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

Given $$A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$$, $$B = \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix}$$.

Find a pattern for powers of $$A$$.

$$A^2 = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix}$$

$$A^3 = A^2 \cdot A = \begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix}\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 7 & -12 \\ 3 & -5 \end{bmatrix}$$

Pattern: $$A^n = \begin{bmatrix} 2n+1 & -4n \\ n & -(2n-1) \end{bmatrix}$$

Verify: $$A^1$$: $$\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$$ ✓, $$A^2$$: $$\begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix}$$ ✓, $$A^3$$: $$\begin{bmatrix} 7 & -12 \\ 3 & -5 \end{bmatrix}$$ ✓

$$A^{15} = \begin{bmatrix} 31 & -60 \\ 15 & -29 \end{bmatrix}$$

$$A^{15} + B = \begin{bmatrix} 31-29 & -60+49 \\ 15-13 & -29+18 \end{bmatrix} = \begin{bmatrix} 2 & -11 \\ 2 & -11 \end{bmatrix}$$

Solve $$(A^{15} + B)\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$:

$$2x - 11y = 0$$

$$2x - 11y = 0$$ (same equation)

So $$2x = 11y$$. From Option 1: $$x = 11, y = 2$$: $$2(11) = 22 = 11(2)$$ ✓

The answer is Option 1: $$x = 11, y = 2$$.