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(C) Given $A = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix}$.First,find $\operatorname{adj}(A) = \begin{bmatrix} 1 & -2 \ 0 & 1 \end{bmatrix}$.Let $M

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$-88$

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(C) Given $A = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix}$.First,find $\operatorname{adj}(A) = \begin{bmatrix} 1 & -2 \ 0 & 1 \end{bmatrix}$.Let $M = \operatorname{adj}(A) = \begin{bmatrix} 1 & -2 \ 0 & 1 \end{bmatrix}$.Then $M^2 = \begin{bmatrix} 1 & -2 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -4 \ 0 & 1 \end{bmatrix}$.By induction,$M^k = \begin{bmatrix} 1 & -2k \ 0 & 1 \end{bmatrix}$.$B = I + M + M^2 + \dots + M^{10} = \sum_{k=0}^{10} M^k$.The sum of the diagonal elements is $\sum_{k=0}^{10} 1 + \sum_{k=0}^{10} 1 = 11 + 11 = 22$.The sum of the off-diagonal elements is $\sum_{k=0}^{10} (-2k) + 0 = -2 	imes \frac{10 	imes 11}{2} = -110$.Thus,$B = \begin{bmatrix} 11 & -110 \ 0 & 11 \end{bmatrix}$.The sum of all elements of $B$ is $11 - 110 + 0 + 11 = -88$.

Let $A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ and $B = I + \operatorname{adj}(A) + (\operatorname{adj} A)^2 + \dots + (\operatorname{adj} A)^{10}$. Then,the sum of all the elements of the matrix $B$ is:

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