If A = [100−1],\begin{bmatrix}[r] 1 & 0 \ 0 & -1 \end{bmatrix},[100−1], find A2 and A3. Also state which of these is equal to A.
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A2=[100−1][100−1]=[1×1+0×01×0+0×(−1)0×1+(−1)×00×0+(−1)×(−1)]=[1001]A^2 = \begin{bmatrix}[r] 1 & 0 \ 0 & -1 \end{bmatrix} \begin{bmatrix}[r] 1 & 0 \ 0 & -1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 \times 1 + 0 \times 0 & 1 \times 0 + 0 \times (-1) \ 0 \times 1 + (-1) \times 0 & 0 \times 0 + (-1) \times (-1) \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix}A2=[100−1][100−1]=[1×1+0×00×1+(−1)×01×0+0×(−1)0×0+(−1)×(−1)]=[1001]
A3=A2×A=[1001][100−1]=[1×1+0×01×0+0×(−1)0×1+1×00×0+1×(−1)][1+00+00+00−1]=[100−1]A^3 = A^2 \times A \\[1em] = \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \begin{bmatrix}[r] 1 & 0 \ 0 & -1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 \times 1 + 0 \times 0 & 1 \times 0 + 0 \times (-1) \ 0 \times 1 + 1 \times 0 & 0 \times 0 + 1 \times (-1) \end{bmatrix} \\[1em] \begin{bmatrix}[r] 1 + 0 & 0 + 0 \ 0 + 0 & 0 -1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 & 0 \ 0 & -1 \end{bmatrix}A3=A2×A=[1001][100−1]=[1×1+0×00×1+1×01×0+0×(−1)0×0+1×(−1)][1+00+00+00−1]=[100−1]
Hence, the matrix A2=[1001] and A3=[100−1].A^2 = \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \text{ and } A^3 = \begin{bmatrix}[r] 1 & 0 \ 0 & -1 \end{bmatrix}.A2=[1001] and A3=[100−1]. Thus, A3 = A.
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