Question: User: 3. Consider the matrices $A = \begin{pmatrix}1 & 2 \\ 0 & 1\end{pmatrix}$ and $B = \begin{pmatrix}1 & 0 \\ 3 & 1\end{pmatrix}$. Determine $A^{-1}$ and $B^{-1}$. Hence, or otherwise, determine $(AB)^{-1}$.
1. **State the problem:** We are given two matrices
$$A = \begin{pmatrix}1 & 2 \\ 0 & 1\end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix}1 & 0 \\ 3 & 1\end{pmatrix}$$
We need to find the inverses $A^{-1}$ and $B^{-1}$, and then find the inverse of the product $(AB)^{-1}$.
2. **Recall the formula for the inverse of a 2x2 matrix:**
For a matrix $M = \begin{pmatrix}a & b \\ c & d\end{pmatrix}$, if $\det(M) = ad - bc \neq 0$, then
$$M^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$
3. **Find $A^{-1}$:**
- Calculate determinant:
$$\det(A) = (1)(1) - (2)(0) = 1$$
- Since determinant is 1, inverse exists.
- Apply formula:
$$A^{-1} = \frac{1}{1} \begin{pmatrix}1 & -2 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & -2 \\ 0 & 1\end{pmatrix}$$
4. **Find $B^{-1}$:**
- Calculate determinant:
$$\det(B) = (1)(1) - (0)(3) = 1$$
- Since determinant is 1, inverse exists.
- Apply formula:
$$B^{-1} = \frac{1}{1} \begin{pmatrix}1 & 0 \\ -3 & 1\end{pmatrix} = \begin{pmatrix}1 & 0 \\ -3 & 1\end{pmatrix}$$
5. **Find $(AB)^{-1}$:**
- Recall the property:
$$(AB)^{-1} = B^{-1} A^{-1}$$
- Calculate $B^{-1} A^{-1}$:
$$B^{-1} A^{-1} = \begin{pmatrix}1 & 0 \\ -3 & 1\end{pmatrix} \begin{pmatrix}1 & -2 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 \cdot 1 + 0 \cdot 0 & 1 \cdot (-2) + 0 \cdot 1 \\ -3 \cdot 1 + 1 \cdot 0 & -3 \cdot (-2) + 1 \cdot 1\end{pmatrix} = \begin{pmatrix}1 & -2 \\ -3 & 7\end{pmatrix}$$
**Final answers:**
$$A^{-1} = \begin{pmatrix}1 & -2 \\ 0 & 1\end{pmatrix}, \quad B^{-1} = \begin{pmatrix}1 & 0 \\ -3 & 1\end{pmatrix}, \quad (AB)^{-1} = \begin{pmatrix}1 & -2 \\ -3 & 7\end{pmatrix}$$