The problem shown involves diagonalizing the matrix $A$ and then finding the matrix $A^4$. The matrix $A$ is given as:

$A = \begin{bmatrix} 6 & -2 \\ 2 & 3 \end{bmatrix}$

Step 1: Finding the Eigenvalues of A We start by finding the eigenvalues, $\lambda$, which satisfy the characteristic equation $\det(A - \lambda I) = 0$.

$A - \lambda I = \begin{bmatrix} 6 - \lambda & -2 \\ 2 & 3 - \lambda \end{bmatrix}$ $\text{Determinant} = (6-\lambda)(3-\lambda) - (-2)(2) = \lambda^2 - 9\lambda + 22$

Solving $\lambda^2 - 9\lambda + 22 = 0$ will give us the eigenvalues.

Step 2: Finding the Eigenvectors of A For each eigenvalue $\lambda$, solve $(A - \lambda I)\vec{v} = 0$ for the eigenvector $\vec{v}$.

Step 3: Diagonalizing A Form the matrix $P$ from the eigenvectors of $A$ and the diagonal matrix $D$ from the eigenvalues.

Step 4: Calculating $A^4$ $A^4 = PDP^{-1}PDP^{-1}PDP^{-1}PDP^{-1} = PD^4P^{-1}$ Here, $D^4$ is simply the diagonal matrix with each eigenvalue raised to the fourth power.

Let's proceed with these calculations now, starting with the eigenvalues.