Let $A$ be a $3\times 3$ diagonalizable matrix whose eigenvalues are $\lambda_1 = -3$, $\lambda_2 = 1$, and $\lambda_3 = 3$. If

$$\mathbf{v_1} = \left[\begin{array}{c}1 & 0 & 0 \end{array}\right], \quad \mathbf{v_2} = \left[\begin{array}{c}1 & 1 & 0 \end{array}\right], \quad \mathbf{v_3} = \left[\begin{array}{c} 0 & 1 & 1 \end{array}\right]$$

are eigenvectors of $A$ corresponding to $\lambda_1$, $\lambda_2$, and $\lambda_3$, respectively, then factor $A$ into a product $XDX^{-1}$ with $D$ diagonal, and use this factorization to find $A^5$.

$A^5 =$

$\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\!\strut\\\end{array}}\right[$ $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\!\strut\\\end{array}}\right]$