Let A = [aᵢⱼ] be a 2 × 2 matrix such that aᵢⱼ ∈ {0, 1} for all i and j. Let the random variable X denote the possible values of the determinant of the matrix A. Then, the variance of X is:

Solution

Maths sir · 3 steps

Count the 16 matrices by their determinant, and symmetry makes the mean zero for free.

1. Total matrices. Each of the 4 entries is 0 or 1, so there are 2⁴ = 16 equally likely matrices. Write A = (a, b; c, d), so X = ad - bc, where ad and bc are each 0 or 1.
2. Count each value. X = 1 needs ad = 1 and bc = 0. That means a = d = 1, and (b, c) is one of 3 pairs that are not both 1, so there are 3 matrices. By the same count X = -1 also has 3 matrices, and the other 16 - 6 = 10 give X = 0.

   X	Number of matrices	P(X)	X² P(X)
   -1	3	3/16	3/16
   0	10	10/16	0
   1	3	3/16	3/16

   Var(X) = E(X²) - (E(X))² = 6/16 - 0² = 3/8

3. So. E(X) = -3/16 + 3/16 = 0 because the distribution is symmetric, so the variance equals E(X²) = 3/8. The answer is (d).

The idea to keep. For a variance question, first build the probability table by honest counting. If the distribution is symmetric about 0, the mean is 0 and the variance is just E(X²).

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