If AX=B,where A=left[begin{array}{ccc}1 & -1 | vedclass

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(A) Given the matrix equation $AX=B$:$\left[\begin{array}{ccc}1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 2 & 1\end{array}ight]\left[\begin{array}{l}x \ y \ z

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$2$

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(A) Given the matrix equation $AX=B$:$\left[\begin{array}{ccc}1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 2 & 1\end{array}ight]\left[\begin{array}{l}x \ y \ z\end{array}ight]=\left[\begin{array}{l}4 \ 0 \ 2\end{array}ight]$Applying row operations $R_2 ightarrow R_2 - 2R_1$ and $R_3 ightarrow R_3 - R_1$:$\left[\begin{array}{ccc}1 & -1 & 1 \ 0 & 3 & -5 \ 0 & 3 & 0\end{array}ight]\left[\begin{array}{l}x \ y \ z\end{array}ight]=\left[\begin{array}{c}4 \ -8 \ -2\end{array}ight]$From the third row,$3y = -2 \Rightarrow y = -\frac{2}{3}$.From the second row,$3y - 5z = -8 \Rightarrow 3(-\frac{2}{3}) - 5z = -8 \Rightarrow -2 - 5z = -8 \Rightarrow -5z = -6 \Rightarrow z = \frac{6}{5}$.From the first row,$x - y + z = 4 \Rightarrow x - (-\frac{2}{3}) + \frac{6}{5} = 4 \Rightarrow x + \frac{2}{3} + \frac{6}{5} = 4 \Rightarrow x + \frac{10+18}{15} = 4 \Rightarrow x + \frac{28}{15} = 4 \Rightarrow x = 4 - \frac{28}{15} = \frac{60-28}{15} = \frac{32}{15}$.Calculating $2x + y - z = 2(\frac{32}{15}) + (-\frac{2}{3}) - \frac{6}{5} = \frac{64}{15} - \frac{10}{15} - \frac{18}{15} = \frac{36}{15} = 2.4$.

If $AX=B$,where $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 2 & 1\end{array}\right]$,$X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$,and $B=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$,then find the value of $2x+y-z$.

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