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-x+y+2 z=0&n | vedclass

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Question

$\Delta=\left|\begin{array}{ccc}-1 & 1 & 2 \ 3 & -a & 5 \ 2 & -2 & -a\end{array}ight|$

$=-1\left(a^{2}+10ight)-1(-3 a

Vedclass · Accepted Answer

$\mathrm{n}\left(\mathrm{S}_{1}\right)=2, \mathrm{n}\left(\mathrm{S}_{2}\right)=0$

Vedclass · Answer

$\Delta=\left|\begin{array}{ccc}-1 & 1 & 2 \ 3 & -a & 5 \ 2 & -2 & -a\end{array}ight|$

$=-1\left(a^{2}+10ight)-1(-3 a-10)+2(-6+2 a)$

$=-a^{2}-10+3 a+10-12+4 a$

$\Delta=-a^{2}+7 a-12$

$\Delta=-\left[a^{2}-7 a+12ight]$

$\Delta=-[(a-3)(a-4)]$

$\Delta_{1}=\left|\begin{array}{ccc}0 & 1 & 2 \ 1 & -a & 5 \ 7 & -2 & -a\end{array}ight|$

$=0-1(-\mathrm{a}-35)+2(-2+7 \mathrm{a})$

$\Rightarrow \mathrm{a}+35-4+14 \mathrm{a}$

$15 \mathrm{a}+31$

Now $\quad \Delta_{1}=15 \mathrm{a}+31$

For inconsistent $\Delta=0 	herefore \mathrm{a}=3, \mathrm{a}=4$

and for $\mathrm{a}=3$ and $4 \quad \Delta_{1} 
eq 0$

$\mathrm{n}\left(\mathrm{S}_{1}ight)=2$

For infinite solution : $\Delta=0$

and $\Delta_{1}=\Delta_{2}=\Delta_{3}=0$

Not possible

$	herefore \mathrm{n}\left(\mathrm{S}_{2}ight)=0$

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