Let A = [ 1, - 2, α; α, 2, - 1 ] and B = [ 2, α; - 1, 2; 4, - 5 ], α ∈ C. Then the absolute value of the sum of all values of α for which det(AB) = 0 is :

Solution

Maths sir · 4 steps

A is not square, so det(AB) cannot be split into det A times det B. Multiply first, then take the determinant.

1. Multiply first. A is 2 × 3 and B is 3 × 2, so AB is 2 × 2. The rule det(AB) = det A · det B needs square matrices, so you have to find AB itself.
2. Find AB. Row 1 of A is (1, -2, α) and row 2 is (α, 2, -1). Taking each row with the columns (2, -1, 4) and (α, 2, -5) gives AB = (4 + 4α, -4 - 4α; 2α - 6, α² + 9).

   det(AB) = (4+4α)(α²+9) + (4+4α)(2α-6) = 4(α+1)(α²+2α+3)

3. Collect the roots. det(AB) = 0 gives α = -1, or α² + 2α + 3 = 0. The quadratic has complex roots, and since α ∈ ℂ both of them count. Their sum is -2.
4. So. The sum of all values is -1 + (-2) = -3, and its absolute value is 3, option (a).

The idea to keep. Before you use det(AB) = det A det B, check that both matrices are square. To add up the roots, take the sum straight from the factors and do not solve the quadratic.

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