To find matrices $P$ and $D$ such that $A = PDP^{-1}$, we need:

1. The eigenvalues of $A$.
2. The eigenvectors corresponding to these eigenvalues.

Finding eigenvalues:

For a triangular matrix, the eigenvalues are the entries on the diagonal. So, the eigenvalues of $A$ are:

• $\lambda_1 = 1$
• $\lambda_2 = 4$
• $\lambda_3 = 6$

Finding eigenvectors:

1. Eigenvector for $\lambda_1 = 1$:

   Solve $(A - I)x = 0$.

   0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$ $$x_3$$ is free. Choose $$x_3 = 1$$, giving $$x_2 = -5/3$$ and $$x_1 = -1/3$$. Eigenvector: $$\begin{pmatrix} -1/3 \\ -5/3 \\ 1 \end{pmatrix}$$.

2. Eigenvector for $\lambda_2 = 4$:

   Solve $(A - 4I)x = 0$.

   -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$ $$x_3$$ is free. Choose $$x_3 = 1$$, giving $$x_2 = 0$$ and $$x_1 = 1$$. Eigenvector: $$\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$$.

3. Eigenvector for $\lambda_3 = 6$:

   Solve $(A - 6I)x = 0$.

   -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$ $$x_2$$ is free. Choose $$x_2 = 1$$, giving $$x_3 = 2/5$$ and $$x_1 = 1/5$$. Eigenvector: $$\begin{pmatrix} 1/5 \\ 1 \\ 2/5 \end{pmatrix}$$.

Construct matrices $D$ and $P$: