If x=alpha, y=beta, z=gamma is the solution f | vedclass

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(C) Using Cramer's rule,we calculate the determinants:$\Delta = \begin{vmatrix} 2 & -1 & 8 \ 3 & 4 & 5 \ 5 & -2 & 7 \end{vmatrix} = 2(28+10) + 1(21-

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$7$

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(C) Using Cramer's rule,we calculate the determinants:$\Delta = \begin{vmatrix} 2 & -1 & 8 \ 3 & 4 & 5 \ 5 & -2 & 7 \end{vmatrix} = 2(28+10) + 1(21-25) + 8(-6-20) = 76 - 4 - 208 = -136$$\Delta_1 = \begin{vmatrix} 13 & -1 & 8 \ 18 & 4 & 5 \ 20 & -2 & 7 \end{vmatrix} = 13(28+10) + 1(126-100) + 8(-36-80) = 494 + 26 - 928 = -408$$\Delta_2 = \begin{vmatrix} 2 & 13 & 8 \ 3 & 18 & 5 \ 5 & 20 & 7 \end{vmatrix} = 2(126-100) - 13(21-25) + 8(60-90) = 52 + 52 - 240 = -136$$\Delta_3 = \begin{vmatrix} 2 & -1 & 13 \ 3 & 4 & 18 \ 5 & -2 & 20 \end{vmatrix} = 2(80+36) + 1(60-90) + 13(-6-20) = 232 - 30 - 338 = -136$Now,$\alpha = \frac{\Delta_1}{\Delta} = \frac{-408}{-136} = 3$,$\beta = \frac{\Delta_2}{\Delta} = \frac{-136}{-136} = 1$,$\gamma = \frac{\Delta_3}{\Delta} = \frac{-136}{-136} = 1$.Thus,$\alpha\beta + \beta\gamma + \gamma\alpha = (3)(1) + (1)(1) + (1)(3) = 3 + 1 + 3 = 7$.

If $x=\alpha, y=\beta, z=\gamma$ is the solution for the system of equations:
$\begin{aligned} 2x-y+8z &= 13 \\ 3x+4y+5z &= 18 \\ 5x-2y+7z &= 20 \end{aligned}$
then $\alpha\beta+\beta\gamma+\gamma\alpha=$

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If $x=\alpha, y=\beta, z=\gamma$ is the solution for the system of equations:$\begin{aligned} 2x-y+8z &= 13 \\ 3x+4y+5z &= 18 \\ 5x-2y+7z &= 20 \end{aligned}$then $\alpha\beta+\beta\gamma+\gamma\alpha=$