If A=$\left(\begin{array}{cc}1& 2\\ 2& -1\end{array}\right)$ ,then which ONE od the following is A⁸ ?

The correct option is:
$\left(\begin{array}{cc}625& 0\\ 0& 625\end{array}\right)$

To find the value of $A^8$, we can first calculate the square of matrix $A$, denoted as $A^2$.

Given the matrix $A$:
$A=\left(\begin{array}{cc}1& 2\\ 2& -1\end{array}\right)$

We compute $A^2=AA$:
$A^2=\left(\begin{array}{cc}1& 2\\ 2& -1\end{array}\right)\left(\begin{array}{cc}1& 2\\ 2& -1\end{array}\right)$

Performing matrix multiplication:
- The top-left element is $(1\times 1)+(2\times 2)=1+4=5$
- The top-right element is $(1\times 2)+(2\times (-1\left)\right)=2-2=0$
- The bottom-left element is $(2\times 1)+(-1\times 2)=2-2=0$
- The bottom-right element is $(2\times 2)+(-1\times (-1\left)\right)=4+1=5$

Thus, we obtain:
$A^2=\left(\begin{array}{cc}5& 0\\ 0& 5\end{array}\right)=5I$
where $I$ is the $2\times 2$ identity matrix.

Using this result, we can raise both sides to the power of 4 to compute $A^8$:
$A^8={\left(A^2\right)}^4={(5I)}^4$

Since any power of the identity matrix $I$ is still $I$:
$A^8=5^4{I}^4=625I$

Writing this back in matrix form:
$A^8=625\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}625& 0\\ 0& 625\end{array}\right)$