Solution

The system of equations can be written in the form \(AX=B\), where \(A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix}\), \(X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\) and \(B = \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix}\). We see that \(|A| = 3(2-3) - (-2)(4+4) + 3(-6-4) = -3 + 16 - 30 = -17 \neq 0\). Hence, A is nonsingular and so its inverse exists. Now \(A_{11} = -1, A_{12} = -8, A_{13} = -10\) \(A_{21} = -5, A_{22} = -6, A_{23} = 1\) \(A_{31} = -1, A_{32} = 9, A_{33} = 7\) So, \(A^{-1} = -\frac{1}{17}\begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix}\) So, \(X = A^{-1}B = -\frac{1}{17}\begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix}\begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix}\) \(\begin{bmatrix} x \\ y \\ z \end{bmatrix} = -\frac{1}{17}\begin{bmatrix} -8-5-4 \\ -64-6+36 \\ -80+1+28 \end{bmatrix} = -\frac{1}{17}\begin{bmatrix} -17 \\ -34 \\ -51 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\) Hence \(x=1, y=2, z=3\).