Apply Cayley – Hamilton theorm to A = $\begin{bmatrix} 1& 2 \\ 2&-1 \end{bmatrix} $ & hence find A$^8$

Subject: Applied Mathematics 4

Topic: Matrix Theory

Difficulty: Medium

A = $\begin{bmatrix} 1& 2 \\ 2&-1 \end{bmatrix} $

For characteristic equation |A - $\lambda$ I| = 0

$$ \begin{vmatrix} 1-\lambda& 2 \\ 2&-1-\lambda \end{vmatrix} = 0 $$

$ (1-\lambda)(-1-\lambda) - 4 = 0 \\ -(1-\lambda)(1+\lambda) - 4 = 0 \\ -(1-\lambda^2) - 4 = 0 \\ \lambda^2 - 5 = 0 \\ \therefore A^2 - 5I = 0 \\ \therefore A^2 = 5I \\ A^8 = (A^2)^4 = (5I)^4 = 5^4I^4 = 625I $