Question

$a\ne b\ne c\ne 1;ax+y+z=0$,
$x+by+z=0,x+y+$ cz $=0$ have non trivial solutions then $a+b+c-$ abc $=\dots \dots \dots \dots \dots$

• A. $0$
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (C)

2

$Determinent=\left|\begin{array}{ccc}1& b& 1\\ 1& 1& c\\ a& 1& 1\end{array}\right|$
$R_1\to R_1-R_2$
$\Delta =\left|\begin{array}{ccc}0& b-1& 1-c\\ 1& 1& c\\ a& 1& 1\end{array}\right|$
$R_2\to R_2-R_3$
$\Delta =\left|\begin{array}{ccc}0& b-1& 1-c\\ 1-a& 0& c-1\\ a& 1& 1\end{array}\right|$
$\Delta =-(b-1)\left(\right(1-a)-a(c-1)+(1-c\left)\right(1-a\left)\right)$
$=(1-a)(1-b)+a(1-c)(1-b)+(1-c)(1-a)$
$=1-a-b+ab+a-ac-ab+abc+1-a-c+ac$
$\Delta =abc-a-b-c+2$
Now for non-trivial solution
$\Delta =0$ $

$\therefore a+b+c-abc=2$