Let M = { A = ( a, b; c, d ):a,b,c,d ∈ { ± 3, ± 2, ± 1,0} }. Define f : M → Z, as f(A) = det(A), for all A∈M, where z is set of all integers. Then the number of A∈M such that f(A) = 15 is equal to _____________.

Solution

Maths sir · 3 steps

With entries at most 3, ad and bc can only be 9 or 6 in size, so 15 must be 9 + 6.

1. Set up. We need ad - bc = 15. Each entry lies in {0, ± 1, ± 2, ± 3}, so a product of two entries can only be 0, ± 1, ± 2, ± 3, ± 4, ± 6 or ± 9.
2. Find the splits. The largest ad can be is 9 and the smallest bc can be is -9. To reach 15 we need ad + (-bc) = 15 using two of these products, and the only ways are 9 + 6 and 6 + 9.

   Case	Pairs (a, d)	Pairs (b, c)	Matrices
   ad = 9, bc = -6	(3, 3), (-3, -3): 2 ways	(2, -3), (-2, 3), (3, -2), (-3, 2): 4 ways	2 × 4 = 8
   ad = 6, bc = -9	(2, 3), (3, 2), (-2, -3), (-3, -2): 4 ways	(3, -3), (-3, 3): 2 ways	4 × 2 = 8

3. So. The total is 8 + 8 = 16 matrices with f(A) = 15.

The common slip: 8. Counting only the case ad = 9, bc = -6 gives 8. The swapped case ad = 6, bc = -9 gives another 8.

The idea to keep. When entries come from a small set, list every value a product can take first. A large target like 15 then forces only one or two splits, and you count ordered pairs in each.

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