Question

Let \(A = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}\), \(x \in \mathbb{R}\) and \(A^4 = [a_{ij}]\). If \(a_{11} = 109\), then \(a_{22}\) is equal to _______ .

Hint:

Compute \(A^2\) first, then \(A^4 = (A^2)^2\), and use \(a_{11} = 109\) to find \(x^2\).

Answer 1

Correct Answer: 10

\(A^2 = \begin{bmatrix} x^2 + 1 & x \\ x & 1 \end{bmatrix}\).
\(A^4 = (A^2)^2\), so \(a_{11} = (x^2 + 1)^2 + x^2\) and \(a_{22} = x^2 + 1\).
Given \((x^2 + 1)^2 + x^2 = 109\). Let \(u = x^2 + 1\): \(u^2 + u - 1 = 109\), so \(u^2 + u - 110 = 0\), \((u + 11)(u - 10) = 0\).
\(u = 10\) (since \(u = x^2 + 1 \ge 1\)).
So \(a_{22} = x^2 + 1 = 10\).
Answer: 10