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(C) Given,$\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \ 2 & -1 & 2 \ -1 & 1 & 5\end{array}ight|$,$\Delta_1=\left|\begin{array}{ccc}5 & 1 & 1 \ 4 &

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$5$

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(C) Given,$\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \ 2 & -1 & 2 \ -1 & 1 & 5\end{array}ight|$,$\Delta_1=\left|\begin{array}{ccc}5 & 1 & 1 \ 4 & -1 & 2 \ 11 & 1 & 5\end{array}ight|$ and $X=\left[\begin{array}{l}\alpha \ 2 \ \beta\end{array}ight]$.By Cramer's rule,$x = \frac{\Delta_1}{\Delta}$.First,calculate $\Delta = 1(-5-2) - 1(10+2) + 1(2-1) = -7 - 12 + 1 = -18$.Next,calculate $\Delta_1 = 5(-5-2) - 1(20-22) + 1(4+11) = 5(-7) - 1(-2) + 15 = -35 + 2 + 15 = -18$.Thus,$\alpha = x = \frac{\Delta_1}{\Delta} = \frac{-18}{-18} = 1$.Now,use the system $AX=B$ where $X = [\alpha, 2, \beta]^T = [1, 2, \beta]^T$:$1(1) + 1(2) + 1(\beta) = 5 \Rightarrow 3 + \beta = 5 \Rightarrow \beta = 2$.Alternatively,using the second row: $2(1) - 1(2) + 2(\beta) = 4 \Rightarrow 2 - 2 + 2\beta = 4 \Rightarrow 2\beta = 4 \Rightarrow \beta = 2$.Therefore,$\alpha^2 + \beta^2 = 1^2 + 2^2 = 1 + 4 = 5$.

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