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 $\{(n, m) : n, m \in \{1, 2, \ldots, 10\}$ and $nA^n + mB^m = I\}$ is _____.

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Aicharya · Accepted Answer

Given, $A=\begin{pmatrix} 2 & -2 \ 1 & -1 \end{pmatrix}$. So, $A^2=\begin{pmatrix} 2 & -2 \ 1 & -1 \end{pmatrix}\begin{pmatrix} 2 & -2 \ 1 & -1 \end{pmatrix}=\begin{pmatrix} 2 & -2 \ 1 & -1 \end{pmatrix}=A$. Also, $B=\begin{pmatrix} -1 & 2 \ -1 & 2 \end{pmatrix}$. $B^2=\begin{pmatrix} -1 & 2 \ -1 & 2 \end{pmatrix}\begin{pmatrix} -1 & 2 \ -1 & 2 \end{pmatrix}=\begin{pmatrix} -1 & 2 \ -1 & 2 \end{pmatrix}=B$. So, $A+B=\begin{pmatrix} 2 & -2 \ 1 & -1 \end{pmatrix}+\begin{pmatrix} -1 & 2 \ -1 & 2 \end{pmatrix}=\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}=I$. Therefore the equation $nA^n+mB^m=I$ is true for $n=1$ and $m=1$ so, only one set is possible.

Let $A = \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix}$ and $B = \begin{pmatrix} -1 & 2 \\ -1 & 2…

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