Question

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

• A. $1$
• B. $2$
• C. $-1$
• D. $-2$

Answer 1

Answer: (B)

$2$

Given:

$a(y+z)=x,b(z+x)=y,c(x+y)=z$

Converting these equations in the matrix form, we get

$\left[\begin{array}{ccc}-1& a& a\\ b& -1& b\\ c& c& -1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

Let's say,

$D=\left|\begin{array}{ccc}-1& a& a\\ b& -1& b\\ c& c& -1\end{array}\right|$
Using matrix operations,

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

Hence, option B.