<p>Let <em>a</em> and <em>b</em> be two real numbers such that \(a > 1\), \(b > 1\). If \(A = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}\), then \(\lim_{n \to \infty} (A^n)^{-1}\) is</p>
Step-by-Step Solution
Key Concept: For a diagonal matrix with entries greater than 1, as n→∞, A^n has entries that grow without bound, making (A^n)^{-1} approach a zero matrix. The inverse of a diagonal matrix with large entries becomes a diagonal matrix with entries approaching zero.
<p><strong>Step 1:</strong> For diagonal matrix A = [[a, 0], [0, b]], compute A^n directly:</p><p>A^n = [[a^n, 0], [0, b^n]]</p><p><strong>Step 2:</strong> Find the inverse of A^n:</p><p>(A^n)^(-1) = [[1/a^n, 0], [0, 1/b^n]]</p><p><strong>Step 3:</strong> Take the limit as n→∞. Since a > 1 and b > 1:</p><p>- 1/a^n → 0 as n→∞</p><p>- 1/b^n → 0 as n→∞</p><p><strong>Step 4:</strong> Therefore:</p><p>lim(n→∞) (A^n)^(-1) = [[0, 0], [0, 0]] = O (zero matrix)</p><p>∴ Answer: B</p>
Correct Answer: B