If `A^(-1)=|{:(1,3,3),(1,4,3),(1,3,4):}|" and B"=[{:(5,0,4),(2,3,2),(1,2,1):}]," then find"(AB)^(-1)`

Answer

To find \((AB)^{-1}\), we can use the property of inverses of matrices, which states that \((AB)^{-1} = B^{-1}A^{-1}\). Given that \(A^{-1}\) and \(B\) are provided, we will first find \(B^{-1}\) and then compute \((AB)^{-1}\).

### Step 1: Write down the given matrices
We have:
\[ A^{-1} = \begin{pmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{pmatrix} \]
\[ B = \begin{pmatrix} 5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1 \end{pmatrix} \]

### Step 2: Find the determinant of matrix \(B\)
To find \(B^{-1}\), we first need to calculate the determinant of \(B\):
\[ \text{det}(B) = 5 \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} - 0 + 4 \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} \]
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:
\[ \text{det}(B) = 5(-1) + 0 + 4(1) = -5 + 4 = -1 \]

### Step 3: Find the adjoint of matrix \(B\)
To find \(B^{-1}\), we also need the adjoint of \(B\). The adjoint is obtained by finding the cofactor matrix and then transposing it.

The cofactor matrix \(C\) is calculated as follows:
- For \(C_{11}\): \(\begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} = -1\)
- For \(C_{12}\): \(-\begin{vmatrix} 2 & 2 \\ 1 & 1 \end{vmatrix} = 0\)
- For \(C_{13}\): \(\begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = 1\)

Continuing this process for all elements gives us the cofactor matrix:
\[ C = \begin{pmatrix} -1 & 0 & 1 \\ -2 & -1 & 2 \\ 6 & -5 & 5 \end{pmatrix} \]
Now, we take the transpose to get the adjoint:
\[ \text{adj}(B) = C^T = \begin{pmatrix} -1 & -2 & 6 \\ 0 & -1 & -5 \\ 1 & 2 & 5 \end{pmatrix} \]

### Step 4: Calculate \(B^{-1}\)
Now we can find \(B^{-1}\):
\[ B^{-1} = \frac{1}{\text{det}(B)} \cdot \text{adj}(B) = -1 \cdot \begin{pmatrix} -1 & -2 & 6 \\ 0 & -1 & -5 \\ 1 & 2 & 5 \end{pmatrix} = \begin{pmatrix} 1 & 2 & -6 \\ 0 & 1 & 5 \\ -1 & -2 & -5 \end{pmatrix} \]

### Step 5: Calculate \((AB)^{-1}\)
Now we can find \((AB)^{-1} = B^{-1}A^{-1}\):
\[ B^{-1} = \begin{pmatrix} 1 & 2 & -6 \\ 0 & 1 & 5 \\ -1 & -2 & -5 \end{pmatrix} \]
\[ A^{-1} = \begin{pmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{pmatrix} \]
Now we multiply \(B^{-1}\) and \(A^{-1}\):
\[ (AB)^{-1} = B^{-1}A^{-1} = \begin{pmatrix} 1 & 2 & -6 \\ 0 & 1 & 5 \\ -1 & -2 & -5 \end{pmatrix} \begin{pmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{pmatrix} \]
Calculating the product:
1. First row:
- \(1*1 + 2*1 + (-6)*1 = 1 + 2 - 6 = -3\)
- \(1*3 + 2*4 + (-6)*3 = 3 + 8 - 18 = -7\)
- \(1*3 + 2*3 + (-6)*4 = 3 + 6 - 24 = -15\)

2. Second row:
- \(0*1 + 1*1 + 5*1 = 0 + 1 + 5 = 6\)
- \(0*3 + 1*4 + 5*3 = 0 + 4 + 15 = 19\)
- \(0*3 + 1*3 + 5*4 = 0 + 3 + 20 = 23\)

3. Third row:
- \(-1*1 + (-2)*1 + (-5)*1 = -1 - 2 - 5 = -8\)
- \(-1*3 + (-2)*4 + (-5)*3 = -3 - 8 - 15 = -26\)
- \(-1*3 + (-2)*3 + (-5)*4 = -3 - 6 - 20 = -29\)

Thus, we have:
\[ (AB)^{-1} = \begin{pmatrix} -3 & -7 & -15 \\ 6 & 19 & 23 \\ -8 & -26 & -29 \end{pmatrix} \]

### Final Answer
\[ (AB)^{-1} = \begin{pmatrix} -3 & -7 & -15 \\ 6 & 19 & 23 \\ -8 & -26 & -29 \end{pmatrix} \]