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

The three equations can be expressed as

$(a-1) x-y-z=0$

$-x+(b-1) y-z=0$

$-x-y+(c-1) z=0$

Expressing this as a determinant, we get

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

If the matrix has a non-trivial solution, then

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

$\Rightarrow(a-1)[(b-1)(c-1)-1]+1[-(c-1)-1]-1[1+b-1]=0$

$\Rightarrow(a-1)[b c-c-b+1-1]+1[-c+1-1]-1[b]=0$

$\Rightarrow(a-1)[b c-b-c]-c-b=0$

$\Rightarrow a b c-a b-a c-b c+b+c-b-c=0$

$\Rightarrow a b+a c+b c=a b c$

Hence proved.