Question

If the equations $a(y+z)=x,b(z+x)=y$ and $c(y+z)=x,$ where $a\ne -1,$ $b\ne -1,$ $c\ne -1,$ admit of nontrivial solutions then ${(1+a)}^{-1}+{(1+b)}^{-1}+{(1+c)}^{-1}$

• A. $2$
• B. $1$
• C. $\frac{1}{2}$
• D. none of these

Answer 1

The correct option is A $2$

Here, $D=\left|\begin{array}{ccc}-1& a& a\\ b& -1& b\\ c& c& -1\end{array}\right|$
$C_2\to C_2-C_1,C_3\to C_3-C_1$
$D=\left|\begin{array}{ccc}-1& 1+a& 1+a\\ b& -(1+b)& 0\\ c& 0& -(1+c)\end{array}\right|$
For non-trivial solution,
$D=0$
$\Rightarrow b(1+a)(1+c)+c(1+a)(1+b)=(1+b)(1+c)$ ....(i)
Consider, $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$
$=\frac{(1+b)(1+c)+(1+a)(1+c)+(1+a)(1+b)}{(1+a)(1+b)(1+c)}$
$=\frac{b(1+a)(1+c)+c(1+a)(1+b)+(1+a)(1+c)+(1+a)(1+b)}{(1+a)(1+b)(1+c)}$ (using (i))
$=\frac{(b+1)(1+a)(1+c)+(1+c)(1+a)(1+b)}{(1+a)(1+b)(1+c)}$
$=1+1=2$