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(A) The given system of equations can be written in the form $AX=B$,where$A=\begin{bmatrix} 1 & -1 & 2 \ 3 & 4 & -5 \ 2 & -1 & 3 \end{bmatrix}$,$X=\

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

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(A) The given system of equations can be written in the form $AX=B$,where$A=\begin{bmatrix} 1 & -1 & 2 \ 3 & 4 & -5 \ 2 & -1 & 3 \end{bmatrix}$,$X=\begin{bmatrix} x \ y \ z \end{bmatrix}$,and $B=\begin{bmatrix} 7 \ -5 \ 12 \end{bmatrix}$.First,calculate the determinant $|A|$:$|A| = 1(12-5) - (-1)(9+10) + 2(-3-8) = 1(7) + 1(19) + 2(-11) = 7 + 19 - 22 = 4 
eq 0$.Since $|A| 
eq 0$,the system has a unique solution $X = A^{-1}B$.The matrix of cofactors $C$ is calculated as:$C_{11} = (12-5) = 7, C_{12} = -(9+10) = -19, C_{13} = (-3-8) = -11$$C_{21} = -(-3+2) = 1, C_{22} = (3-4) = -1, C_{23} = -(-1+2) = -1$$C_{31} = (5-8) = -3, C_{32} = -(-5-6) = 11, C_{33} = (4+3) = 7$Thus,$adj(A) = C^T = \begin{bmatrix} 7 & 1 & -3 \ -19 & -1 & 11 \ -11 & -1 & 7 \end{bmatrix}$.$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{4} \begin{bmatrix} 7 & 1 & -3 \ -19 & -1 & 11 \ -11 & -1 & 7 \end{bmatrix}$.$X = A^{-1}B = \frac{1}{4} \begin{bmatrix} 7 & 1 & -3 \ -19 & -1 & 11 \ -11 & -1 & 7 \end{bmatrix} \begin{bmatrix} 7 \ -5 \ 12 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 49-5-36 \ -133+5+132 \ -77+5+84 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 8 \ 4 \ 12 \end{bmatrix} = \begin{bmatrix} 2 \ 1 \ 3 \end{bmatrix}$.Therefore,$x=2, y=1, z=3$.

Solve the system of linear equations using the matrix method: $x-y+2z=7$,$3x+4y-5z=-5$,$2x-y+3z=12$.

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