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

Vedclass · Accepted Answer

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

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(C) The given system of equations can be written in the form $AX=B$,where$A=\begin{bmatrix} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{bmatrix}$,$X=\begin{bmatrix} x \ y \ z \end{bmatrix}$,and $B=\begin{bmatrix} 4 \ 0 \ 2 \end{bmatrix}$.First,calculate the determinant $|A|$:$|A| = 1(1+3) - (-1)(2+3) + 1(2-1) = 4 + 5 + 1 = 10 
eq 0$.Since $|A| 
eq 0$,the matrix $A$ is non-singular and $A^{-1}$ exists.Next,find the cofactor matrix $C$:$C_{11} = (1+3) = 4, C_{12} = -(2+3) = -5, C_{13} = (2-1) = 1$$C_{21} = -(-1-1) = 2, C_{22} = (1-1) = 0, C_{23} = -(1+1) = -2$$C_{31} = (3-1) = 2, C_{32} = -(-3-2) = 5, C_{33} = (1+2) = 3$Thus,$adj(A) = C^T = \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3 \end{bmatrix}$.$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3 \end{bmatrix}$.Now,$X = A^{-1}B = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \ 0 \ 2 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 16+0+4 \ -20+0+10 \ 4+0+6 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 20 \ -10 \ 10 \end{bmatrix} = \begin{bmatrix} 2 \ -1 \ 1 \end{bmatrix}$.Therefore,$x=2, y=-1, z=1$.

Solve the system of linear equations using the matrix method: $x-y+z=4$,$2x+y-3z=0$,and $x+y+z=2$.

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