Let $M = \begin{bmatrix} 0 & 1 & a \\ 1 & 2 & 3 \\ 3 & b & 1 \end{bmatrix}$ and $(\operatorname{adj} M) = \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix}$ where $a$ and $b$ are real numbers. Which of the following options is/are correct ?

M = 0 1 a 1 2 3 3 b 1 , adj M = -1 1 -1 8 -6 2 -5 3 -1

$a, b \in \mathbb{R}$

$\text{To find: the correct option(s) among } (1)\text{–}(4)$

$\text{Key identity: } M\,(\operatorname{adj} M) = (\det M)\,I_3$

$(M\operatorname{adj}M)_{11} = 0(-1)+1(8)+a(-5) = 8-5a$

$(M\operatorname{adj}M)_{12} = 0(1)+1(-6)+a(3) = -6+3a$

$(M\operatorname{adj}M)_{13} = 0(-1)+1(2)+a(-1) = 2-a$

$\text{off-diagonal} = 0:\quad -6+3a = 0\ \text{ and }\ 2-a = 0 \ \Rightarrow\ a = 2$

$(M\operatorname{adj}M)_{31} = 3(-1)+b(8)+1(-5) = 8b-8 = 0 \ \Rightarrow\ b = 1$

$(M\operatorname{adj}M)_{33} = 3(-1)+b(2)+1(-1) = 2b-4 = \det M = -2$

$\det M = 8-5a = -2,\qquad a=2,\ b=1$

$a + b = 3 \ \checkmark \ \Rightarrow\ (1)\ \text{TRUE}$

$\det(\operatorname{adj}N) = (\det N)^{2}\quad (N\ 3\times3)$

$\det(M^2) = (\det M)^2 = 4,\quad \det(\operatorname{adj}M^2) = 4^2 = 16 \ne 81$

$\Rightarrow\ (2)\ \text{FALSE}$

$\operatorname{adj}M = (\det M)\,M^{-1} = -2M^{-1}$

$(\operatorname{adj}M)^{-1} = -\tfrac12 M = \operatorname{adj}(M^{-1})$

$\Rightarrow\ (\operatorname{adj}M)^{-1} + \operatorname{adj}(M^{-1}) = -M \ \checkmark$

$\beta + 2\gamma = 1;\quad \alpha + 2\beta + 3\gamma = 2;\quad 3\alpha + \beta + \gamma = 3$

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

$\alpha - \beta + \gamma = 1 - (-1) + 1 = 3 \ \checkmark \ \Rightarrow\ (4)\ \text{TRUE}$

$\boxed{\text{Correct options: } (1),\ (3),\ (4)}$