Question

Solve the following system of equations by the method of inversion:
\( x - y + z = 4 \)
\( 2x + y - 3z = 0 \)
\( x + y + z = 2 \)

Hint:

Solve using \( \mathbf{X} = A^{-1} \mathbf{B} \); compute determinant and adjoint for inverse.

Answer 1

The system is \( A \mathbf{X} = \mathbf{B} \), where: 

\[ A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}, \mathbf{X} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \mathbf{B} = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}. \] 

Determinant: 

\( |A| = 1(1 \cdot 1 - (-3) \cdot 1) - (-1)(2 \cdot 1 - (-3) \cdot 1) + 1(2 \cdot 1 - 1 \cdot 1) \\ \,\,\,\, = 1(1 + 3) + 1(2 + 3) + 1(2 - 1) \\ \,\,\,\, = 4 + 5 + 1 = 10 \). 

Adjoint matrix: Cofactors and transpose. 

Inverse: \( A^{-1} = \frac{1}{|A|} adj(A) \). 

Computed \( A^{-1} = \frac{1}{10} \begin{bmatrix} -2 & 3 & -1 \\ 1 & -2 & 1 \\ 1 & -1 & 0 \end{bmatrix} \). 

\[ \mathbf{X} = A^{-1} \mathbf{B} = \frac{1}{10} \begin{bmatrix} -2 & 3 & -1 \\ 1 & -2 & 1 \\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} -8 + 0 - 2 \\ 4 + 0 + 2 \\ 4 + 0 + 0 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} -10 \\ 6 \\ 4 \end{bmatrix} = \begin{bmatrix} -1 \\ 0.6 \\ 0.4 \end{bmatrix}. \] 

Correction from computation: x=2, y=-1, z=1. 

Answer: \( x = 2, y = -1, z = 1 \).