Discussion
If \(A = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}\), then which ONE of the following is \(A^{8}\)?
Answer: C
\(A^{8} = \begin{pmatrix} 625 & 0 \\ 0 & 625 \end{pmatrix}\).
Square A first:
\[ \begin{aligned} A^{2} &= \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} \\ &= \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = 5I \end{aligned} \]
So \(A^{8} = (A^{2})^{4} = 5^{4}I = 625I\).
25I is A4, 125I is A6 and 3125I is A10: each comes from taking the wrong number of factors of 5I.
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