Let M = [ cases 0 & 1 & a 1 & 2 & 3 3 & b & 1 cases ] and adjM = [...

Given, M = \left[\begin{cases}0 & 1 & a \\ 1 & 2 & 3 \\ 3 & b & 1\end{cases}\right] ...(i)
adj\left(M\right) = \left[\begin{cases}- 1 & 1 & - 1 \\ 8 & - 6 & 2 \\ - 5 & 3 & - 1\end{cases}\right] ...(ii)
Option $( A )$ :
Now from matrix $M$
${\left(adjM\right)}_{11} = 2 - 3b$
As given $2 - 3b = - 1 ⇒ b = 1$

Similarly, $a = 2$

$a + b = 3$
Using the values of $a and$b$in matrix$M ⇒ | M | = - 2$<br />$| adj\left({M}^{2}\right) | = | {M}^{2}{|}^{2} = | M{|}^{4} = 16$<br />Now,${M}^{- 1} = \dfrac{adjM}{\left|M\right|}$<br />$adjM = \left|M\right|{M}^{- 1}$<br />$adj ( M ) = - 2{M}^{- 1}$<br />${\left(adj\left(M\right)\right)}^{- 1} = - \dfrac{1}{2}{\left({M}^{- 1}\right)}^{- 1}$\left({\left(KA\right)}^{- 1} = \dfrac{1}{K}{A}^{- 1}\right)$
${( adj ( M ) )}^{- 1} = \dfrac{- M}{2}$
and $adj ( {M}^{- 1} ) = {( {M}^{- 1} )}^{- 1} | {M}^{- 1} |$ \left(adjA = \left|A\right|{A}^{1}\right)$<br />$adj\left({M}^{- 1}\right) = \dfrac{M}{\left|M\right|}$\left(\left|{A}^{- 1}\right| = \dfrac{1}{\left|A\right|}\right)$
$adj\left({M}^{- 1}\right) = - \dfrac{M}{2}$ ...(iii)
From (iii) and (iv)
$( adj{\left(M\right)}^{- 1} + \left(adj\left({M}^{- 1}\right)\right) = - M$
Option $( D )$ :
MX = B\left(whenx = \left[$\begin{array}{ll}\alpha \\ \beta \\ \gamma\end{array}$\right]andB = \left[$\begin{array}{ll}1 \\ 2 \\ 3\end{array}$\right]\right)
$X = {M}^{- 1} ( B )$
$X = \dfrac{adj\left(M\right)}{\left|M\right|}B$
\left[$$\begin{array}{ll}\alpha \\ \beta \\ \gamma\end{array}$$\right] = - \dfrac{1}{2}\left[$\begin{cases}- 1 & 1 & - 1 \\ 8 & - 6 & 2 \\ - 5 & 3 & - 1\end{cases}$\right]\left[$$\begin{array}{ll}1 \\ 2 \\ 3\end{array}$$\right]$<br />$= - \dfrac{1}{2}\left[$$\begin{array}{ll}- 2 \\ 2 \\ - 2\end{array}$$\right]<br />\left[$\begin{array}{ll}\alpha \\ \beta \\ \gamma\end{array}$\right] = \left[$\begin{array}{ll}1 \\ - 1 \\ 1\end{array}$\right]
$∵ \alpha = 1\beta = - 1\gamma = 1$
$\alpha - \beta + \gamma = 3$