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(C) Let $A = \begin{bmatrix} 1 & 0 & 2 \ -1 & 1 & -2 \ 0 & 2 & 1 \end{bmatrix}$. The adjoint of a matrix is the transpose of its cofactor matrix, i.

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$[4 \quad 1]$

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(C) Let $A = \begin{bmatrix} 1 & 0 & 2 \ -1 & 1 & -2 \ 0 & 2 & 1 \end{bmatrix}$. The adjoint of a matrix is the transpose of its cofactor matrix, i.e.,$\operatorname{adj}(A) = [C_{ij}]^T$.The cofactors $C_{ij}$ are calculated as follows:$C_{11} = +\begin{vmatrix} 1 & -2 \ 2 & 1 \end{vmatrix} = 1 - (-4) = 5$$C_{12} = -\begin{vmatrix} -1 & -2 \ 0 & 1 \end{vmatrix} = -(-1 - 0) = 1$$C_{13} = +\begin{vmatrix} -1 & 1 \ 0 & 2 \end{vmatrix} = -2 - 0 = -2$$C_{21} = -\begin{vmatrix} 0 & 2 \ 2 & 1 \end{vmatrix} = -(0 - 4) = 4$$C_{22} = +\begin{vmatrix} 1 & 2 \ 0 & 1 \end{vmatrix} = 1 - 0 = 1$$C_{23} = -\begin{vmatrix} 1 & 0 \ 0 & 2 \end{vmatrix} = -(2 - 0) = -2$$C_{31} = +\begin{vmatrix} 0 & 2 \ 1 & -2 \end{vmatrix} = 0 - 2 = -2$$C_{32} = -\begin{vmatrix} 1 & 2 \ -1 & -2 \end{vmatrix} = -(-2 - (-2)) = 0$$C_{33} = +\begin{vmatrix} 1 & 0 \ -1 & 1 \end{vmatrix} = 1 - 0 = 1$The cofactor matrix is $\begin{bmatrix} 5 & 1 & -2 \ 4 & 1 & -2 \ -2 & 0 & 1 \end{bmatrix}$.Taking the transpose, $\operatorname{adj}(A) = \begin{bmatrix} 5 & 4 & -2 \ 1 & 1 & 0 \ -2 & -2 & 1 \end{bmatrix}$.Comparing this with the given matrix $\begin{bmatrix} 5 & a & -2 \ 1 & 1 & 0 \ -2 & -2 & b \end{bmatrix}$, we get $a = 4$ and $b = 1$.Therefore, $[a \quad b] = [4 \quad 1]$.

If $\operatorname{adj}\begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 5 & a & -2 \\ 1 & 1 & 0 \\ -2 & -2 & b \end{bmatrix}$, then $[a \quad b]$ is equal to

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