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(C) First, we find the general form of $A^n$. Given $A = \begin{bmatrix} 3 & -4 \ 1 & -1 \end{bmatrix}$, we observe $A^2 = \begin{bmatrix} 3 & -4 \

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$x=11, y=2$

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(C) First, we find the general form of $A^n$. Given $A = \begin{bmatrix} 3 & -4 \ 1 & -1 \end{bmatrix}$, we observe $A^2 = \begin{bmatrix} 3 & -4 \ 1 & -1 \end{bmatrix} \begin{bmatrix} 3 & -4 \ 1 & -1 \end{bmatrix} = \begin{bmatrix} 5 & -8 \ 2 & -3 \end{bmatrix}$.By induction, $A^n = \begin{bmatrix} 2n+1 & -4n \ n & -2n+1 \end{bmatrix}$.For $n=15$, $A^{15} = \begin{bmatrix} 2(15)+1 & -4(15) \ 15 & -2(15)+1 \end{bmatrix} = \begin{bmatrix} 31 & -60 \ 15 & -29 \end{bmatrix}$.Now, $A^{15}+B = \begin{bmatrix} 31 & -60 \ 15 & -29 \end{bmatrix} + \begin{bmatrix} -29 & 49 \ -13 & 18 \end{bmatrix} = \begin{bmatrix} 2 & -11 \ 2 & -11 \end{bmatrix}$.The equation $(A^{15}+B)\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}$ becomes $\begin{bmatrix} 2 & -11 \ 2 & -11 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}$.This simplifies to the linear equation $2x - 11y = 0$, or $2x = 11y$.Checking the options, for $x=11$ and $y=2$, we get $2(11) = 22$ and $11(2) = 22$. Thus, $x=11, y=2$ is the correct solution.

For the matrices $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix}$, if $(A^{15}+B)\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$, then among the following which one is true?

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