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?

Question

Accepted Answer

C. 1. Find the characteristic equation of $A$ using its trace and determinant. For $A=\begin{bmatrix}3&-4\1&-1\end{bmatrix}$: trace $=3-1=2$, and $\det(A)=3(-1)-(-4)(1)=1$.

$$A^2-2A+I=0$$

2. This means $(A-I)^2=0$, i.e. $A-I$ is **nilpotent**. This special structure gives a simple closed formula for every power of $A$, provable by induction from $A^2=2A-I$.

$$A^n=nA-(n-1)I$$

3. Apply this formula for $n=15$.

$$A^{15}=15A-14I$$

4. Add matrix $B$ to this result to get the combined matrix.

$$A^{15}+B=\begin{pmatrix}2&-11\1&-11\end{pmatrix}$$

5. Set up the equation $(A^{15}+B)(x,y)^T=(0,0)^T$ using this result. Using the first row:

$$2x-11y=0\Rightarrow2x=11y$$

6. Check which option satisfies this relation: for $x=11, y=2$: $2(11)=22$ and $11(2)=22$. ✓ These match.

Hence, the answer is **C ($x=11, y=2$)**.