Let A and B be two 3 × 3 non-zero real matrices such that AB is a zero matrix. Then

Solution

Maths sir · 4 steps

B is non-zero, so one of its columns is a non-zero X with AX = 0, and that one vector forces |A| = 0.

1. Read AB column-wise. The j-th column of AB is A times the j-th column of B. Since AB = O, every column bⱼ of B satisfies Abⱼ = 0.
2. A non-zero solution. B is non-zero, so at least one column b is non-zero. Then X = b is a non-trivial solution of AX = 0, so |A| = 0. A homogeneous system with one non-zero solution has infinitely many, because kX works for every real k.
3. B is singular too. If B were invertible, multiplying AB = O on the right by B⁻¹ would give A = O. That contradicts A being non-zero, so |B| = 0.

   |adj(A)| = |A|ⁿ⁻¹ = |A|² = 0

4. So. adj(A) is singular, B is singular, and AX = 0 has non-trivial solutions. Only option (b) holds: AX = 0 has infinitely many solutions.

The idea to keep. If AB = O with both A and B non-zero, then both A and B are singular. Each non-zero column of B is a non-trivial solution of AX = 0, so that system has infinitely many solutions.

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