Given two matrices $A$ and $D$, we are tasked to find matrices $B$ and $C$ such that:

$AB = D \quad \text{and} \quad CA = D$

The matrix $A$ and $D$ are defined as:

$A = \begin{bmatrix} -1 & -3 & 1 \\ -3 & -4 & -4 \\ 2 & 2 & 3 \end{bmatrix}$

$D = \begin{bmatrix} 5 & 0 & 2 \\ 3 & 3 & 3 \\ 0 & -2 & 1 \end{bmatrix}$

We will now solve the two matrix equations separately:

1. Finding matrix $B$ such that $AB = D$:

To find matrix $B$, we need to solve the system:

$AB = D$

This can be rewritten as:

$B = A^{-1}D$

So, we first need to calculate the inverse of $A$, $A^{-1}$, and then multiply it by $D$ to get matrix $B$.

2. Finding matrix $C$ such that $CA = D$:

Similarly, to find matrix $C$, we solve the system:

$CA = D$

This can be rewritten as:

$C = DA^{-1}$

Thus, we need to compute $A^{-1}$ once and then use it for both $B$ and $C$.

I will now proceed with calculating $A^{-1}$ and then find the matrices $B$ and $C$.### Results:

1. The inverse of matrix $A$ is:

-1.333 & 3.667 & 5.333 \\ 0.333 & -1.667 & -2.333 \\ 0.667 & -1.333 & -1.667 \end{bmatrix}$$ 2. The matrix $$B$$ such that $$AB = D$$ is: $$B = \begin{bmatrix} 4.333 & 0.333 & 13.667 \\ -3.333 & -0.333 & -6.667 \\ -0.667 & -0.667 & -4.333 \end{bmatrix}$$ 3. The matrix $$C$$ such that $$CA = D$$ is: $$C = \begin{bmatrix} -5.333 & 15.667 & 23.333 \\ -1.000 & 2.000 & 4.000 \\ 0.000 & 2.000 & 3.000 \end{bmatrix}$$ Would you like more details on how these matrices were calculated or any additional clarifications? --- Here are five related questions for further exploration: 1. How do you calculate the inverse of a 3x3 matrix manually? 2. What are the conditions under which a matrix has an inverse? 3. Can you explain why matrix multiplication is not commutative? 4. What is the geometrical interpretation of matrix multiplication? 5. How do you solve a system of linear equations using matrices? **Tip**: Always check if a matrix is invertible by ensuring its determinant is non-zero!