A is a 2 times 2 matrix such that A begin{bma | vedclass

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(D) Let $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.Given $A \begin{bmatrix} 1 \ -1 \end{bmatrix} = \begin{bmatrix} -1 \ 2 \end{bmatrix}$,we

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$5$

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(D) Let $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.Given $A \begin{bmatrix} 1 \ -1 \end{bmatrix} = \begin{bmatrix} -1 \ 2 \end{bmatrix}$,we get:$a - b = -1$ $(1)$$c - d = 2$ $(2)$Given $A^2 \begin{bmatrix} 1 \ -1 \end{bmatrix} = \begin{bmatrix} 1 \ 0 \end{bmatrix}$,we can write this as $A \left( A \begin{bmatrix} 1 \ -1 \end{bmatrix} ight) = \begin{bmatrix} 1 \ 0 \end{bmatrix}$.Substituting $(1)$,we get $A \begin{bmatrix} -1 \ 2 \end{bmatrix} = \begin{bmatrix} 1 \ 0 \end{bmatrix}$.This gives the system:$-a + 2b = 1$ $(3)$$-c + 2d = 0$ $(4)$Solving $(1)$ and $(3)$: From $(1)$,$a = b - 1$. Substituting into $(3)$: $-(b - 1) + 2b = 1 \Rightarrow -b + 1 + 2b = 1 \Rightarrow b = 0$. Then $a = -1$.Solving $(2)$ and $(4)$: From $(2)$,$c = d + 2$. Substituting into $(4)$: $-(d + 2) + 2d = 0 \Rightarrow -d - 2 + 2d = 0 \Rightarrow d = 2$. Then $c = 4$.The matrix $A = \begin{bmatrix} -1 & 0 \ 4 & 2 \end{bmatrix}$.The sum of the elements is $a + b + c + d = -1 + 0 + 4 + 2 = 5$.

$A$ is a $2 \times 2$ matrix such that $A \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$ and $A^2 \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$. The sum of the elements of $A$ is:

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