Let S={M=[aᵢⱼ], aᵢⱼ ∈{0,1,2}, 1 ≤ i, j ≤ 2} be a sample space and A={M ∈ S: M is invertible } be an event. Then P(A) is equal to :

Solution

Maths sir · 3 steps

Count the singular matrices instead: ad = bc happens only when both products land on the same value.

1. Total outcomes. Write M = (a, b; c, d). Each of the 4 entries has 3 choices, so n(S) = 3⁴ = 81.
2. Singular condition. M is not invertible exactly when ad - bc = 0, that is ad = bc. Sort the 9 ordered pairs from {0,1,2} by their product. The number of singular matrices for each product value is the square of that pair count, because (a,d) and (b,c) are chosen independently.

   Product	Ordered pairs giving it	Count	Singular matrices
   0	(0,0),(0,1),(0,2),(1,0),(2,0)	5	5² = 25
   1	(1,1)	1	1² = 1
   2	(1,2),(2,1)	2	2² = 4
   4	(2,2)	1	1² = 1

   n(singular) = 25 + 1 + 4 + 1 = 31, n(A) = 81 - 31 = 50

3. So. P(A) = 50/81, which is option (c).

The idea to keep. For 2 × 2 matrices with entries from a small set, group the pairs by product value. Then the count of ad = bc is the sum of squares of those group sizes.

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