Given `A=[(0,-1,2),(2,-2,0)]andB-[(0,1),(1,0),(1,1)]`. Find the product AB and also find `(AB)^(-1)`

Answer

To solve the problem, we need to find the product of matrices \( A \) and \( B \), and then find the inverse of the resulting matrix \( AB \).

### Step 1: Define the matrices
Let:
\[ A = \begin{pmatrix} 0 & -1 & 2 \\ 2 & -2 & 0 \end{pmatrix} \]
\[ B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{pmatrix} \]

### Step 2: Multiply matrices \( A \) and \( B \)
To find the product \( AB \), we perform matrix multiplication. The resulting matrix will be a \( 2 \times 2 \) matrix since \( A \) is \( 2 \times 3 \) and \( B \) is \( 3 \times 2 \).

The element in the first row and first column of \( AB \) is calculated as:
\[ (0 \cdot 0) + (-1 \cdot 1) + (2 \cdot 1) = 0 - 1 + 2 = 1 \]

The element in the first row and second column of \( AB \) is:
\[ (0 \cdot 1) + (-1 \cdot 0) + (2 \cdot 1) = 0 + 0 + 2 = 2 \]

The element in the second row and first column of \( AB \) is:
\[ (2 \cdot 0) + (-2 \cdot 1) + (0 \cdot 1) = 0 - 2 + 0 = -2 \]

The element in the second row and second column of \( AB \) is:
\[ (2 \cdot 1) + (-2 \cdot 0) + (0 \cdot 1) = 2 + 0 + 0 = 2 \]

Thus, the product \( AB \) is:
\[ AB = \begin{pmatrix} 1 & 2 \\ -2 & 2 \end{pmatrix} \]

### Step 3: Find the inverse of matrix \( AB \)
To find the inverse of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), we use the formula:
\[ (AB)^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]

For our matrix \( AB = \begin{pmatrix} 1 & 2 \\ -2 & 2 \end{pmatrix} \):
- \( a = 1 \)
- \( b = 2 \)
- \( c = -2 \)
- \( d = 2 \)

First, we calculate the determinant:
\[ \text{det}(AB) = ad - bc = (1)(2) - (2)(-2) = 2 + 4 = 6 \]

Now we can find the inverse:
\[ (AB)^{-1} = \frac{1}{6} \begin{pmatrix} 2 & -2 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} \frac{2}{6} & \frac{-2}{6} \\ \frac{2}{6} & \frac{1}{6} \end{pmatrix} = \begin{pmatrix} \frac{1}{3} & -\frac{1}{3} \\ \frac{1}{3} & \frac{1}{6} \end{pmatrix} \]

### Final Result
The product \( AB \) is:
\[ AB = \begin{pmatrix} 1 & 2 \\ -2 & 2 \end{pmatrix} \]
And the inverse \( (AB)^{-1} \) is:
\[ (AB)^{-1} = \begin{pmatrix} \frac{1}{3} & -\frac{1}{3} \\ \frac{1}{3} & \frac{1}{6} \end{pmatrix} \]