Question

If the equations $a(y+z)=x$, $b(z+x)=y$, $c(x+y)=z$ have non-trivial solutions, then the value of $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}is$

• A. $1$
• B. $2$
• C. $-1$
• D. $-2$

Answer 1

Answer: (B)

$\left|\begin{array}{ccc}-1& a& a\\ b& -1& b\\ c& c& -1\end{array}\right|=0$
Applying $C_2\to C2-C_1$ and $C_3\to C_3-C_1$, we get
$\left|\begin{array}{ccc}-1& a+1& a+1\\ b& -\left(b+1\right)& 0\\ c& 0& -\left(1+c\right)\end{array}\right|=0$
Taking $\left(a+1\right)$, $\left(b+1\right)$ and $\left(c+1\right)$ common from $R_1$, $R_2$ and $R_3$ respectively.
$\left|\begin{array}{ccc}-\frac{1}{a+1}& 1& 1\\ \frac{b}{b+1}& -1& 0\\ \frac{c}{c+1}& 0& -1\end{array}\right|=0$
Expanding along $C_1$, we get
$-\frac{1}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=0$
$\Rightarrow -\frac{1}{a+1}+1-\frac{1}{b+1}+1-\frac{1}{c+1}=0$
$\therefore \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2$