FM June 2023 p12 q01
4206
Let \(A = \begin{pmatrix} 3 & 0 \\ 1 & 1 \end{pmatrix}\).
(a) Prove by mathematical induction that, for all positive integers \(n\),
\(2A^n = \begin{pmatrix} 2 \times 3^n & 0 \\ 3^n - 1 & 2 \end{pmatrix}.\)
(b) Find, in terms of \(n\), the inverse of \(A^n\).
Solution
(a) Base case: For \(n = 1\),
\(2A = \begin{pmatrix} 6 & 0 \\ 2 & 2 \end{pmatrix} = \begin{pmatrix} 2 \times 3^1 & 0 \\ 3^1 - 1 & 2 \end{pmatrix}.\)
Assume true for \(n = k\), so \(2A^k = \begin{pmatrix} 2 \times 3^k & 0 \\ 3^k - 1 & 2 \end{pmatrix}\).
Then for \(n = k+1\),
\(2A^{k+1} = \begin{pmatrix} 2 \times 3^k & 0 \\ 3^k - 1 & 2 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 \times 3^{k+1} & 0 \\ 3^{k+1} - 3 + 2 & 2 \end{pmatrix}.\)
Thus, true for \(n = k+1\). By induction, true for all positive integers.
(b) \(\det A^n = \det \begin{pmatrix} 3^n & 0 \\ \frac{1}{2}(3^n - 1) & 1 \end{pmatrix} = 3^n\).
\(A^{-n} = 3^{-n} \begin{pmatrix} 1 & 0 \\ \frac{1}{2}(1 - 3^n) & 3^n \end{pmatrix}\).
Log in to record attempts.
