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

Question

If $x=\alpha, y=\beta, z=\gamma$ is the solution, for the system of equations $\begin{aligned} 2 x-y+8 z & =13 \ 3 x+4 y+5 z & =18 \ 5 x-2 y+7 z & =20 \end{aligned}$ then $\alpha \beta+\beta \gamma+\gamma \alpha=$, 1, 0, 7, -3

MARKS App · Accepted Answer

According to Cramer's rule, $\begin{aligned} & x=\alpha=\frac{\Delta_1}{\Delta}, y=\beta=\frac{\Delta_2}{\Delta} \text { and } z=\gamma=\frac{\Delta_3}{\Delta} \ & \text {where } \begin{aligned} \Delta & =\left|\begin{array}{ccc} 2 & -1 & 8 \ 3 & 4 & 5 \ 5 & -2 & 7 \end{array}\right| \ & =2(28+10)+1(21-25)+8(-6-20) \ & =76-4-208=76-212=-136 \end{aligned} \end{aligned}$ $\begin{aligned} \Delta_1 & =\left|\begin{array}{ccc} 13 & -1 & 8 \ 18 & 4 & 5 \ 20 & -2 & 7 \end{array}\right| \ & =13(28+10)+1(126-100)+8(-36-80) \ & =(13 \times 38)+26-(8 \times 116) \ & =494+26-928=-408 \ \Delta_2 & =\left|\begin{array}{lll} 2 & 13 & 8 \ 3 & 18 & 5 \ 5 & 20 & 7 \end{array}\right| \ & =2(126-100)-13(21-25)+8(60-90) \ & =(2 \times 26)+(13 \times 4)-(8 \times 30) \ & =52+52-240=-136 \ \Delta_3 & =\left|\begin{array}{ccc} 2 & -1 & 13 \ 3 & 4 & 18 \ 5 & -2 & 20 \end{array}\right| \ & =2(80+36)+1(60-90)+13(-6-20) \ & =(2 \times 116)-30-(13 \times 26) \ & =232-30-338=-136 \end{aligned}$ So, $\alpha=3, \beta=1, \gamma=1$ $\therefore \quad \alpha \beta+\beta \gamma+\gamma \alpha=3+1+3=7$ Hence, option (c) is correct.

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