Let a, b, c, d in arithmetic progression with common difference λ. If | x + a - c, x + b, x + a; x - 1, x + c, x + b; x - b + d, x + d, x + c | = 2, then value of λ² is equal to ________________.

Solution

Maths sir · 4 steps

Write every entry in terms of λ first, then a single column subtraction leaves a column of λ's.

1. Use the AP. Take b = a + λ, c = a + 2λ, d = a + 3λ. Then a - c = -2λ and d - b = 2λ, so the first column is x - 2λ, x - 1, x + 2λ.
2. Column operation. Apply C₂ → C₂ - C₃. The new entries are b - a, c - b, d - c, and each one equals λ. Then apply R₁ → R₁ - R₂ and R₃ → R₃ - R₂, which puts zeros in column 2 everywhere except the middle.

   Δ = |1 - 2λ, 0, -λ; x - 1, λ, x + b; 1 + 2λ, 0, λ|

3. Expand along C2. Only the middle entry λ survives, and its sign is +. Its minor is (1 - 2λ)λ - (-λ)(1 + 2λ) = λ - 2λ² + λ + 2λ² = 2λ. So Δ = λ × 2λ = 2λ², and every x has dropped out.
4. So. Setting 2λ² = 2 gives λ² = 1.

The idea to keep. When the entries come from an AP, subtract neighbouring columns or rows. The common difference shows up as a constant column, and the variable x usually drops out completely.

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