# Consider the system of linear equations

Question:

Consider the system of linear equations

$-x+y+2 z=0$

$3 x-a y+5 z=1$

$2 x-2 y-a z=7$

Let $S_{1}$ be the set of all a $\in \mathbf{R}$ for which the system is inconsistent and $\mathrm{S}_{2}$ be the set of all $\mathrm{a} \in \mathbf{R}$ for which the system has infinitely many solutions. If $n\left(S_{1}\right)$ and $n\left(S_{2}\right)$ denote the number of elements in $S_{1}$ and $\mathrm{S}_{2}$ respectively, then

1. $n\left(S_{1}\right)=2, n\left(S_{2}\right)=2$

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

3. $n\left(S_{1}\right)=2, n\left(S_{2}\right)=0$

4. $n\left(S_{1}\right)=0, n\left(S_{2}\right)=2$

Correct Option: , 3

Solution:

$\Delta=\left|\begin{array}{ccc}-1 & 1 & 2 \\ 3 & -\mathrm{a} & 5 \\ 2 & -2 & -\mathrm{a}\end{array}\right|$

$=-1\left(a^{2}+10\right)-1(-3 a-10)+2(-6+2 a)$

$=-a^{2}-10+3 a+10-12+4 a$

$\Delta=-a^{2}+7 a-12$

$\Delta=-\left[a^{2}-7 a+12\right]$

$\Delta=-[(a-3)(a-4)]$

$\Delta_{1}=\left|\begin{array}{ccc}0 & 1 & 2 \\ 1 & -\mathrm{a} & 5 \\ 7 & -2 & -\mathrm{a}\end{array}\right|$

$=0-1(-a-35)+2(-2+7 a)$

$\Rightarrow a+35-4+14 a$

$15 a+31$

Now

$\Delta_{1}=15 a+31$

For inconsistent $\Delta=0 \therefore \mathrm{a}=3, \mathrm{a}=4$ and for $\mathrm{a}=3$ and $4 \quad \Delta_{1} \neq 0$

$n\left(S_{1}\right)=2$

For infinite solution : $\Delta=0$ and $\Delta_{1}=\Delta_{2}=\Delta_{3}=0$

Not possible

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

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