Let A=[aᵢⱼ] be a square matrix of order 2 with entries either 0 or 1 . Let E be the event that A is an invertible matrix. Then the probability P(E) is :

Solution

Maths sir · 3 steps

Do not list all 16 matrices: split det A = ad - bc into the two products ad and bc and count those.

1. Sample space. Write A = (a, b; c, d). Each of the 4 entries is 0 or 1, so there are 2⁴ = 16 equally likely matrices.
2. When is A invertible. A is invertible exactly when det A = ad - bc ≠ 0. Each product ad and bc is 0 or 1. A product is 1 in only 1 of its 4 cases (both entries 1), and it is 0 in the other 3.

   ad	bc	det A	Number of matrices
   1	0	1, invertible	1 × 3 = 3
   0	1	-1, invertible	3 × 1 = 3
   1	1	0, singular	1 × 1 = 1
   0	0	0, singular	3 × 3 = 9

3. So. There are 3 + 3 = 6 invertible matrices out of 16, so P(E) = 6/16 = 3/8, which is option (a).

The idea to keep. For a 2 × 2 matrix with 0 and 1 entries, think of det A = ad - bc as two independent products. Count the cases where they differ, and check that all your counts add up to the total.

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