The value of | (a + 1)(a + 2), a + 2, 1; (a + 2)(a + 3), a + 3, 1; (a + 3)(a + 4), a + 4, 1 | is :

Solution

Maths sir · 4 steps

Subtract neighbouring rows: the 1s in column 3 turn into 0s and the a-terms cancel out.

1. Row operations. Apply R₂ → R₂ - R₁ and R₃ → R₃ - R₂. These do not change the value of the determinant, and they turn the last column into 1, 0, 0.
2. New entries. In column 1, (a+2)(a+3) - (a+1)(a+2) = (a+2)[(a+3)-(a+1)] = 2(a+2). In the same way, (a+3)(a+4) - (a+2)(a+3) = 2(a+3). Column 2 gives 1 in both rows.

   Δ = |(a+1)(a+2), a+2, 1; 2(a+2), 1, 0; 2(a+3), 1, 0|

3. Expand along column 3. Only the top 1 survives. Its cofactor sign is +, so Δ = 2(a+2) × 1 - 1 × 2(a+3) = 2a + 4 - 2a - 6.
4. So. Δ = -2 for every value of a, which is option (a).

The idea to keep. When a column is all 1s, subtract neighbouring rows to make two zeros there. With products of consecutive terms, the differences collapse to constants.

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