Question

If a, b, c are non-zero real numbers and if the system of equations
(a − 1) x = y + z
(b − 1) y = z + x
(c − 1) z = x + y
has a non-trivial solution, then prove that ab + bc + ca = abc.

Answer 1

The three equations can be expressed as

$$\begin{gathered} \left(a-1\right)x-y-z=0 \\ -x+\left(b-1\right)y-z=0 \\ -x-y+\left(c-1\right)z=0 \end{gathered}$$

Expressing this as a determinant, we get

$∆=\left|\begin{array}{ccc}\left(a-1\right)& -1& -1\\ -1& \left(b-1\right)& -1\\ -1& -1& \left(c-1\right)\end{array}\right|$

If the matrix has a non-trivial solution, then

$\left|\begin{array}{ccc}\left(a-1\right)& -1& -1\\ -1& \left(b-1\right)& -1\\ -1& -1& \left(c-1\right)\end{array}\right|=0$

$$\begin{gathered} \Rightarrow \left(a-1\right)\left[\left(b-1\right)\left(c-1\right)-1\right]+1\left[-\left(c-1\right)-1\right]-1\left[1+b-1\right]=0 \\ \Rightarrow \left(a-1\right)\left[bc-c-b+1-1\right]+1\left[-c+1-1\right]-1\left[b\right]=0 \\ \Rightarrow \left(a-1\right)\left[bc-b-c\right]-c-b=0 \\ \Rightarrow abc-ab-ac-bc+b+c-b-c=0 \\ \Rightarrow ab+ac+bc=abc \end{gathered}$$

Hence proved.