Question

Given matrix:
Let \[ A = \begin{bmatrix} 2 & 1 & -3 \\ 3 & 2 & 1 \\ 1 & 2 & -1 \end{bmatrix} \] Find \( A^{-1} \) and hence solve the following system of equations:

System of equations:
\[ 2x + y - 3z = 13, \quad 3x + 2y + z = 4, \quad x + 2y - z = 8 \]

Hint:

Always verify the determinant before finding the inverse of a matrix. If \( |A| = 0 \), the inverse does not exist.

Answer 1

Step 1: Compute \( A^{-1} \)
The given matrix \( A \) is: \[ A = \begin{bmatrix} 2 & 1 & -3 \\ 3 & 2 & 1 \\ 1 & 2 & -1 \end{bmatrix} \] The determinant of \( A \) is: \[ |A| = 2(2 \cdot -1 - 1 \cdot 2) - 1(3 \cdot -1 - 1 \cdot 1) + (-3)(3 \cdot 2 - 2 \cdot 1) = -16 \] Since \( |A| \neq 0 \), \( A^{-1} \) exists. The adjoint of \( A \) is: \[ \text{adj}(A) = \begin{bmatrix} -4 & -5 & 7 \\ 4 & 1 & -11 \\ 4 & -3 & 1 \end{bmatrix} \] Thus: \[ A^{-1} = \frac{1}{|A|} \text{adj}(A) = -\frac{1}{16} \begin{bmatrix} -4 & -5 & 7 \\ 4 & 1 & -11 \\ 4 & -3 & 1 \end{bmatrix} \]
Step 2: Represent the system in matrix form
The system of equations can be written as: \[ AX = B, \quad \text{where} \quad A = \begin{bmatrix} 2 & 1 & -3 \\ 3 & 2 & 1 \\ 1 & 2 & -1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 13 \\ 4 \\ 8 \end{bmatrix} \] To solve for \( X \): \[ X = A^{-1} B \]
Step 3: Solve for \( X \)
\[ X = -\frac{1}{16} \begin{bmatrix} -4 & -5 & 7 \\ 4 & 1 & -11 \\ 4 & -3 & 1 \end{bmatrix} \begin{bmatrix} 13 \\ 4 \\ 8 \end{bmatrix} \] \[ X = -\frac{1}{16} \begin{bmatrix} -16 \\ -32 \\ 48 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix} \]
Step 4: Final result
The solution to the system of equations is: \[ x = 1, \quad y = 2, \quad z = -3 \]