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

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(D) Given the matrix equation: $\begin{bmatrix} 1 & 1 & 1 \ 1 & -2 & -2 \ 1 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin

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$\begin{bmatrix} 1 \ 2 \ -3 \end{bmatrix}$

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(D) Given the matrix equation: $\begin{bmatrix} 1 & 1 & 1 \ 1 & -2 & -2 \ 1 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 0 \ 3 \ 4 \end{bmatrix}$.This corresponds to the system of linear equations:$1) x + y + z = 0$$2) x - 2y - 2z = 3$$3) x + 3y + z = 4$Subtracting equation $(1)$ from equation $(3)$:$(x + 3y + z) - (x + y + z) = 4 - 0$$2y = 4 \implies y = 2$.Substitute $y = 2$ into equations $(1)$ and $(2)$:$x + 2 + z = 0 \implies x + z = -2$ $(4)$$x - 2(2) - 2z = 3 \implies x - 2z = 7$ $(5)$Subtracting equation $(5)$ from equation $(4)$:$(x + z) - (x - 2z) = -2 - 7$$3z = -9 \implies z = -3$.Substitute $z = -3$ into equation $(4)$:$x - 3 = -2 \implies x = 1$.Thus,the solution is $\begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 1 \ 2 \ -3 \end{bmatrix}$.

If $\begin{bmatrix} 1 & 1 & 1 \\ 1 & -2 & -2 \\ 1 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \\ 4 \end{bmatrix}$,then $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is equal to

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