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(D) The given system of equations can be written in the form $AX = B$,where $A = \begin{bmatrix} 2 & 3 & 3 \ 1 & -2 & 1 \ 3 & -1 & -2 \end{bmatrix}$

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

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(D) The given system of equations can be written in the form $AX = B$,where $A = \begin{bmatrix} 2 & 3 & 3 \ 1 & -2 & 1 \ 3 & -1 & -2 \end{bmatrix}$,$X = \begin{bmatrix} x \ y \ z \end{bmatrix}$,and $B = \begin{bmatrix} 5 \ -4 \ 3 \end{bmatrix}$.First,we calculate the determinant of $A$:$|A| = 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6) = 2(5) - 3(-5) + 3(5) = 10 + 15 + 15 = 40 
eq 0$.Since $|A| 
eq 0$,the system has a unique solution given by $X = A^{-1}B$.The matrix of cofactors is calculated as:$C_{11} = 5, C_{12} = 5, C_{13} = 5$$C_{21} = 3, C_{22} = -13, C_{23} = 11$$C_{31} = 9, C_{32} = 1, C_{33} = -7$Thus,$adj(A) = \begin{bmatrix} 5 & 3 & 9 \ 5 & -13 & 1 \ 5 & 11 & -7 \end{bmatrix}$.$A^{-1} = \frac{1}{40} \begin{bmatrix} 5 & 3 & 9 \ 5 & -13 & 1 \ 5 & 11 & -7 \end{bmatrix}$.Now,$X = A^{-1}B = \frac{1}{40} \begin{bmatrix} 5 & 3 & 9 \ 5 & -13 & 1 \ 5 & 11 & -7 \end{bmatrix} \begin{bmatrix} 5 \ -4 \ 3 \end{bmatrix} = \frac{1}{40} \begin{bmatrix} 25 - 12 + 27 \ 25 + 52 + 3 \ 25 - 44 - 21 \end{bmatrix} = \frac{1}{40} \begin{bmatrix} 40 \ 80 \ -40 \end{bmatrix} = \begin{bmatrix} 1 \ 2 \ -1 \end{bmatrix}$.Therefore,$x = 1, y = 2, z = -1$.

Solve the system of linear equations using the matrix method: $2x + 3y + 3z = 5$,$x - 2y + z = -4$,$3x - y - 2z = 3$.

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