Given,

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

To Prove :- $A^8 = 625I$

Proof :-

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

The characteristic equation of A will be |A - λI| = 0

$\implies \lambda^2 – 5 = 0$

$\implies \lambda = \pm \sqrt{5}$

Now, we have to prove that -

$\implies A^2 - 5I = 0$

$ \implies A^2 – 5I = 0$

$\implies A^2 = 5I$

Now, Taking the LHS we have,

$\implies A^2 = A \cdot A = \begin{bmatrix} \ 1 & 2 \\ \ 2 & -1 \\ \end{bmatrix} . \begin{bmatrix} \ 1 & 2 \\ \ 2 & -1 \\ \end{bmatrix}$

$\implies A^2= \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} $

$\implies A^2= 5\begin {bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $

$\implies A^2 = 5I$ = RHS

Now, According to the question,

$\implies A^4 = A^2 \cdot A^2 = 25I $

$\implies A^8 = A^4 \cdot A^4 = ? $

$\implies A^8 = 25I \cdot 25I$

$\implies A^8 = 25\begin{bmatrix} \ 1 & 0 \\ \ 0 & 1 \\ \end{bmatrix} \cdot 25\begin{bmatrix} \ 1 & 0 \\ \ 0 & 1 \\ \end{bmatrix}$

$\implies A^8 = 25 \times 25 \cdot \begin{bmatrix} \ 1 & 0 \\ \ 0 & 1 \\ \end{bmatrix} $

$\implies A^8 = 625 \cdot \begin{bmatrix} \ 1 & 0 \\ \ 0 & 1 \\ \end{bmatrix} $

$ \ \ \therefore\ A^8 = 625 I\ Proved.$