Let A and B be two square matrices of order 3 such that |A|=3 and |B|=2. Then |Aᵀ A(adj(2 A))⁻¹(adj(4 B))(adj(AB))⁻¹ AAᵀ| is equal to :

Solution

Maths sir · 2 steps

The determinant of a product is the product of determinants, so take the determinant of each factor.

1. Tools needed. For order 3: |Mᵀ| = |M|, |M⁻¹| = 1/|M|, |kM| = k³|M| and |adj M| = |M|². Also |MN| = |M||N|, so every factor can be handled on its own.

   Factor	Determinant
   Aᵀ, A, A, Aᵀ	3 each, so 3⁴ = 81 together
   (adj(2A))⁻¹	|2A| = 8 × 3 = 24, so 1/24² = 1/576
   adj(4B)	|4B| = 64 × 2 = 128, so 128² = 16384
   (adj(AB))⁻¹	|AB| = 6, so 1/6² = 1/36

   (81 × 16384)/(576 × 36) = (81 × 16384)/20736 = (81 × 16384)/(81 × 256) = 64

2. So. The determinant equals 64, which is option (c).

The idea to keep. Break a long matrix product into single factors. For order n, use |kM| = kⁿ|M| and |adj M| = |M|ⁿ⁻¹. Never multiply the matrices themselves.

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