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To solve the given matrix problem, we will follow these steps:

You are given two matrices:
\[
A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 3 & -1 \\ 0 & -1 & 3 \end{pmatrix}, 
B = \begin{pmatrix} 4 & 2 & 1 \\ -3 & 0 & 5 \\ -1 & 1 & 6 \end{pmatrix}.
\]

We are required to:
1. Compute the product \( AB \) and \( BA \).
2. Find the inverses \( A^{-1} \) and \( B^{-1} \).
3. Use the results to compute \((AB)^{-1}\) and \((BA)^{-1}\).

Let's start with the matrix multiplication \( AB \) and \( BA \).

### Step 1: Multiply \( A \) and \( B \) to find \( AB \)
The product of two matrices \( A \) and \( B \) is obtained by the formula:
\[
AB = \begin{pmatrix} a_{11}b_{11} + a_{12}b_{21} + a_{13}b_{31} & a_{11}b_{12} + a_{12}b_{22} + a_{13}b_{32} & a_{11}b_{13} + a_{12}b_{23} + a_{13}b_{33} \\
                     a_{21}b_{11} + a_{22}b_{21} + a_{23}b_{31} & a_{21}b_{12} + a_{22}b_{22} + a_{23}b_{32} & a_{21}b_{13} + a_{22}b_{23} + a_{23}b_{33} \\
                     a_{31}b_{11} + a_{32}b_{21} + a_{33}b_{31} & a_{31}b_{12} + a_{32}b_{22} + a_{33}b_{32} & a_{31}b_{13} + a_{32}b_{23} + a_{33}b_{33} \end{pmatrix}
\]

### Step 2: Multiply \( B \) and \( A \) to find \( BA \)
Similarly, you will find the matrix \( BA \) using the same multiplication procedure but reversing the order.

### Step 3: Calculate inverses
Once we compute the products \( AB \) and \( BA \), we'll proceed to find the inverses \( A^{-1} \) and \( B^{-1} \) using the formula for the inverse of a 3x3 matrix. The inverse of a matrix \( A \) is given by:
\[
A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A),
\]
where \(\text{det}(A)\) is the determinant of \(A\), and \(\text{adj}(A)\) is the adjugate of \(A\).

I'll calculate each of these values for you now.Here are the results of the calculations:

1. **Product \( AB \):**
   \[
   AB = \begin{pmatrix} -3 & 3 & 17 \\ 0 & 3 & 11 \\ 0 & 3 & 13 \end{pmatrix}
   \]

2. **Product \( BA \):**
   \[
   BA = \begin{pmatrix} 8 & 13 & 5 \\ -3 & -11 & 12 \\ 1 & -5 & 16 \end{pmatrix}
   \]

3. **Inverse of \( A \) (\( A^{-1} \)):**
   \[
   A^{-1} = \begin{pmatrix} -1.333 & 1.167 & 0.833 \\ 1 & -0.5 & -0.5 \\ 0.333 & -0.167 & 0.167 \end{pmatrix}
   \]

4. **Inverse of \( B \) (\( B^{-1} \)):**
   \[
   B^{-1} = \begin{pmatrix} -1.667 & -3.667 & 3.333 \\ 4.333 & 8.333 & -7.667 \\ -1 & -2 & 2 \end{pmatrix}
   \]

5. **Inverse of \( AB \) (\( (AB)^{-1} \)):**
   \[
   (AB)^{-1} = \begin{pmatrix} -0.333 & -0.667 & 1 \\ 0 & 2.167 & -1.833 \\ 0 & -0.5 & 0.5 \end{pmatrix}
   \]

6. **Inverse of \( BA \) (\( (BA)^{-1} \)):**
   \[
   (BA)^{-1} = \begin{pmatrix} 6.444 & 12.944 & -11.722 \\ -3.333 & -6.833 & 6.167 \\ -1.444 & -2.944 & 2.722 \end{pmatrix}
   \]

Would you like further details on any step? Here are some related questions:

1. How do you calculate the determinant of a 3x3 matrix?
2. What is the significance of matrix inverses in linear algebra?
3. Can two matrices always be multiplied together? Why or why not?
4. How do you interpret matrix multiplication geometrically?
5. How can the properties of inverse matrices be used in solving systems of linear equations?

**Tip:** The inverse of a matrix is unique only if the matrix is square and its determinant is non-zero.

If A = [[1, 2, 1], [2, 3, -1], [0, -1, 3]] and B = [[4, 2, 1], [-3, 0, 5], [-1, 1, 6]], find the following: AB, BA, A^-1, B^-1, (AB)^-1, (BA)^-1.

Learn how to multiply matrices A and B and find their inverses in this detailed solution. The problem includes matrix multiplication, inversion, and the use of determinants to compute inverse matrices.

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