If a, b, c are non - zero real numbers and if the system of...

Given, $\left(a - 1\right)x = y + z ⇒ \left(a - 1\right)x - y - z = 0$

\left(b - 1\right)y = z + x ⇒ - x + \left(b - 1\right)y - z = 0$<br /><br />\left(c - 1\right)z = y + x = - x - y + \left(c - 1\right)z = 0$

For a non-trivial solution

\left|$\begin{cases}a - 1 & - 1 & - 1 \\ - 1 & b - 1 & - 1 \\ - 1 & - 1 & c - 1\end{cases}$\right| = 0$<br /><br />$⇒ \left|$\begin{cases}1 - a & 1 & 1 \\ 1 & 1 - b & 1 \\ 1 & 1 & 1 - c\end{cases}$\right| = 0$<br /><br />$⇒ \left(1 - a\right)\left[\left(1 - b\right)\left(1 - c\right) - 1\right] - 1\left[\left(1 - c\right) - 1\right] + \left[1 - \left(1 - b\right)\right] = 0$<br /><br />$⇒ 1 - \left(a + b + c\right) + \left(ab + bc + ca\right) - abc - 1 + \left(a + b + c\right) = 0$<br /><br />$⇒ ab + bc + ca = abc