The problem asks to compute the matrix product $CD$, where $C = \begin{bmatrix} 9 & 0 & 4 \\ 1 & 13 & -1 \\ 9 & 0 & 4 \end{bmatrix}$ and $D = \begin{bmatrix} 4 & 4 & -8 \\ -1 & -1 & 2 \\ -9 & -9 & 18 \end{bmatrix}$.

1. Problem Description

The problem asks to compute the matrix product $CD$, where $C = \begin{bmatrix} 9 & 0 & 4 \\ 1 & 13 & -1 \\ 9 & 0 & 4 \end{bmatrix}$ and $D = \begin{bmatrix} 4 & 4 & -8 \\ -1 & -1 & 2 \\ -9 & -9 & 18 \end{bmatrix}$.

2. Solution Steps

To compute the matrix product $CD$, we multiply each row of $C$ by each column of $D$.

First row of $C$ times first column of $D$: $(9)(4) + (0)(-1) + (4)(-9) = 36 + 0 - 36 = 0$.

First row of $C$ times second column of $D$: $(9)(4) + (0)(-1) + (4)(-9) = 36 + 0 - 36 = 0$.

First row of $C$ times third column of $D$: $(9)(-8) + (0)(2) + (4)(18) = -72 + 0 + 72 = 0$.

Second row of $C$ times first column of $D$: $(1)(4) + (13)(-1) + (-1)(-9) = 4 - 13 + 9 = 0$.

Second row of $C$ times second column of $D$: $(1)(4) + (13)(-1) + (-1)(-9) = 4 - 13 + 9 = 0$.

Second row of $C$ times third column of $D$: $(1)(-8) + (13)(2) + (-1)(18) = -8 + 26 - 18 = 0$.

Third row of $C$ times first column of $D$: $(9)(4) + (0)(-1) + (4)(-9) = 36 + 0 - 36 = 0$.

Third row of $C$ times second column of $D$: $(9)(4) + (0)(-1) + (4)(-9) = 36 + 0 - 36 = 0$.

Third row of $C$ times third column of $D$: $(9)(-8) + (0)(2) + (4)(18) = -72 + 0 + 72 = 0$.

Therefore, the matrix $CD$ is $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.

3. Final Answer

$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$