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

Answer

To find \((AB)^{-1}\) where \(A = \begin{pmatrix} 0 & 1 \\ 2 & 3 \\ 1 & -1 \end{pmatrix}\) and \(B = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 1 & 0 \end{pmatrix}\), we will follow these steps:

### Step 1: Multiply Matrices A and B
We first need to compute the product \(AB\).

\[ AB = A \cdot B = \begin{pmatrix} 0 & 1 \\ 2 & 3 \\ 1 & -1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 & 1 \\ 2 & 1 & 0 \end{pmatrix} \]

To multiply the matrices, we calculate each element of the resulting matrix:

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

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

- First row, third column:
\[ 0 \cdot 1 + 1 \cdot 0 = 0 + 0 = 0 \]

- Second row, first column:
\[ 2 \cdot 1 + 3 \cdot 2 = 2 + 6 = 8 \]

- Second row, second column:
\[ 2 \cdot 2 + 3 \cdot 1 = 4 + 3 = 7 \]

- Second row, third column:
\[ 2 \cdot 1 + 3 \cdot 0 = 2 + 0 = 2 \]

- Third row, first column:
\[ 1 \cdot 1 + (-1) \cdot 2 = 1 - 2 = -1 \]

- Third row, second column:
\[ 1 \cdot 2 + (-1) \cdot 1 = 2 - 1 = 1 \]

- Third row, third column:
\[ 1 \cdot 1 + (-1) \cdot 0 = 1 + 0 = 1 \]

Thus, the product \(AB\) is:

\[ AB = \begin{pmatrix} 2 & 1 & 0 \\ 8 & 7 & 2 \\ -1 & 1 & 1 \end{pmatrix} \]

### Step 2: Calculate the Determinant of AB
Next, we need to find the determinant of the matrix \(AB\).

For a \(3 \times 3\) matrix \(\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}\), the determinant is calculated as:

\[ \text{det}(AB) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Substituting the values from \(AB\):

\[ \text{det}(AB) = 2(7 \cdot 1 - 2 \cdot 1) - 1(8 \cdot 1 - 2 \cdot (-1)) + 0(8 \cdot 1 - 7 \cdot (-1)) \]

Calculating each term:

- First term:
\[ 2(7 - 2) = 2 \cdot 5 = 10 \]

- Second term:
\[ -1(8 + 2) = -1 \cdot 10 = -10 \]

- Third term:
\[ 0 \cdot \text{(anything)} = 0 \]

Thus, the determinant is:

\[ \text{det}(AB) = 10 - 10 + 0 = 0 \]

### Step 3: Determine if the Inverse Exists
Since the determinant of \(AB\) is 0, the inverse \((AB)^{-1}\) does not exist.

### Final Answer
\((AB)^{-1}\) does not exist.