If the system of linear equations x-2 y | vedclass

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Question

$\mathrm{D}=\left|\begin{array}{ccc}1 & -2 & 1 \ 2 & \alpha & 3 \ 3 & -1 & \beta\end{array}ight|$

$=1(\alpha \beta+3)+2

Vedclass · Accepted Answer

$58$

Vedclass · Answer

$\mathrm{D}=\left|\begin{array}{ccc}1 & -2 & 1 \ 2 & \alpha & 3 \ 3 & -1 & \beta\end{array}ight|$

$=1(\alpha \beta+3)+2(2 \beta-9)+1(-2-3 \alpha)$

$=\alpha \beta+3+4 \beta-18-2-3 \alpha$

For infinite solutions $\mathrm{D}=0, \mathrm{D}_1=0, \mathrm{D}_2=0$ and

$D_3=0$

 $D=0$

 $\alpha \beta-3 \alpha+4 \beta=17 \ldots....($1$)$

$D_1=\left|\begin{array}{ccc}-4 & -2 & 1 \ 5 & \alpha & 3 \ 3 & -1 & \beta\end{array}ight|=0$

$D_2=\left|\begin{array}{ccc}1 & -4 & 1 \ 2 & 5 & 3 \ 3 & 3 & \beta\end{array}ight|=0$

$\Rightarrow 1(5 \beta-9)+4(2 \beta-9)+1(6-15)=0$

$13 \beta-9-36-9=0$

$13 \beta=54, \beta=\frac{54}{13} 	ext { put in }(1)$

$\frac{54}{13} \alpha-3 \alpha+4\left(\frac{54}{13}ight)=17$

$54 \alpha-39 \alpha+216=221$

$15 \alpha=5 \quad \alpha=\frac{1}{3}$

$	ext { Now, } 12 \alpha+13 \beta=12 \cdot \frac{1}{3}+13 \cdot \frac{54}{13}$

$=4+54=58$

If the system of linear equations $x-2 y+z=-4 $ ; $2 x+\alpha y+3 z=5 $ ; $3 x-y+\beta z=3$ has infinitely many solutions, then $12 \alpha+13 \beta$ is equal to

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