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

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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.

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