For the system x-y+z=4, 2x+y-3z=0, x+y+z=2,th | vedclass

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(B) The given system of linear equations is: $x - y + z = 4$ $2x + y - 3z = 0$ $x + y + z = 2$ In matrix form,this is $AX = B$,where $A = \begin{bmatr

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$2, -1, 1$

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(B) The given system of linear equations is: $x - y + z = 4$ $2x + y - 3z = 0$ $x + y + z = 2$ In matrix form,this is $AX = B$,where $A = \begin{bmatrix} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{bmatrix}, X = \begin{bmatrix} x \ y \ z \end{bmatrix}, B = \begin{bmatrix} 4 \ 0 \ 2 \end{bmatrix}$ First,calculate the determinant $|A|$: $|A| = 1(1 + 3) - (-1)(2 + 3) + 1(2 - 1) = 1(4) + 1(5) + 1(1) = 4 + 5 + 1 = 10$ Since $|A| 
eq 0$,the system has a unique solution $X = A^{-1}B$. Find the adjoint of $A$: $C_{11} = 4, C_{12} = -5, C_{13} = 1$ $C_{21} = 2, C_{22} = 0, C_{23} = -2$ $C_{31} = 2, C_{32} = 5, C_{33} = 3$ $adj(A) = \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3 \end{bmatrix}$ $X = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \ 0 \ 2 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 16 + 0 + 4 \ -20 + 0 + 10 \ 4 + 0 + 6 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 20 \ -10 \ 10 \end{bmatrix} = \begin{bmatrix} 2 \ -1 \ 1 \end{bmatrix}$ Thus,$x = 2, y = -1, z = 1$.

For the system $x-y+z=4, 2x+y-3z=0, x+y+z=2$,the values of $x, y, z$ respectively are given by

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