5. Let \( A=\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right] \) and \( B=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right] \), verify that \( (A B)^{-1}=B^{-1} A^{-1} \)

To verify that \( (AB)^{-1} = B^{-1} A^{-1} \), we will first compute the product \( AB \), then find its inverse \( (AB)^{-1} \), and finally compute \( B^{-1} \) and \( A^{-1} \) and their product \( B^{-1} A^{-1} \). ### Step 1: Compute \( AB \) Given: \[ A = \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} \] Now, we compute the product \( AB \): \[ AB = \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} \] Calculating the entries: - First row, first column: \( 1 \cdot 1 + 1 \cdot 0 = 1 \) - First row, second column: \( 1 \cdot 2 + 1 \cdot 3 = 2 + 3 = 5 \) - Second row, first column: \( 0 \cdot 1 + 2 \cdot 0 = 0 \) - Second row, second column: \( 0 \cdot 2 + 2 \cdot 3 = 0 + 6 = 6 \) Thus, \[ AB = \begin{bmatrix} 1 & 5 \\ 0 & 6 \end{bmatrix} \] ### Step 2: Compute \( (AB)^{-1} \) To find the inverse of \( AB \), we use the formula for the inverse of a \( 2 \times 2 \) matrix: \[ \text{If } M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \text{ then } M^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] For our matrix \( AB = \begin{bmatrix} 1 & 5 \\ 0 & 6 \end{bmatrix} \): - \( a = 1, b = 5, c = 0, d = 6 \) - The determinant \( ad - bc = 1 \cdot 6 - 5 \cdot 0 = 6 \) Thus, \[ (AB)^{-1} = \frac{1}{6} \begin{bmatrix} 6 & -5 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{5}{6} \\ 0 & \frac{1}{6} \end{bmatrix} \] ### Step 3: Compute \( A^{-1} \) and \( B^{-1} \) **For \( A \)**: \[ A^{-1} = \frac{1}{1 \cdot 2 - 1 \cdot 0} \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{1}{2} \\ 0 & \frac{1}{2} \end{bmatrix} \] **For \( B \)**: \[ B^{-1} = \frac{1}{1 \cdot 3 - 2 \cdot 0} \begin{bmatrix} 3 & -2 \\ 0 & 1 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 3 & -2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{2}{3} \\ 0 & \frac{1}{3} \end{bmatrix} \] ### Step 4: Compute \( B^{-1} A^{-1} \) Now we compute \( B^{-1} A^{-1} \): \[ B^{-1} A^{-1} = \begin{bmatrix} 1 & -\frac{2}{3} \\ 0 & \frac{1}{3} \end{bmatrix} \begin{bmatrix} 1 & -\frac{1}{2} \\ 0 & \frac{1}{2} \end{bmatrix} \] Calculating the entries: - First row, first column: \( 1 \cdot 1 + (-\frac{2}{3}) \cdot 0 = 1 \) - First row, second column: \( 1 \cdot (-\frac{1}{2}) + (-\frac{2}{3}) \cdot \frac{1}{2} = -\frac{1}{2} - \frac{1}{3} = -\frac{3}{6} - \frac{2}{6} = -\frac{5}{6} \) - Second row, first column: \( 0 \cdot 1 + \frac{1}{3} \cdot 0 = 0 \) - Second row, second column: \( 0 \cdot (-\frac{1}{2}) + \frac{1}{3} \cdot \frac{1}{2} = 0 + \frac{1}{6} = \frac{1}{6} \) Thus, \[ B^{-1} A^{-1} = \begin{bmatrix} 1 & -\frac{5}{6} \\ 0 & \frac{1}{6} \end{bmatrix} \] ### Conclusion We have: \[ (AB)^{-1} = \begin{bmatrix} 1 & -\frac{5}{6} \\ 0 & \frac{1}{6} \end{bmatrix} \quad \text{and} \quad B^{-1} A^{-1} = \begin{bmatrix} 1 & -\frac{5}{6} \\ 0 & \frac{1}{6} \end{bmatrix} \] Thus, we conclude that: \[ (AB)^{-1} = B^{-1} A^{-1} \] This verifies the statement.