Let M = begin{bmatrix} 0 & 1 & a 1 & 2 & 3 | vedclass

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(A) Given $(\operatorname{adj} M)_{11} = 2 - 3b = -1 \Rightarrow b = 1$.Also,$(\operatorname{adj} M)_{22} = -3a = -6 \Rightarrow a = 2$.Thus,$a+b = 2+

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$1, 3, 4$

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(A) Given $(\operatorname{adj} M)_{11} = 2 - 3b = -1 \Rightarrow b = 1$.Also,$(\operatorname{adj} M)_{22} = -3a = -6 \Rightarrow a = 2$.Thus,$a+b = 2+1 = 3$. So,$(1)$ is correct.Now,$\operatorname{det} M = \begin{vmatrix} 0 & 1 & 2 \ 1 & 2 & 3 \ 3 & 1 & 1 \end{vmatrix} = 0(2-3) - 1(1-9) + 2(1-6) = 8 - 10 = -2$.$\operatorname{det}(\operatorname{adj} M^2) = (\operatorname{det}(\operatorname{adj} M))^2 = ((\operatorname{det} M)^2)^2 = ((-2)^2)^2 = 16 
eq 81$. So,$(2)$ is incorrect.Since $M^{-1} = \frac{\operatorname{adj} M}{\operatorname{det} M}$,we have $\operatorname{adj} M = -2M^{-1}$.Then $(\operatorname{adj} M)^{-1} = ( -2M^{-1} )^{-1} = -\frac{1}{2}M$.Also,$\operatorname{adj}(M^{-1}) = \operatorname{det}(M^{-1}) (M^{-1})^{-1} = \frac{1}{\operatorname{det} M} M = -\frac{1}{2}M$.Thus,$(\operatorname{adj} M)^{-1} + \operatorname{adj}(M^{-1}) = -\frac{1}{2}M - \frac{1}{2}M = -M$. So,$(3)$ is correct.For $M \begin{bmatrix} \alpha \ \beta \ \gamma \end{bmatrix} = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix}$,we have $\begin{bmatrix} \alpha \ \beta \ \gamma \end{bmatrix} = M^{-1} \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix} = \frac{\operatorname{adj} M}{-2} \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} -1 & 1 & -1 \ 8 & -6 & 2 \ -5 & 3 & -1 \end{bmatrix} \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} -2 \ 2 \ -2 \end{bmatrix} = \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix}$.So $\alpha = 1, \beta = -1, \gamma = 1$. Then $\alpha - \beta + \gamma = 1 - (-1) + 1 = 3$. So,$(4)$ is correct.

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