If A=1/(5 ! 6 ! 7 !)[5 !, 6 !, 7 !; 6 !, 7 !, 8 !; 7 !, 8 !, 9 !], then |adj(adj(2 A))| is equal to :

Solution

Maths sir · 4 steps

Factor out a factorial per row to get |A|, then two adjoints raise |2A| to the power 4.

1. Factor the rows. Take 5! out of row 1, 6! out of row 2 and 7! out of row 3. Since 6!/5! = 6, 7!/5! = 42, 8!/6! = 56 and 9!/7! = 72, the rows become (1, 6, 42), (1, 7, 56) and (1, 8, 72). Subtracting row 1 from the other two leaves (0, 1, 14) and (0, 2, 30), so this determinant is 30 - 28 = 2.
2. Value of det A. The factor 1/(5! 6! 7!) in front is meant to cancel the three factorials pulled out of the rows, which gives |A| = 2. That is the reading behind the official key. Read strictly as a scalar times a 3 × 3 matrix, the factor would be cubed and no option would match.
3. Scale by 2. For a 3 × 3 matrix, |kA| = k³|A|. So |2A| = 8 × 2 = 16 = 2⁴.

   |adj(adjB)| = |B|^((n-1)²) = 16⁴ = 2¹⁶

4. So. With B = 2A and n = 3, the exponent is (n-1)² = 4, which gives |adj(adj(2A))| = 2¹⁶. The answer is option (d).

The idea to keep. For an n × n matrix, |adjB| = |B|ⁿ⁻¹, and taking the adjoint twice gives |B|^((n-1)²). Also remember that |kA| = kⁿ|A|, not k|A|.

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