Let A=pmatrix2&-21&-1pmatrix and B=pmatrix-1&2-1&2pmatrix.

Question

Let $A=\begin{pmatrix}2&-2\1&-1\end{pmatrix}$ and $B=\begin{pmatrix}-1&2\-1&2\end{pmatrix}$. Then the number of elements in the set $\{(n,m): n,m\in\{1,\dots,10\},\ nA^n+mB^m=I\}$ is\_\_\_

Answer: 1

Accepted Answer

Direct computation shows $A^2=A$ and $B^2=B$ (both idempotent), so $A^k=A$ and $B^k=B$ for all $k\ge1$.

So $nA^n+mB^m=nA+mB=\begin{pmatrix}2n-m&-2n+2m\n-m&-n+2m\end{pmatrix}=I$ gives $2n-m=1,\ -n+m=0,\ -n+2m=1$, i.e. $m=n$ and $2n-n=1\Rightarrow n=1$.

So $(m,n)=(1,1)$ is the unique solution: 1 element.

Solutions · 1