If begin{bmatrix} 2 & 1 & 1 0 & 3 & -1 1 & | vedclass

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(D) The given system of linear equations is: $2x + y + z = 1$ $(i)$ $3y - z = 1$ $(ii)$ $x - y + z = 0$ $(iii)$ Subtracting equation $(iii)$ from equa

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$\begin{bmatrix} 1 \ 0 \ -1 \end{bmatrix} + K \begin{bmatrix} -2 \ 1 \ 3 \end{bmatrix}, K \in R$

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(D) The given system of linear equations is: $2x + y + z = 1$ $(i)$ $3y - z = 1$ $(ii)$ $x - y + z = 0$ $(iii)$ Subtracting equation $(iii)$ from equation $(i)$, we get: $(2x + y + z) - (x - y + z) = 1 - 0$ $x + 2y = 1 \Rightarrow x = 1 - 2y$ $(iv)$ From equation $(ii)$, $z = 3y - 1$. Let $y = K$, where $K \in R$. Then $x = 1 - 2K$ and $z = 3K - 1$. Thus, the solution vector is: $\begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 1 - 2K \ K \ 3K - 1 \end{bmatrix} = \begin{bmatrix} 1 \ 0 \ -1 \end{bmatrix} + K \begin{bmatrix} -2 \ 1 \ 3 \end{bmatrix}$ Therefore, the correct option is $D$.

If $\begin{bmatrix} 2 & 1 & 1 \\ 0 & 3 & -1 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$, then $\begin{bmatrix} x \\ y \\ z \end{bmatrix} =$

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