For a 3 × 3 matrix M, let trace (M) denote the sum of all the diagonal elements of M. Let A be a 3 × 3 matrix such that |A|=1/2 and trace (A)=3. If B=adj(adj(2 A)), then the value of |B|+ trace (B) equals :

Solution

Maths sir · 4 steps

Do not compute any adjoint here: one formula turns adj of adj into a plain multiple of 2A.

1. Adjoint twice. For any n × n matrix X, adj(adj X) = |X|ⁿ⁻² X. With n = 3 this becomes |X| X.

   B = adj(adj(2A)) = |2A| · 2A

2. Find the scalar. For a 3 × 3 matrix, |2A| = 2³ |A| = 8 × 1/2 = 4. So B = 4 × 2A = 8A.
3. Determinant and trace. |B| = |8A| = 8³ |A| = 512 × 1/2 = 256. The trace scales by 8, so trace(B) = 8 × 3 = 24.
4. So. |B| + trace(B) = 256 + 24 = 280, which is option (d).

The idea to keep. Pulling a scalar out of a determinant gives kⁿ, but pulling it out of a trace gives only k. Remember adj(adj X) = |X|ⁿ⁻²X and never expand an adjoint by hand.

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