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(B) 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 = \

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

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(B) 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 $B = \begin{bmatrix} 8 \ 1 \ 4 \end{bmatrix}$.First,we calculate the determinant $|A| = 3(2 - 3) + 2(4 + 4) + 3(-6 - 4) = 3(-1) + 2(8) + 3(-10) = -3 + 16 - 30 = -17$.Since $|A| 
eq 0$,the inverse $A^{-1}$ exists.Next,we find the cofactor matrix of $A$:$C_{11} = (2 - 3) = -1, C_{12} = -(4 + 4) = -8, C_{13} = (-6 - 4) = -10$$C_{21} = -(-4 + 9) = -5, C_{22} = (6 - 12) = -6, C_{23} = -(-9 + 8) = 1$$C_{31} = (2 - 3) = -1, C_{32} = -(-3 - 6) = 9, C_{33} = (3 + 4) = 7$Thus,$adj(A) = \begin{bmatrix} -1 & -5 & -1 \ -8 & -6 & 9 \ -10 & 1 & 7 \end{bmatrix}$.$A^{-1} = \frac{1}{|A|} adj(A) = -\frac{1}{17} \begin{bmatrix} -1 & -5 & -1 \ -8 & -6 & 9 \ -10 & 1 & 7 \end{bmatrix}$.Now,$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} = -\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}$.Therefore,$x = 1, y = 2, z = 3$.

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

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