Let $A=\left(\begin{array}{ll}2 & -2 \ 1 & -1\end{array} ight)$ and $B=\left(\begin{array}{ll}-1 &

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Let $A=\left(\begin{array}{ll}2 & -2 \ 1 & -1\end{array}ight)$ and $B=\left(\begin{array}{ll}-1 & 2 \ -1 & 2\end{array}ight)$. Then the number of elements in the set $\{(n, m): n, m \in\{1,2, \ldots \ldots .10\}$ and $\left.n A^{n}+m B^{m}=Iight\}$ is

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(1) : $A=\left[\begin{array}{ll}2 & -2 \ 1 & -1\end{array}ight]$\ $\Rightarrow A^{2}=\left[\begin{array}{ll}2 & -2 \ 1 & -1\end{array}ight]\left[\begin{array}{ll}2 & -2 \ 1 & -1\end{array}ight]=\left[\begin{array}{ll}2 & -2 \ 1 & -1\end{array}ight]=A$\ $	herefore \quad A^{3}=A^{2}=A \ldots$\ So, $\quad A^{k}=A$, for each value of $k, k \in N$\ Now, $B=\left[\begin{array}{ll}-1 & 2 \ -1 & 2\end{array}ight] \Rightarrow B^{3}=B^{2}=B \ldots .$.\ So, $B^{k}=B$ for each value of $k, k \in N$\ $	herefore n A^{n}+m B^{m}=I \Rightarrow n A+m B=I$\ $\Rightarrow\left[\begin{array}{cc}2 n & -2 n \ n & -n\end{array}ight]+\left[\begin{array}{ll}-m & 2 m \ -m & 2 m\end{array}ight]=\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}ight]$\ $\Rightarrow \quad 2 n-m=1,2 m-2 n=0, n-m=0,2 m$ $-n=1 \Rightarrow m=n=1$\ So, only 1 element exists.\