If `A = [(2,4),(1,3)] and B = [(1,1),(0,1)]` then find `(A^(-1)B^(-1))`

Answer

To find \( A^{-1}B^{-1} \) where \( A = \begin{pmatrix} 2 & 4 \\ 1 & 3 \end{pmatrix} \) and \( B = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \), we can use the property that \( A^{-1}B^{-1} = (BA)^{-1} \).

### Step-by-Step Solution:

1. **Calculate the product \( BA \)**:
\[ B = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \quad A = \begin{pmatrix} 2 & 4 \\ 1 & 3 \end{pmatrix} \]
To find \( BA \), we multiply the matrices:
\[ BA = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 4 \\ 1 & 3 \end{pmatrix} \]

- For the first row, first column:
\[ 1 \cdot 2 + 1 \cdot 1 = 2 + 1 = 3 \]
- For the first row, second column:
\[ 1 \cdot 4 + 1 \cdot 3 = 4 + 3 = 7 \]
- For the second row, first column:
\[ 0 \cdot 2 + 1 \cdot 1 = 0 + 1 = 1 \]
- For the second row, second column:
\[ 0 \cdot 4 + 1 \cdot 3 = 0 + 3 = 3 \]

Therefore,
\[ BA = \begin{pmatrix} 3 & 7 \\ 1 & 3 \end{pmatrix} \]

2. **Calculate the determinant of \( BA \)**:
The determinant of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by \( ad - bc \).
\[ \text{det}(BA) = (3)(3) - (7)(1) = 9 - 7 = 2 \]

3. **Find the adjoint of \( BA \)**:
The adjoint of \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is \( \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \).
\[ \text{adj}(BA) = \begin{pmatrix} 3 & -7 \\ -1 & 3 \end{pmatrix} \]

4. **Calculate the inverse of \( BA \)**:
The inverse of a matrix \( M \) is given by \( M^{-1} = \frac{1}{\text{det}(M)} \text{adj}(M) \).
\[ (BA)^{-1} = \frac{1}{2} \begin{pmatrix} 3 & -7 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} \frac{3}{2} & -\frac{7}{2} \\ -\frac{1}{2} & \frac{3}{2} \end{pmatrix} \]

5. **Conclusion**:
Therefore,
\[ A^{-1}B^{-1} = (BA)^{-1} = \begin{pmatrix} \frac{3}{2} & -\frac{7}{2} \\ -\frac{1}{2} & \frac{3}{2} \end{pmatrix} \]