entries of square matrix $A$ are integers, therefore all co-factors should also be integers.

If det $A = \pm 1\,\,$ then ${A^{ - 1}}\,\,$ exists. Also all entries of ${A^{ - 1}}$ are integers.

The given equations are

$x - 2y + 3z = - 1$

$ - x + y - 2z = k$

$x - 3y + 4z = 1$

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

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

If $k \ne 3$, the system has no solutions.

Hence, Statement - 1 is True Statement - 2 is correct explanation for Statement - 1.

$\left| {{A^2}} \right| = 25 \Rightarrow {\left| A \right|^2} = 25$

$ \Rightarrow {\left( {25\alpha } \right)^2} = 25 \Rightarrow \left| \alpha \right| = {1 \over 5}$