The number of singular matrices of order 2 , whose elements are from the set {2,3,6,9}, is __________.

Solution

Maths sir · 4 steps

Singular means ad = bc, so count how many ways each diagonal product can be matched by the other diagonal.

1. Condition. For (a, b; c, d) to be singular we need ad - bc = 0, that is ad = bc. Entries may repeat, so each of a, b, c, d has 4 choices.
2. Count by product. List every ordered pair from {2,3,6,9} by its product. If n ordered pairs give the same product, then (a,d) and (b,c) can be chosen in n × n = n² ways. Watch the one coincidence: 2 × 9 = 3 × 6 = 18.

   Products	Ordered pairs for each	Matrices
   4, 9, 36, 81	1 each, e.g. (2,2)	4 × 1² = 4
   6, 12, 27, 54	2 each, e.g. (2,3), (3,2)	4 × 2² = 16
   18	4: (2,9), (9,2), (3,6), (6,3)	4² = 16

3. Check. The ordered pairs add up to 4 × 1 + 4 × 2 + 4 = 16 = 4², so no pair has been missed.
4. So. The number of singular matrices is 4 + 16 + 16 = 36.

The common slip: 28. Missing that 2 × 9 = 3 × 6 treats 18 as two separate groups of 2. That counts 4 + 4 instead of 16 for product 18 and gives 28.

The idea to keep. To count matrices with ad = bc, group the ordered pairs by their product and add n² for each group. Always look for different pairs that give the same product.

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