Question

If $a, \, b$ and $c$ are non-zero real numbers and if the system of equations $\left(a - 1\right)x=y+z, \, \left(b - 1\right)y=z+x$ and $\left(c - 1\right)z=x+y$ have a non-trivial solution, then $\frac{3}{2 a}+\frac{3}{2 b}+\frac{3}{2 c}$ is equal to

Answer 1

System of equations can be written as
$\left(a - 1\right)x-y-z=0$
$x-\left(b - 1\right)y+z=0$
$x+y-\left(c - 1\right)z=0$
Using Cramer's rule, for non-trivial solution $\Delta =0$
$\begin{vmatrix} \left(a - 1\right) & -1 & -1 \\ 1 & -\left(b - 1\right) & 1 \\ 1 & 1 & -\left(c - 1\right) \end{vmatrix}=0$
$\Rightarrow \left(a - 1\right)\left[\left(b - 1\right) \left(c - 1\right) - 1\right]-\left(- 1\right)\left[- \left(c - 1\right) - 1\right]-1\left[1 + b - 1\right]=0$
$\Rightarrow \left(a - 1\right)\left[b c - b - c\right]-c-b=0$
$\Rightarrow abc-ab-ac-bc+b+c-c-b=0$
$\Rightarrow ab+bc+ca=abc\Rightarrow \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)=1$