Matrix $A$ Power Calculation

We are given the matrix $A = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}$. Our task is to find the value of $A^8$. We will achieve this by computing the powers of the matrix $A$ sequentially.

Calculating $A^2$

First, we compute $A^2$ by multiplying the matrix $A$ by itself:

$ A^2 = A \times A = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix} $

Performing the matrix multiplication step-by-step:

$ A^2 = \begin{pmatrix} (1 \times 1 + 2 \times 2) & (1 \times 2 + 2 \times -1) \\ (2 \times 1 + -1 \times 2) & (2 \times 2 + -1 \times -1) \end{pmatrix} $

$ A^2 = \begin{pmatrix} 1 + 4 & 2 - 2 \\ 2 - 2 & 4 + 1 \end{pmatrix} $

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

We observe that the resulting matrix $A^2$ is $5$ times the identity matrix $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. So, we can write $A^2 = 5I$. This observation simplifies the calculation of higher powers.

Calculating $A^4$

Next, we find $A^4$ by squaring the matrix $A^2$:

$ A^4 = (A^2)^2 = (5I)^2 $

Using the properties of scalar multiplication with matrices, specifically $(cM)^n = c^n M^n$, and the fact that $I^n = I$ for any positive integer $n$:

$ A^4 = 5^2 \times I^2 = 25 \times I $

$ A^4 = 25I = \begin{pmatrix} 25 & 0 \\ 0 & 25 \end{pmatrix} $

Calculating $A^8$

Finally, we compute $A^8$ by squaring the matrix $A^4$:

$ A^8 = (A^4)^2 = (25I)^2 $

Applying the same properties as in the previous step:

$ A^8 = 25^2 \times I^2 = 625 \times I $

$ A^8 = 625I = \begin{pmatrix} 625 & 0 \\ 0 & 625 \end{pmatrix} $

Thus, the matrix $A^8$ is found to be $\begin{pmatrix} 625 & 0 \\ 0 & 625 \end{pmatrix}$.