Given `A=[{:(5,0,4),(2,3,2),(1,2,1):}]`, `B^(-1)=[{:(1,2,3),(1,4,3),(1,3,4):}]`. Compute `(AB)^(-1)`.

Answer

To compute \((AB)^{-1}\), we can use the property of inverses that states:

\[ (AB)^{-1} = B^{-1}A^{-1} \]

Given:
- \( A = \begin{pmatrix} 5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1 \end{pmatrix} \)
- \( B^{-1} = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{pmatrix} \)

### Step 1: Calculate the Determinant of A

To find \( A^{-1} \), we first need to calculate the determinant of \( A \).

\[ \text{det}(A) = 5 \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} - 0 + 4 \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} \]

Calculating the 2x2 determinants:

\[ \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} = (3)(1) - (2)(2) = 3 - 4 = -1 \]

\[ \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = (2)(2) - (3)(1) = 4 - 3 = 1 \]

Now substituting back:

\[ \text{det}(A) = 5(-1) + 4(1) = -5 + 4 = -1 \]

### Step 2: Calculate the Adjoint of A

The adjoint of \( A \) is the transpose of the cofactor matrix. We calculate the cofactors:

1. For \( A_{11} \):
\[ C_{11} = \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} = -1 \quad \Rightarrow \quad C_{11} = -1 \]
2. For \( A_{12} \):
\[ C_{12} = -\begin{vmatrix} 2 & 2 \\ 1 & 1 \end{vmatrix} = -0 = 0 \]
3. For \( A_{13} \):
\[ C_{13} = \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = 1 \]
4. For \( A_{21} \):
\[ C_{21} = -\begin{vmatrix} 0 & 4 \\ 2 & 1 \end{vmatrix} = -(-8) = 8 \]
5. For \( A_{22} \):
\[ C_{22} = \begin{vmatrix} 5 & 4 \\ 1 & 1 \end{vmatrix} = 5 - 4 = 1 \]
6. For \( A_{23} \):
\[ C_{23} = -\begin{vmatrix} 5 & 0 \\ 1 & 2 \end{vmatrix} = -10 \]
7. For \( A_{31} \):
\[ C_{31} = \begin{vmatrix} 0 & 4 \\ 3 & 2 \end{vmatrix} = 0 - 12 = -12 \]
8. For \( A_{32} \):
\[ C_{32} = -\begin{vmatrix} 5 & 4 \\ 2 & 2 \end{vmatrix} = -2 \]
9. For \( A_{33} \):
\[ C_{33} = \begin{vmatrix} 5 & 0 \\ 2 & 3 \end{vmatrix} = 15 \]

The cofactor matrix is:

\[ \text{Cof}(A) = \begin{pmatrix} -1 & 0 & 1 \\ 8 & 1 & -10 \\ -12 & -2 & 15 \end{pmatrix} \]

Taking the transpose to get the adjoint:

\[ \text{adj}(A) = \begin{pmatrix} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{pmatrix} \]

### Step 3: Calculate \( A^{-1} \)

Using the formula:

\[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) = -1 \cdot \text{adj}(A) \]

Thus:

\[ A^{-1} = \begin{pmatrix} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{pmatrix} \]

### Step 4: Compute \( (AB)^{-1} \)

Now we can compute:

\[ (AB)^{-1} = B^{-1} A^{-1} \]

Substituting the matrices:

\[ B^{-1} = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{pmatrix}, \quad A^{-1} = \begin{pmatrix} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{pmatrix} \]

Now we multiply these two matrices:

\[ (AB)^{-1} = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{pmatrix} \begin{pmatrix} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{pmatrix} \]

Calculating the product:

1. First row:
- \( 1 \cdot 1 + 2 \cdot 0 + 3 \cdot (-1) = 1 + 0 - 3 = -2 \)
- \( 1 \cdot (-8) + 2 \cdot (-1) + 3 \cdot 10 = -8 - 2 + 30 = 20 \)
- \( 1 \cdot 12 + 2 \cdot 2 + 3 \cdot (-15) = 12 + 4 - 45 = -29 \)

2. Second row:
- \( 1 \cdot 1 + 4 \cdot 0 + 3 \cdot (-1) = 1 + 0 - 3 = -2 \)
- \( 1 \cdot (-8) + 4 \cdot (-1) + 3 \cdot 10 = -8 - 4 + 30 = 18 \)
- \( 1 \cdot 12 + 4 \cdot 2 + 3 \cdot (-15) = 12 + 8 - 45 = -25 \)

3. Third row:
- \( 1 \cdot 1 + 3 \cdot 0 + 4 \cdot (-1) = 1 + 0 - 4 = -3 \)
- \( 1 \cdot (-8) + 3 \cdot (-1) + 4 \cdot 10 = -8 - 3 + 40 = 29 \)
- \( 1 \cdot 12 + 3 \cdot 2 + 4 \cdot (-15) = 12 + 6 - 60 = -42 \)

Thus, we have:

\[ (AB)^{-1} = \begin{pmatrix} -2 & 20 & -29 \\ -2 & 18 & -25 \\ -3 & 29 & -42 \end{pmatrix} \]

### Final Answer

\[ (AB)^{-1} = \begin{pmatrix} -2 & 20 & -29 \\ -2 & 18 & -25 \\ -3 & 29 & -42 \end{pmatrix} \]