Let A = begin{bmatrix} 3 & -4 1 & -1 end{bma | vedclass

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(D) Step $1$: Find $A^{-1}$. Given $A = \begin{bmatrix} 3 & -4 \ 1 & -1 \end{bmatrix}$, $\det(A) = (3)(-1) - (-4)(1) = -3 + 4 = 1$. Thus, $A^{-1} = \

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$(5, 3)$

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(D) Step $1$: Find $A^{-1}$. Given $A = \begin{bmatrix} 3 & -4 \ 1 & -1 \end{bmatrix}$, $\det(A) = (3)(-1) - (-4)(1) = -3 + 4 = 1$. Thus, $A^{-1} = \frac{1}{1} \begin{bmatrix} -1 & 4 \ -1 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 \ -1 & 3 \end{bmatrix}$.Step $2$: Calculate $(A^{-1})^2$. $(A^{-1})^2 = \begin{bmatrix} -1 & 4 \ -1 & 3 \end{bmatrix} \begin{bmatrix} -1 & 4 \ -1 & 3 \end{bmatrix} = \begin{bmatrix} (-1)(-1) + (4)(-1) & (-1)(4) + (4)(3) \ (-1)(-1) + (3)(-1) & (-1)(4) + (3)(3) \end{bmatrix} = \begin{bmatrix} -3 & 8 \ -2 & 5 \end{bmatrix}$.Step $3$: Calculate $(A^{-1})^2 + B$. $\begin{bmatrix} -3 & 8 \ -2 & 5 \end{bmatrix} + \begin{bmatrix} 6 & -13 \ 5 & -10 \end{bmatrix} = \begin{bmatrix} 3 & -5 \ 3 & -5 \end{bmatrix}$.Step $4$: Solve the system $\begin{bmatrix} 3 & -5 \ 3 & -5 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}$. This implies $3x - 5y = 0$, or $3x = 5y$. Among the options, only $(5, 3)$ satisfies $3(5) = 5(3)$, i.e., $15 = 15$.

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