Let for A = [ 1, 2, 3; α, 3, 1; 1, 1, 2 ],|A| = 2. If |2 adj (2 adj (2A))| = 32ⁿ, then 3n + α is equal to

Solution

Maths sir · 4 steps

Work outside-in with determinants only: each 2 multiplies by 2³, and each adj squares the determinant.

1. Find alpha. Expand along the first row: |A| = 1(6-1) - 2(2α - 1) + 3(α - 3) = -α - 2. Setting -α - 2 = 2 gives α = -4.
2. Two rules. For a 3 × 3 matrix M, |kM| = k³|M| and |adj M| = |M|². You never need the actual adjoint, only its determinant.

   |2A| = 2³ × 2 = 2⁴ → |adj(2A)| = 2⁸ → |2 adj(2A)| = 2¹¹ → |adj(·)| = 2²² → |2 adj(·)| = 2²⁵

3. Match powers. 32ⁿ = 2⁵ⁿ, so 5n = 25 and n = 5.
4. So. 3n + α = 15 + (-4) = 11, which is option (a).

The idea to keep. For an n × n matrix, |kM| = kⁿ|M| and |adj M| = |M|ⁿ⁻¹. Peel nested adj and scalar layers one at a time, tracking only powers of 2.

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