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

Answer 1

Answer: (C)

$\Delta =\left|\begin{array}{ccc}-1& 1& 2\\ 3& -a& 5\\ 2& -2& -a\end{array}\right|$
$=-1\left(a^2+10\right)-1(-3a-10)+2(-6+2a)$
$=-a^2-10+3a+10-12+4a$
$\Delta =-a^2+7a-12$
$\Delta =-\left[a^2-7a+12\right]$
$\Delta =-[(a-3)(a-4)]$
${\Delta }_1=\left|\begin{array}{ccc}0& 1& 2\\ 1& -a& 5\\ 7& -2& -a\end{array}\right|$
$=0-1(-a-35)+2(-2+7a)$
$\Rightarrow a+35-4+14a$
$15a+31$
Now ${\Delta }_1=15a+31$
For inconsistent $\Delta =0$
$\therefore a=3,a=4$ and for $a=3$ and $4$
${\Delta }_1\ne 0$
$n\left(S_1\right)=2$
For infinite solution : $\Delta =0$
and ${\Delta }_1={\Delta }_2={\Delta }_3=0$
Not possible
$\therefore n\left(S_2\right)=0$