FM June 2023 p11 q01
4199
Let \(\mathbf{A} = \begin{pmatrix} 3 & 0 \\ 1 & 1 \end{pmatrix}\).
(a) Prove by mathematical induction that, for all positive integers \(n\),
\(2\mathbf{A}^n = \begin{pmatrix} 2 \times 3^n & 0 \\ 3^n - 1 & 2 \end{pmatrix}.\)
(b) Find, in terms of \(n\), the inverse of \(\mathbf{A}^n\).
Solution
(a) Base case: For \(n = 1\),
\(2\mathbf{A} = \begin{pmatrix} 6 & 0 \\ 2 & 2 \end{pmatrix} = \begin{pmatrix} 2 \times 3^1 & 0 \\ 3^1 - 1 & 2 \end{pmatrix}.\)
Assume true for \(n = k\), so
\(2\mathbf{A}^k = \begin{pmatrix} 2 \times 3^k & 0 \\ 3^k - 1 & 2 \end{pmatrix}.\)
Then for \(n = k + 1\),
\(2\mathbf{A}^{k+1} = \begin{pmatrix} 2 \times 3^{k+1} & 0 \\ 3^{k+1} - 3 + 2 & 2 \end{pmatrix}.\)
Thus, true for \(n = k + 1\). By induction, true for all positive integers.
(b) \(\det \mathbf{A}^n = \det \begin{pmatrix} 3^n & 0 \\ \frac{1}{2}(3^n - 1) & 1 \end{pmatrix} = 3^n\).
\(\mathbf{A}^{-n} = 3^{-n} \begin{pmatrix} 1 & 0 \\ \frac{1}{2}(1 - 3^n) & 3^n \end{pmatrix}\).
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