Question

(b) Given \[ A = \begin{bmatrix} 2 & 3 & 2 \\ 1 & 1 & 2 \\ 1 & 1 & 2 \end{bmatrix}, \text{ find } A^{-1}. \text{ Use it to solve the following system of equations:} \] \[ x + y + z = 1 \] \[ 2x + 3y + 2z = 2 \] \[ x + y + 2z = 4. \]

Hint:

To determine if a matrix is invertible, calculate its determinant. If the determinant is zero, the matrix is singular and does not have an inverse. For such cases, consider other methods like row reduction or substitution to solve the system.

Answer 1

The given matrix \( A \) is: \[ A = \begin{bmatrix} 2 & 3 & 2 \\ 1 & 1 & 2 \\ 1 & 1 & 2 \end{bmatrix}. \] To solve the system of equations using \( A^{-1} \), we must first determine whether \( A \) is invertible. 
Step 1: Compute the determinant of \( A \). The determinant of \( A \) is: \[ |A| = \begin{vmatrix} 2 & 3 & 2 \\ 1 & 1 & 2 \\ 1 & 1 & 2 \end{vmatrix}. \] Using cofactor expansion along the first row: \[ |A| = 2 \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} - 3 \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} + 2 \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix}. \] Simplifying the \( 2 \times 2 \) determinants: \[ |A| = 2(1 \cdot 2 - 1 \cdot 2) - 3(1 \cdot 2 - 1 \cdot 2) + 2(1 \cdot 1 - 1 \cdot 1). \] \[ |A| = 2(0) - 3(0) + 2(0) = 0. \] 
Step 2: Check invertibility of \( A \). Since \( |A| = 0 \), the matrix \( A \) is singular, meaning it does not have an inverse.
Step 3: Conclusion. Because \( A \) is not invertible, the system of equations cannot be solved using \( A^{-1} \). 
Final Answer: \[ \boxed{\text{The matrix } A \text{ is singular, so } A^{-1} \text{ does not exist. The system cannot be solved using the inverse method.}} \]