Let A be a matrix of order 3 × 3 and det (A) = 2. Then det (det (A) adj (5 adj (A³))) is equal to _____________.

Solution

Maths sir · 4 steps

Peel the layers from the inside out, using two rules: |kB| = k³|B| and |adj B| = |B|².

1. Two rules. For any 3 × 3 matrix B and scalar k: |kB| = k³|B| and |adj B| = |B|³⁻¹ = |B|². Also |A³| = |A|³.
2. Inside layers. |A³| = 2³ = 8, so |adj(A³)| = 8² = 64. Then |5 adj(A³)| = 5³ × 64 = 125 × 64 = 8000.
3. Outer adjoint. |adj(5 adj(A³))| = 8000² = 64 × 10⁶. The outside factor det(A) = 2 is just a scalar multiplying a 3 × 3 matrix, so it comes out as 2³.

   2³ × 64 × 10⁶ = 8 × 64 × 10⁶ = 512 × 10⁶

4. So. The value is 512 × 10⁶, which is option (a).

The idea to keep. Any scalar in front of a 3 × 3 matrix comes out of the determinant cubed, and every adjoint squares the determinant. Work from the innermost bracket outwards.

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