Let A = [2, 2+p, 2+p+q; 4, 6+2p, 8+3p+2q; 6, 12+3p, 20+6p+3q] .
If det(adj(adj(3A))) = 2ᵐ · 3ⁿ , m, n ∈ ℕ , then m + n is equal to

Solution

Maths sir · 4 steps

Clear the first column to get |A| = 8, then apply the adj-of-adj power rule to 3A.

1. Find det A. Do R₂ → R₂ - 2R₁ and R₃ → R₃ - 3R₁. The first column becomes 2, 0, 0, with R₂ = (0, 2, 4+p) and R₃ = (0, 6, 14+3p).
2. p and q vanish. Expand along the first column: |A| = 2[2(14+3p) - 6(4+p)] = 2(28 + 6p - 24 - 6p) = 2 × 4 = 8. The value does not depend on p or q.
3. Scale by 3. A is 3 × 3, so each of the 3 rows picks up a factor 3: |3A| = 3³ |A| = 27 × 8 = 216 = 2³ · 3³.

   |adj(adj B)| = |B|^((n-1)²) = |B|⁴ (n = 3)

4. So. |adj(adj(3A))| = (2³ · 3³)⁴ = 2¹² · 3¹², so m = n = 12 and m + n = 24, option (c).

The idea to keep. For an n × n matrix, |kA| = kⁿ|A|, |adj B| = |B|ⁿ⁻¹ and |adj(adj B)| = |B|^((n-1)²). Get the base determinant first, then just raise powers.

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