Question

Consider the system of linear equations
$-x+y+2z=0$
$3x-ay+5z=1$
$2x-2y-az=7$
Let $S_1$ be the set of all $a\in R$ for which system is inconsistent and $S_2$ be the set of all $a\in R$ 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 $S_2$ respectively, then

• A. $n\left(S_1\right)=2,n\left(S_2\right)=0$
• B. $n\left(S_1\right)=1,n\left(S_2\right)=0$
• C. $n\left(S_1\right)=2,n\left(S_2\right)=2$
• D. $n\left(S_1\right)=0,n\left(S_2\right)=2$

Answer 1

The correct option is A $n\left(S_1\right)=2,n\left(S_2\right)=0$

$\Delta =\left|\begin{array}{ccc}-1& 1& 2\\ 3& -a& 5\\ 2& -2& a\end{array}\right|=0$
$\Rightarrow -(a^2+10)-(-3a-10)+2(-6+2a)=0$
$\Rightarrow -a^2+7a-12=0$
$\Rightarrow (a-3)(a-4)=0$
$\Rightarrow a=3$ or $a=4$
${\Delta }_1=\left|\begin{array}{ccc}0& 1& 2\\ 1& -a& 5\\ 7& -2& -a\end{array}\right|=a+35-4+14a=15a+31$
${\Delta }_2=\left|\begin{array}{ccc}-1& 0& 2\\ 3& 1& 5\\ 2& 7& -a\end{array}\right|=a+35+38=a+73$
${\Delta }_3=\left|\begin{array}{ccc}-1& 1& 0\\ 3& -a& 1\\ 2& -2& 7\end{array}\right|=7a-2+19=7a+17$
Case - I
For ${}^{\prime }{\infty }^{\prime }$ solutions $\Delta ={\Delta }_1={\Delta }_2={\Delta }_3=0$
Here there are no common values of ${}^{\prime }{a}^{\prime }$
$\Rightarrow n\left(S_2\right)=0$
Case - II
For inconsistent system $\Delta =0$ and one of ${\Delta }_1,{\Delta }_2,{\Delta }_3\ne 0$
$\therefore a=3,a=4$ both cases hold good
$\Rightarrow n\left(S_1\right)=2$