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

Answer

To solve the problem, we need to find the product of matrices A and B, denoted as AB, and then find the inverse of that product, denoted as (AB)^(-1).

### Step 1: Define the matrices A and B
Given:
\[ A = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} \]

### Step 2: Calculate the product AB
To find the product of two matrices, we use the formula for matrix multiplication. The element at position (i, j) in the resulting matrix is obtained by taking the dot product of the i-th row of the first matrix and the j-th column of the second matrix.

\[ AB = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} \]

Calculating each element of the resulting matrix:

- First row, first column:
\[ 2 \cdot 1 + 3 \cdot 3 = 2 + 9 = 11 \]

- First row, second column:
\[ 2 \cdot 0 + 3 \cdot 1 = 0 + 3 = 3 \]

- Second row, first column:
\[ 1 \cdot 1 + 2 \cdot 3 = 1 + 6 = 7 \]

- Second row, second column:
\[ 1 \cdot 0 + 2 \cdot 1 = 0 + 2 = 2 \]

Thus, the product AB is:
\[ AB = \begin{pmatrix} 11 & 3 \\ 7 & 2 \end{pmatrix} \]

### Step 3: Find the determinant of AB
The determinant of a 2x2 matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is calculated as:
\[ \text{det}(AB) = ad - bc \]
For our matrix:
\[ \text{det}(AB) = (11)(2) - (3)(7) = 22 - 21 = 1 \]

### Step 4: Find the adjoint of AB
The adjoint of a 2x2 matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is given by:
\[ \text{adj}(AB) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]
Applying this to our matrix:
\[ \text{adj}(AB) = \begin{pmatrix} 2 & -3 \\ -7 & 11 \end{pmatrix} \]

### Step 5: Calculate the inverse of AB
The inverse of a matrix is given by:
\[ (AB)^{-1} = \frac{\text{adj}(AB)}{\text{det}(AB)} \]
Since \(\text{det}(AB) = 1\):
\[ (AB)^{-1} = \text{adj}(AB) = \begin{pmatrix} 2 & -3 \\ -7 & 11 \end{pmatrix} \]

### Final Answer
Thus, the final results are:
\[ AB = \begin{pmatrix} 11 & 3 \\ 7 & 2 \end{pmatrix}, \quad (AB)^{-1} = \begin{pmatrix} 2 & -3 \\ -7 & 11 \end{pmatrix} \]