Let A = begin{bmatrix} 2 & -1 0 & 2 end{bmat | vedclass

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(C) Given $A = \begin{bmatrix} 2 & -1 \ 0 & 2 \end{bmatrix}$.First,we find $\operatorname{adj} A$. For a $2 	imes 2$ matrix $\begin{bmatrix} a & b \

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$-7$

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(C) Given $A = \begin{bmatrix} 2 & -1 \ 0 & 2 \end{bmatrix}$.First,we find $\operatorname{adj} A$. For a $2 	imes 2$ matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$,the adjoint is $\begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Thus,$\operatorname{adj} A = \begin{bmatrix} 2 & 1 \ 0 & 2 \end{bmatrix}$.The expression for $B$ is given by the binomial expansion: $B = (I - \operatorname{adj} A)^{5}$.Calculate $I - \operatorname{adj} A = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 1 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \ 0 & -1 \end{bmatrix}$.Let $M = \begin{bmatrix} -1 & -1 \ 0 & -1 \end{bmatrix}$. We need to find $M^{5}$.$M^{2} = \begin{bmatrix} -1 & -1 \ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & -1 \ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix}$.$M^{3} = M^{2} \cdot M = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & -1 \ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & -3 \ 0 & -1 \end{bmatrix}$.$M^{4} = M^{2} \cdot M^{2} = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 4 \ 0 & 1 \end{bmatrix}$.$M^{5} = M^{4} \cdot M = \begin{bmatrix} 1 & 4 \ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & -1 \ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & -5 \ 0 & -1 \end{bmatrix}$.The sum of all elements of matrix $B$ is $(-1) + (-5) + 0 + (-1) = -7$.

Let $A = \begin{bmatrix} 2 & -1 \\ 0 & 2 \end{bmatrix}$. If $B = I - {}^{5}C_{1} (\operatorname{adj} A) + {}^{5}C_{2} (\operatorname{adj} A)^{2} - \dots - {}^{5}C_{5} (\operatorname{adj} A)^{5}$,then the sum of all elements of the matrix $B$ is

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