Let A be a 3 × 3 matrix such that |adj(adj(adj A))|=12⁴. Then |A⁻¹ adj A| is equal to

Solution

Maths sir · 4 steps

Each adj squares the determinant of a 3 by 3 matrix, so three adjs give the power 2³ = 8.

1. One adjoint. For an n × n matrix, |adj A| = |A|ⁿ⁻¹. Here n = 3, so each adjoint squares the determinant: |adj A| = |A|².
2. Three adjoints. Apply the rule three times. |adj(adj A)| = (|A|²)² = |A|⁴, and one more adjoint gives (|A|⁴)² = |A|⁸.

   |A|⁸ = 12⁴ ⇒ |A|² = 12 ⇒ |A| = 2√3

3. Find the determinant. Use |PQ| = |P| |Q| and |A⁻¹| = 1/|A|. Then |A⁻¹ adj A| = 1/|A| × |A|² = |A|.
4. So. |A⁻¹ adj A| = |A| = 2√3, which is option (b).

The idea to keep. For a 3 by 3 matrix, every adj raises the determinant to the power 2. So k adjoints in a row give |A|^(2ᵏ). The powers multiply; they do not add.

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