Given, (a−1)x=y+z⇒(a−1)x−y−z=0
(b−1)y=z+x⇒−x+(b−1)y−z=0
(c−1)z=y+x=−x−y+(c−1)z=0
For a non-trivial solution
∣a−1−1−1−1b−1−1−1−1c−1∣=0
⇒∣1−a1111−b1111−c∣=0
⇒(1−a)[(1−b)(1−c)−1]−1[(1−c)−1]+[1−(1−b)]=0
⇒1−(a+b+c)+(ab+bc+ca)−abc−1+(a+b+c)=0
⇒ab+bc+ca=abc