Let A and B be two 3 × 3 matrices such that AB = I and |A| = 1/8. Then |adj (B adj(2A))| is equal to

Solution

Maths sir · 4 steps

Get |B| from AB = I, then use |adj M| = |M|² twice, from the inside out.

1. Find |B|. Take determinants of AB = I: |A| |B| = 1. Since |A| = 1/8, we get |B| = 8.

   |adj M| = |M|ⁿ⁻¹ = |M|² (n = 3), |kM| = k³|M|

2. Inner adjoint. |2A| = 2³ × 1/8 = 1, so |adj(2A)| = |2A|² = 1.
3. Outer adjoint. Let M = B adj(2A). Then |M| = |B| × |adj(2A)| = 8 × 1 = 8, and |adj M| = |M|² = 64.
4. So. |adj(B adj(2A))| = 64, which is option (c).

The idea to keep. For an n × n matrix, a scalar comes out as kⁿ and an adjoint gives |M|ⁿ⁻¹. Peel nested adjoints from the inside out using these two rules.

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