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 $a \in R$ 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)

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