JEE Main 2020 · 5 September · evening · Single correct · +4 / −1
If a + x = b + y = c + z + 1, where a, b, c, x, y, z
are non-zero distinct real numbers, then
| x, a + y, x + a; y, b + y, y + b; z, c + y, z + c | is equal to :
- (a) y(b – a)
- (b) y(a – b)
- (c) y(a – c)
- (d) 0
Answer and solution
Answer (b) y(a – b)
Solution
Maths sir · 4 steps
Subtract columns to strip away the sums, then use the given condition to make one column almost constant.
- Clean the columns. Do C₃ → C₃ - C₁, so the third column becomes a, b, c. Then do C₂ → C₂ - C₃, so the second column becomes y, y, y. Take y out of C₂.
Δ = y|x, 1, a; y, 1, b; z, 1, c|
- Use the condition. Let a + x = b + y = k, so c + z = k - 1. Do C₁ → C₁ + C₃: the first column becomes k, k, k-1. Now C₁ → C₁ - k C₂ turns it into 0, 0, -1.
- Expand. Only the bottom entry of C₁ survives. Its sign is (-1)³⁺¹ = +1 and its minor is |1, a; 1, b| = b - a. So Δ = y × (-1)(b - a).
- So. Δ = y(a - b), which is option (b).
The idea to keep. When the entries are sums, subtract columns first to split them apart. Then use the given relation to make a column constant, so it vanishes against a column of 1s.
Answer from the NTA final answer key. Solution by PrepWiser.