Let A = begin{bmatrix} 2 & -2 1 & -1 end{bma | vedclass

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(A) First,calculate $A^2$: $A^2 = \begin{bmatrix} 2 & -2 \ 1 & -1 \end{bmatrix} \begin{bmatrix} 2 & -2 \ 1 & -1 \end{bmatrix} = \begin{bmatrix} 4-2

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$1$

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(A) First,calculate $A^2$: $A^2 = \begin{bmatrix} 2 & -2 \ 1 & -1 \end{bmatrix} \begin{bmatrix} 2 & -2 \ 1 & -1 \end{bmatrix} = \begin{bmatrix} 4-2 & -4+2 \ 2-1 & -2+1 \end{bmatrix} = \begin{bmatrix} 2 & -2 \ 1 & -1 \end{bmatrix} = A$. Since $A^2 = A$,it follows that $A^n = A$ for all $n \geq 1$. Next,calculate $B^2$: $B^2 = \begin{bmatrix} -1 & 2 \ -1 & 2 \end{bmatrix} \begin{bmatrix} -1 & 2 \ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1-2 & -2+4 \ 1-2 & -2+4 \end{bmatrix} = \begin{bmatrix} -1 & 2 \ -1 & 2 \end{bmatrix} = B$. Since $B^2 = B$,it follows that $B^m = B$ for all $m \geq 1$. The given equation $nA^n + mB^m = I$ becomes $nA + mB = I$. Substituting the matrices: $n \begin{bmatrix} 2 & -2 \ 1 & -1 \end{bmatrix} + m \begin{bmatrix} -1 & 2 \ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$. This results in the system of equations: $2n - m = 1$ $-2n + 2m = 0 \implies n = m$. Substituting $n = m$ into $2n - m = 1$,we get $2n - n = 1$,so $n = 1$. Thus,$n = 1$ and $m = 1$. The only pair $(n, m)$ is $(1, 1)$,so there is $1$ element in the set.

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

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