The probability that a randomly chosen 2 × 2 matrix with all the entries from the set of first 10 primes, is singular, is equal to :

Solution

Maths sir · 4 steps

Singular means ad = bc, and with primes that forces the pair {b, c} to be the same as {a, d}.

1. Count all matrices. Each of the 4 entries can be any of the 10 primes, so there are 10⁴ matrices in all.

   |a, b; c, d| = 0 ⇔ ad = bc

2. Use unique factorisation. A product of two primes splits into primes in only one way. So ad = bc holds exactly when b and c are a and d, in some order.
3. Count by cases. If a = d (10 choices), then bc = a² forces b = c = a, so there is 1 matrix for each choice, 10 in all. If a ≠ d (10 × 9 = 90 choices), then (b, c) is (a, d) or (d, a), which gives 90 × 2 = 180 matrices.
4. So. There are 10 + 180 = 190 singular matrices, so the probability is 190/10⁴ = 19/10³, which is option (c).

The idea to keep. When the entries are primes, an equation like ad = bc becomes a matching problem: the same primes must appear on both sides. Count the equal and unequal cases separately.

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