For the matrices A=pmatrix3&-41&-1pmatrix and B=pmatrix-29&49-13&18pmatrix, if (A^15+B)pma…

Question

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

Accepted Answer

The characteristic equation of $A$ is $(3-\lambda)(-1-\lambda)+4=0 \Rightarrow \lambda^2-2\lambda+1=0$, so by Cayley--Hamilton $A^2=2A-I$.

Using this recursively: $A^3=3A-2I$, $A^4=5A-4I$, and in general $A^n=nA-(n-1)I$, so $A^{15}=15A-14I$.

Then $A^{15}+B=15A-14I+B$, which works out to a matrix proportional to $\begin{pmatrix}2&-11\1&-11\end{pmatrix}$ (up to the given $B$), so the condition $(A^{15}+B)\begin{pmatrix}x\y\end{pmatrix}=0$ reduces to $2x=11y$.

Among the options, only $x=11,\ y=2$ satisfies $2(11)=11(2)=22$. So the answer is C.

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