If the solution for the system of equations $x+2 y-z=3$, $3 x-y+2 z=1$ and $2 x-2 y+3 z=2$ is $(\alpha, \beta, \gamma)$, then $\alpha^2+\beta^2+\gamma^2=$

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Aicharya · Accepted Answer

$\left[\begin{array}{ccc}1 & 2 & -1 \ 3 & -1 & 2 \ 2 & -2 & 3\end{array}ight]\left[\begin{array}{l}x \ y \ z\end{array}ight]=\left[\begin{array}{l}3 \ 1 \ 2\end{array}ight]$ Augmented matrix is: $\begin{aligned} & {\left[\begin{array}{ccccc} 1 & 2 & -1 & : & 3 \ 3 & -1 & 2 & : & 1 \ 2 & -2 & 3 & : & 2 \end{array}ight]} \ & R_2 ightarrow R_2-3 R_1 	ext { \& } R_3 ightarrow R_3-2 R_1 \ & \sim\left[\begin{array}{ccccc} 1 & 2 & -1 & : & 3 \ 0 & -7 & 5 & : & -8 \ 0 & -6 & 5 & : & -4 \end{array}ight] \ & R_3 ightarrow 7 R_3-6 R_2 \ & \sim\left[\begin{array}{ccccc} 1 & 2 & -1 & : & 3 \ 0 & -7 & 5 & : & -8 \ 0 & 0 & 5 & : & 20 \end{array}ight] \end{aligned}$ $\because \operatorname{Rank}(A: B)=\operatorname{Rank}(A)=3$ $	herefore \quad$ Solution exists and can be given by : $\begin{aligned} & 	herefore\left[\begin{array}{ccc} 1 & 2 & -1 \ 0 & -7 & 5 \ 0 & 0 & 5 \end{array}ight]\left[\begin{array}{l} x \ y \ z \end{array}ight]=\left[\begin{array}{c} 3 \ -8 \ 20 \end{array}ight] \ & \Rightarrow\left[\begin{array}{l} x+2 y-z \ 0-7 y+5 z \ 0+0+5 z \end{array}ight]=\left[\begin{array}{c} 3 \ -8 \ 20 \end{array}ight] \ & 	herefore \quad x+2 y-z=3 \ & -7 y+5 z=-8 \ & 5 z=20 \Rightarrow z=4 \ & 	herefore \quad-7 y+20=-8 \Rightarrow y=4 \ & x+8-4=3 \Rightarrow x=1 \ & 	herefore \alpha=1, \beta=4, \gamma=4 \ & 	herefore \quad \alpha^2+\beta^2+\gamma^2=1+16+16=33 \ & \end{aligned}$

If the solution for the system of equations $x+2 y-z=3$, $3 x-y+2 z=1$ and $2 x-2 y+3 z=2$ is $(\alpha,…

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