Question

If $a,b,c$ are non - zero real numbers and if the system of equations
$(a-1)x=y+z$
$(b-1)y=z+x$
$(c-1)z=x+y$
has a non-trivial solution, then $ab+bc+ca$ is equal to

• A. $a+b+c$
• B. $abc$
• C. $1$
• D. $-1$

Answer 1

Answer: (B)

$abc$

$\left|\begin{array}{ccc}a-1& -1& -1\\ -1& b-1& -1\\ -1& -1& c-1\end{array}\right|$

$R_3\to R_3-R_2$

$\left|\begin{array}{ccc}a-1& -1& -1\\ -1& b-1& -1\\ 0& -b& c\end{array}\right|$

$R_2\to R_2-R_1$

$\left|\begin{array}{ccc}a-1& -1& -1\\ -a& b& 0\\ 0& -b& c\end{array}\right|$

$\Rightarrow \left|\begin{array}{ccc}a-1& -1& -1\\ -a& b& 0\\ 0& -b& c\end{array}\right|=0$

$(a-1)[bc-0]+1[-ac]-1\left[ab\right]=0$

$abc-bc-ac-ab=0$

$ab+ac+bc=abc$.