For α, β ∈ ℝ and a natural number n, let Aᵣ=|r, 1, n²/2+α; 2 r, 2, n²-β; 3 r-2, 3, n(3 n-1)/2|. Then 2 A₁₀-A₈ is

Solution

Maths sir · 4 steps

Split column 1 into r times column 2 plus a constant column, and the r disappears completely.

1. Split column 1. Column 1 is (r, 2r, 3r-2), which equals r × (1, 2, 3) plus (0, 0, -2). So Aᵣ splits into the sum of two determinants.
2. First part vanishes. In the first determinant, column 1 is r times column 2. Two proportional columns make a determinant zero, so only the second determinant survives.

   Aᵣ = |0, 1, n²/2+α; 0, 2, n²-β; -2, 3, n(3n-1)/2| = -2[(n²-β) - 2(n²/2+α)]

3. Expand along column 1. The bracket is n² - β - n² - 2α = -2α - β. So Aᵣ = -2(-2α-β) = 4α + 2β, the same value for every r.
4. So. A₁₀ = A₈ = 4α + 2β, so 2A₁₀ - A₈ = 2(4α+2β) - (4α+2β) = 4α + 2β, which is option (a).

The idea to keep. When a variable sits in a determinant, look for a column that is the variable times another column plus a constant. Split it, and the variable part usually dies by proportional columns.

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