Let $$A =[a_{ij}]$$ be a 2$$\times$$2 matrix such that $$a_{ij} \in \left\{0,1\right\}$$ 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 :

Matrix $$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$$ where each $$a_{ij} \in \{0, 1\}$$.

Total matrices = $$2^4 = 16$$.

$$\det(A) = a_{11}a_{22} - a_{12}a_{21}$$.

Possible values of det(A): -1, 0, 1.

Let us count each case:

det = 1: $$a_{11}a_{22} = 1$$ and $$a_{12}a_{21} = 0$$. So $$a_{11}=a_{22}=1$$ and at least one of $$a_{12}, a_{21}$$ is 0. Number of ways = 1 × 3 = 3. But we need $$a_{11}a_{22} - a_{12}a_{21} = 1$$, which means $$a_{11}a_{22}=1, a_{12}a_{21}=0$$: 3 cases.

det = -1: $$a_{11}a_{22} = 0$$ and $$a_{12}a_{21} = 1$$. So $$a_{12}=a_{21}=1$$ and at least one of $$a_{11}, a_{22}$$ is 0. Number of ways = 3 × 1 = 3.

det = 0: Remaining cases = 16 - 3 - 3 = 10.

Now computing variance:

$$E(X) = \frac{1}{16}[3(1) + 10(0) + 3(-1)] = 0$$

$$E(X^2) = \frac{1}{16}[3(1) + 10(0) + 3(1)] = \frac{6}{16} = \frac{3}{8}$$

$$Var(X) = E(X^2) - [E(X)]^2 = \frac{3}{8} - 0 = \frac{3}{8}$$

The correct answer is Option 3: $$\frac{3}{8}$$.