Question

If \( A = \begin{bmatrix} 0 & 2 & 1 \\ -2 & -1 & -2 \\ 1 & -1 & 0 \end{bmatrix} \), find \( A^{-1} \) and use it to solve the following system of equations :
\( -2y + z = 7, 2x - y - z = 8, x - 2y = 10 \)

Hint:

Always check if the coefficients of the system match the matrix \( A \) or its transpose \( A^T \).
\( (A^T)^{-1} = (A^{-1})^T \).

Answer 1

Concept:

• Matrix Inverse: \( A^{-1} = \frac{1}{|A|} \text{adj}(A) \).
• System Solution: \( X = A^{-1}B \).

Step 1: Calculate the determinant \( |A| \)
\[ |A| = 0(-1 - 2) - 2(0 - (-2)) + 1(2 - (-1)) \] \[ |A| = 0 - 2(2) + 1(3) = -4 + 3 = -1 \] Since \( |A| \neq 0 \), \( A^{-1} \) exists.

Step 2: Find the adjoint of \( A \)
Cofactors:
\( A_{11}=-1, A_{12}=-2, A_{13}=3, A_{21}=-1, A_{22}=-1, A_{23}=2, A_{31}=-3, A_{32}=-2, A_{33}=4 \) \[ \text{adj}(A) = \begin{bmatrix} -1 & -2 & 3 \\ -1 & -1 & 2 \\ -3 & -2 & 4 \end{bmatrix}^T = \begin{bmatrix} -1 & -1 & -3 \\ -2 & -1 & -2 \\ 3 & 2 & 4 \end{bmatrix} \] \[ A^{-1} = \frac{1}{-1} \begin{bmatrix} -1 & -1 & -3 \\ -2 & -1 & -2 \\ 3 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 3 \\ 2 & 1 & 2 \\ -3 & -2 & -4 \end{bmatrix} \]

Step 3: Solve the system
Note the system matrix \( C = \begin{bmatrix} 0 & -2 & 1 \\ 2 & -1 & -1 \\ 1 & -2 & 0 \end{bmatrix} \). This is \( A^T \).
Since \( (A^T)^{-1} = (A^{-1})^T \), we use the transpose of \( A^{-1} \). \[ \begin{bmatrix} x y z \end{bmatrix} = \begin{bmatrix} 1 & 2 & -3 \\ 1 & 1 & -2 \\ 3 & 2 & -4 \end{bmatrix} \begin{bmatrix} 7 8 10 \end{bmatrix} = \begin{bmatrix} 7 + 16 - 30 \\ 7 + 8 - 20 \\ 21 + 16 - 40 \end{bmatrix} = \begin{bmatrix} -7 -5 -3 \end{bmatrix} \] Solution: \( x = -7, y = -5, z = -3 \).