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

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(D) Given $B = I - {}^{3}C_{1}(\operatorname{adj} A) + {}^{3}C_{2}(\operatorname{adj} A)^{2} - {}^{3}C_{3}(\operatorname{adj} A)^{3}$.Using the binomi

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$-5$

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(D) Given $B = I - {}^{3}C_{1}(\operatorname{adj} A) + {}^{3}C_{2}(\operatorname{adj} A)^{2} - {}^{3}C_{3}(\operatorname{adj} A)^{3}$.Using the binomial expansion $(I - X)^{3} = I - {}^{3}C_{1}X + {}^{3}C_{2}X^{2} - {}^{3}C_{3}X^{3}$,we get $B = (I - \operatorname{adj} A)^{3}$.For $A = \begin{bmatrix} 2 & -1 \ 0 & 2 \end{bmatrix}$,the adjoint is $\operatorname{adj} A = \begin{bmatrix} 2 & 1 \ 0 & 2 \end{bmatrix}$.Then $I - \operatorname{adj} A = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 1 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \ 0 & -1 \end{bmatrix}$.Now,$B = \begin{bmatrix} -1 & -1 \ 0 & -1 \end{bmatrix}^{3} = \begin{bmatrix} -1 & -1 \ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & -1 \ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & -1 \ 0 & -1 \end{bmatrix}$.Calculating the square: $\begin{bmatrix} -1 & -1 \ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & -1 \ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix}$.Calculating the cube: $\begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & -1 \ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & -3 \ 0 & -1 \end{bmatrix}$.The sum of all elements of matrix $B$ is $(-1) + (-3) + 0 + (-1) = -5$.

Let $A = \begin{bmatrix} 2 & -1 \\ 0 & 2 \end{bmatrix}$. If $B = I - {}^{3}C_{1}(\operatorname{adj} A) + {}^{3}C_{2}(\operatorname{adj} A)^{2} - {}^{3}C_{3}(\operatorname{adj} A)^{3}$,then the sum of all elements of the matrix $B$ is

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