Question

If the system of linear equations
$2x+2y+3z=a$
$3x-y+5z=b$
$x-3y+2z=c$
where $a,b,c$ are non-zero real numbers, has more than one solution, then:

• A. $b+c-a=0$
• B. $b-c+a=0$
• C. $b-c-a=0$
• D. $a+b+c=0$

Answer 1

The correct option is C $b-c-a=0$

Since, system of linear equations has more than one solution
$\therefore {\Delta }_1=0={\Delta }_2=0={\Delta }_3$
${\Delta }_1=\left|\begin{array}{ccc}2& 2& a\\ 3& -1& b\\ 1& -3& c\end{array}\right|=0$
$\Rightarrow 2(-c+3b)-2(3c-b)+a(-9+1)=0$
$\Rightarrow -8c+8b-8a=0$
$\Rightarrow b-c-a=0$

$\underset{–}{\text{Alternate approach:}}$
Adding first and third equation, we get second equation.
Hence, $a+c=b$ or $b-c-a=0$