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
\[\left( a - 1 \right)x - y - z = 0\] 
\[ - x + \left( b - 1 \right)y - z = 0\] 
\[ - x - y + \left( c - 1 \right)z = 0\]
Expressing this as a determinant, we get
\[∆ = \begin{vmatrix}\left( a - 1 \right) & - 1 & - 1 \\ - 1 & \left( b - 1 \right) & - 1 \\ - 1 & - 1 & \left( c - 1 \right)\end{vmatrix}\]
If the matrix has a non-trivial solution, then

\[\begin{vmatrix}\left( a - 1 \right) & - 1 & - 1 \\ - 1 & \left( b - 1 \right) & - 1 \\ - 1 & - 1 & \left( c - 1 \right)\end{vmatrix} = 0\]
\[\Rightarrow \left( a - 1 \right)\left[ \left( b - 1 \right)\left( c - 1 \right) - 1 \right] + 1\left[ - \left( c - 1 \right) - 1 \right] - 1\left[ 1 + b - 1 \right] = 0\] 
\[ \Rightarrow \left( a - 1 \right)\left[ bc - c - b + 1 - 1 \right] + 1\left[ - c + 1 - 1 \right] - 1\left[ b \right] = 0\] 
\[ \Rightarrow \left( a - 1 \right)\left[ bc - b - c \right] - c - b = 0\] 
\[ \Rightarrow abc - ab - ac - bc + b + c - b - c = 0\] 
\[ \Rightarrow ab + ac + bc = abc\]
Hence proved.