4)(NC) Let: \( A=\left(\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & -1 & -1\end{array}\right) \) and \( B=\left(\begin{array}{ccc}1 & 0 & 1 \\ 4 & -2 & -2 \\ -3 & 2 & 1\end{array}\right) \) Show that \( A B=k I \) for some constant \( k \), where \( I \) is the \( 3 \times 3 \) identity matrix. Hence obtain: i) The inverse matrix \( A^{-1} \) ii) The matrix \( A^{2} B \)

To show that \( AB = kI \) for some constant \( k \), we first calculate the matrix product \( AB \): \[ AB = \left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & -1 & -1 \end{array}\right) \left(\begin{array}{ccc} 1 & 0 & 1 \\ 4 & -2 & -2 \\ -3 & 2 & 1 \end{array}\right) \] Now, we will compute the product: 1. The element in the first row, first column of \( AB \): \[ 1 \cdot 1 + 1 \cdot 4 + 1 \cdot (-3) = 1 + 4 - 3 = 2 \] 2. The element in the first row, second column of \( AB \): \[ 1 \cdot 0 + 1 \cdot (-2) + 1 \cdot 2 = 0 - 2 + 2 = 0 \] 3. The element in the first row, third column of \( AB \): \[ 1 \cdot 1 + 1 \cdot (-2) + 1 \cdot 1 = 1 - 2 + 1 = 0 \] 4. The element in the second row, first column of \( AB \): \[ 1 \cdot 1 + 2 \cdot 4 + 3 \cdot (-3) = 1 + 8 - 9 = 0 \] 5. The element in the second row, second column of \( AB \): \[ 1 \cdot 0 + 2 \cdot (-2) + 3 \cdot 2 = 0 - 4 + 6 = 2 \] 6. The element in the second row, third column of \( AB \): \[ 1 \cdot 1 + 2 \cdot (-2) + 3 \cdot 1 = 1 - 4 + 3 = 0 \] 7. The element in the third row, first column of \( AB \): \[ 1 \cdot 1 + (-1) \cdot 4 + (-1) \cdot (-3) = 1 - 4 + 3 = 0 \] 8. The element in the third row, second column of \( AB \): \[ 1 \cdot 0 + (-1) \cdot (-2) + (-1) \cdot 2 = 0 + 2 - 2 = 0 \] 9. The element in the third row, third column of \( AB \): \[ 1 \cdot 1 + (-1) \cdot (-2) + (-1) \cdot 1 = 1 + 2 - 1 = 2 \] Putting all the results together, we find: \[ AB = \left(\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right) = 2I \] Thus, \( k = 2 \), and we have \( AB = 2I \). Now we can compute the inverse \( A^{-1} \) using the relationship \( AB = 2I \): \[ A^{-1} = \frac{1}{2} B \] Calculating \( A^{-1} \): \[ A^{-1} = \frac{1}{2} \left(\begin{array}{ccc} 1 & 0 & 1 \\ 4 & -2 & -2 \\ -3 & 2 & 1 \end{array}\right) = \left(\begin{array}{ccc} \frac{1}{2} & 0 & \frac{1}{2} \\ 2 & -1 & -1 \\ -\frac{3}{2} & 1 & \frac{1}{2} \end{array}\right) \] Next, we will calculate \( A^2 B \): First, calculate \( A^2 \): \[ A^2 = A \cdot A = \left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & -1 & -1 \end{array