Let A = begin{bmatrix} 1 & -1 & 1 2 & 1 & -3 | vedclass

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(B) Given the matrix equation $AX = B$,we have: $\begin{bmatrix} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3

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

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(B) Given the matrix equation $AX = B$,we have: $\begin{bmatrix} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = \begin{bmatrix} 4 \ 0 \ 2 \end{bmatrix}$ Applying row operations $R_2 ightarrow R_2 - 2R_1$ and $R_3 ightarrow R_3 - R_1$: $\begin{bmatrix} 1 & -1 & 1 \ 0 & 3 & -5 \ 0 & 2 & 0 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = \begin{bmatrix} 4 \ -8 \ -2 \end{bmatrix}$ From the third row,$2x_2 = -2$,which gives $x_2 = -1$. Substituting $x_2 = -1$ into the second row equation $3x_2 - 5x_3 = -8$: $3(-1) - 5x_3 = -8 \Rightarrow -3 - 5x_3 = -8 \Rightarrow -5x_3 = -5 \Rightarrow x_3 = 1$. Substituting $x_2 = -1$ and $x_3 = 1$ into the first row equation $x_1 - x_2 + x_3 = 4$: $x_1 - (-1) + 1 = 4 \Rightarrow x_1 + 1 + 1 = 4 \Rightarrow x_1 = 2$. Thus,$X = \begin{bmatrix} 2 \ -1 \ 1 \end{bmatrix}$.

Let $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$ such that $AX = B$,then $X =$

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