#### provide solution for RD Sharma maths class 12 chapter determinant exercise 5.5 question 5

Answer: $ab+bc+ca=abc$

Hint: The system has non-trivial solution implies that the determinant is zero.

Given:  $(a-1) x = y + z$

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

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

Solution:

Take x, y and z to the LHS:

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

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

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

Let  $A = [a-1 -1 -1 -1 b-1 -1 -1 -1 c-1 ]$

We know that, $|A|=0$

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

$=>(a-1)[bc-c-b+1-1]+1[-c+1-1]-1[b] = 0$

$=>(a-1)[bc-c-b]-c-b = 0$

$=>abc-ac-ab-bc+c+b-c-b=0$

$=>abc-ac-ab-bc+c+b-c-b$

$=>ab+bc+ca=abc$

Hence proved.