Let the determinant of a square matrix A of order m be m-n, where m and n
satisfy 4 m+n=22 and 17 m+4 n=93.
If det(n adj(adj(m A)))=3ᵃ 5ᵇ 6ᶜ then a+b+c is equal to :

Solution

Maths sir · 5 steps

Find m and n first, then use the adj-of-adj power rule and match the prime factors carefully.

1. Solve for m, n. Multiply 4m + n = 22 by 4 to get 16m + 4n = 88. Subtract it from 17m + 4n = 93 to get m = 5, and then n = 2. So A is of order 5 and |A| = m - n = 3.
2. Scale by m. For order 5, |5A| = 5⁵|A| = 3 × 5⁵. Call B = 5A.

   |adj(adj B)| = |B|^((m-1)²) = |B|¹⁶

3. Multiply by n. Multiplying an order-5 matrix by n = 2 multiplies its determinant by 2⁵. So the determinant is 2⁵ × (3 × 5⁵)¹⁶ = 2⁵ × 3¹⁶ × 5⁸⁰.
4. Match the form. The only source of the factor 2 is 6ᶜ, so c = 5. This 6⁵ already contains 3⁵, so a = 16 - 5 = 11. The power of 5 gives b = 80.
5. So. a + b + c = 11 + 80 + 5 = 96, which is option (a).

The idea to keep. For order m: |kA| = kᵐ|A|, |adjA| = |A|ᵐ⁻¹, and |adj(adjA)| = |A|^((m-1)²). When matching 3ᵃ 6ᶜ, remember that 6ᶜ carries a factor 3ᶜ.

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