Given AB = 3 & 2 1 & 1 and B = 1 & 1 2 & 0 .

To find $A^{-1}$, we use the property that if $AB = C$, then $A^{-1} = BC^{-1}$.
Given $AB = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 1 \\ 2 & 0 \end{bmatrix}$.
Let $C = AB = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$.

Step 1: Calculate the determinant of $C$.
For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is $ad - bc$.

$$\det(C) = (3)(1) - (2)(1) = 3 - 2 = 1$$

Step 2: Calculate the inverse of $C$.
For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the inverse is $\frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

$$C^{-1} = \frac{1}{1} \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix}$$

Step 3: Calculate $A^{-1}$ using the formula $A^{-1} = BC^{-1}$.

$$A^{-1} = \begin{bmatrix} 1 & 1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix}$$

Step 4: Perform the matrix multiplication.

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