If α ≠ a, β ≠ b, γ ≠ c and |α, b, c; a, β, c; a, b, γ|=0, then a/(α-a)+b/(β-b)+γ/(γ-c) is equal to :

Solution

Maths sir · 5 steps

Subtract row 1 from the other rows, then divide by the three differences, and the terms almost cancel.

1. Create zeros. Do R₂ → R₂ - R₁ and R₃ → R₃ - R₁. The rows become (a-α, β-b, 0) and (a-α, 0, γ-c), and the value of the determinant does not change.
2. Expand along row 1. Expanding with these zeros and writing a-α = -(α-a), the condition determinant = 0 becomes:

   α(β-b)(γ-c) + b(α-a)(γ-c) + c(α-a)(β-b) = 0

3. Divide through. None of the differences is zero, so divide by (α-a)(β-b)(γ-c). This gives α/(α-a) + b/(β-b) + c/(γ-c) = 0.
4. Match the asked form. Use a/(α-a) = α/(α-a) - 1 and γ/(γ-c) = c/(γ-c) + 1. The required sum is (α/(α-a) + b/(β-b) + c/(γ-c)) - 1 + 1 = 0 - 1 + 1.
5. So. The expression equals 0, which is option (d).

The idea to keep. When each row differs from the others in only one entry, subtract one row from the rest to get zeros. Then use a/(α-a) = α/(α-a) - 1 to rewrite the terms in the form the question asks for.

Answer from the NTA final answer key. Solution by PrepWiser.