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(A) The system of equations can be written in matrix form $AX=B$ as: $\left[\begin{array}{ccc}1 & 2 & -1 \ 3 & -1 & 2 \ 2 & -2 & 3\end{array}ight]

Vedclass · Accepted Answer

$33$

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(A) The system of equations can be written in matrix form $AX=B$ as: $\left[\begin{array}{ccc}1 & 2 & -1 \ 3 & -1 & 2 \ 2 & -2 & 3\end{array}ight]\left[\begin{array}{l}x \ y \ z\end{array}ight]=\left[\begin{array}{l}3 \ 1 \ 2\end{array}ight]$The augmented matrix is: $\left[\begin{array}{ccccc}1 & 2 & -1 & : & 3 \ 3 & -1 & 2 & : & 1 \ 2 & -2 & 3 & : & 2\end{array}ight]$Applying row operations $R_2 ightarrow R_2-3R_1$ and $R_3 ightarrow R_3-2R_1$:$\sim\left[\begin{array}{ccccc}1 & 2 & -1 & : & 3 \ 0 & -7 & 5 & : & -8 \ 0 & -6 & 5 & : & -4\end{array}ight]$Applying $R_3 ightarrow 7R_3-6R_2$:$\sim\left[\begin{array}{ccccc}1 & 2 & -1 & : & 3 \ 0 & -7 & 5 & : & -8 \ 0 & 0 & 5 & : & 20\end{array}ight]$Since $\operatorname{Rank}(A:B)=\operatorname{Rank}(A)=3$,the system has a unique solution.From the row-echelon form:$5z=20 \Rightarrow z=4$$-7y+5(4)=-8 \Rightarrow -7y=-28 \Rightarrow y=4$$x+2(4)-4=3 \Rightarrow x+4=3 \Rightarrow x=-1$Wait,re-calculating: $x+8-4=3 \Rightarrow x+4=3 \Rightarrow x=-1$. Let's re-check the system: $x+2y-z=3 \Rightarrow -1+8-4 = 3$ (Correct).So,$\alpha=-1, \beta=4, \gamma=4$.Then $\alpha^2+\beta^2+\gamma^2 = (-1)^2 + 4^2 + 4^2 = 1 + 16 + 16 = 33$.

List-$I$	List-$II$
$(A)$ $\lambda=8, \mu \neq 15$	$1$. Infinitely many solutions
$(B)$ $\lambda \neq 8, \mu \in R$	$2$. No solution
$(C)$ $\lambda=8, \mu=15$	$3$. Unique solution

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

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