Let A and B be two square matrices of order 2. If det (A) = 2, det (B) = 3 and det ( (det 5(det A)B)A² ) = 2ᵃ3ᵇ5ᶜ for some a, b, c, ∈ N, then a + b + c is equal to :

Solution

Maths sir · 4 steps

Work from the inside out: every det(k M) for a 2×2 matrix pulls out k squared, not k.

1. The rule to use. For a square matrix M of order n, det(kM) = kⁿ det M. Here n = 2, so each scalar comes out squared. Also det(A²) = (det A)².
2. Inner determinant. 5 det A = 5 × 2 = 10, so the inner part is det(10B) = 10² × 3 = 300. This is just a number.

   det(300 A²) = 300² × (det A)² = 90000 × 4 = 360000

3. Prime factorise. 300 = 2² × 3 × 5², so 300² × 4 = 2⁴ × 3² × 5⁴ × 2² = 2⁶ × 3² × 5⁴. So a = 6, b = 2, c = 4.
4. So. a + b + c = 6 + 2 + 4 = 12, which is option (b).

The idea to keep. A scalar outside a determinant comes out as kⁿ, where n is the order. Peel nested determinants from the inside, and each one becomes a plain number.

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