Let A be a 3 × 3 matrix having entries from the set -1, 0, 1. The number of all such matrices A having sum of all the entries equal to 5, is ___________.

Solution

Maths sir · 3 steps

Every -1 you put in needs one extra +1 to cancel it, so list the cases by how many -1s there are.

1. Set up counts. Say the matrix has p entries equal to 1, q entries equal to -1 and r zeros. Then p + q + r = 9, and the sum condition gives p - q = 5.
2. Possible cases. Since p = q + 5, we need 2q + 5 ≤ 9, so q can only be 0, 1 or 2. In each case, choose which positions get which value: 9!/(p! q! r!) ways.

   -1s (q)	1s (p)	0s (r)	Matrices
   0	5	4	9!/(5! 4!) = 126
   1	6	2	9!/(6! 1! 2!) = 252
   2	7	0	9!/(7! 2!) = 36

3. So. Adding the three cases, 126 + 252 + 36 = 414 matrices.

The common slip: 126. Ignoring the -1 entries and using only five 1s and four 0s gives just 126, missing the 252 and 36 from the other cases.

The idea to keep. For a fixed sum with entries -1, 0, 1, fix the number of -1s first. The number of 1s and 0s then follows, and each case is a multinomial count n!/(p! q! r!).

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