Solve the following system of equations by matrix method: $3x - 2y + 3z = 8$, $2x + y - z = 1$, $4x - 3y + 2z = 4$.

$x = 1, y = 2, z = 3$

The correct answer is Option (1) → $x = 1, y = 2, z = 3$ ##

The system of equations can be written in the form $AX = B$, where

$A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \text{ and } B = \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix} \text{}$

We see that

$|A| = 3(2 - 3) + 2(4 + 4) + 3(-6 - 4) = -17 \neq 0 \text{}$

Hence, $A$ is nonsingular and so its inverse exists. Now

$A_{11} = -1, A_{12} = -8, A_{13} = -10$

$A_{21} = -5, A_{22} = -6, A_{23} = 1$

$A_{31} = -1, A_{32} = 9, A_{33} = 7$

Therefore $A^{-1} = -\frac{1}{17} \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix}$

So $X = A^{-1}B = -\frac{1}{17} \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix} \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix}$

i.e. $\begin{bmatrix} x \\ y \\ z \end{bmatrix} = -\frac{1}{17} \begin{bmatrix} -17 \\ -34 \\ -51 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

Hence $x = 1, y = 2$ and $z = 3$.