Question

If \[ A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix} \] then find \( A^{-1} \). Hence, solve the system of linear equations: \[ x - 2y = 10, \] \[ 2x - y - z = 8, \] \[ -2y + z = 7. \]

Hint:

For \( A^{-1} \), use \( A^{-1} = \frac{1}{\det A} \text{Adj}(A) \) and verify by computing \( A A^{-1} = I \).

Answer 1

The problem requires finding the inverse of matrix \( A \) and then using it to solve a system of linear equations.

1. System Representation:
The system is expressed in matrix form as \( AX = B \), where:

\( A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} \)

2. Matrix Inverse Calculation:
The inverse \( A^{-1} \) is computed using the adjoint method, following the formula \( A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A) \).

The determinant of \( A \) is calculated as:

\[|A| = 1 \cdot ((-1)(1) - (-2)(-1)) - 2 \cdot ((-2)(1) - (-2)(0)) + 0 \cdot (\text{any}) \\= 1(-1 - 2) - 2(-2) = -3 + 4 = 1\]
The determinant is 1.
The adjugate matrix is computed, resulting in:
\[A^{-1} = \begin{bmatrix}-3 & -2 & -4 \\-2 & 1 & 2 \\-1 & -2 & -3\end{bmatrix}\]

3. Solution Derivation:
The solution vector \( X \) is obtained by multiplying \( A^{-1} \) by \( B \):

\[X = A^{-1}B = \begin{bmatrix}-3 & -2 & -4 \\-2 & 1 & 2 \\-1 & -2 & -3\end{bmatrix}\begin{bmatrix}10 \\ 8 \\ 7\end{bmatrix}\]
The element-wise multiplication yields:
\[x = -3(10) + (-2)(8) + (-4)(7) = -30 - 16 - 28 = -74 \\y = -2(10) + 1(8) + 2(7) = -20 + 8 + 14 = 2 \\z = -1(10) + (-2)(8) + (-3)(7) = -10 - 16 - 21 = -47\]

Result:
The solution to the system is: \( x = -74, \, y = 2, \, z = -47 \)