JEE Mathematics 2025 Question with Solution

Given: $A = [a_{ij}]$ is a $2 \times 2$ matrix with each entry from $\{0,1\}$.

Find: The variance of the random variable $X$, where $X$ is the determinant of $A$.

The determinant is

$$|A| = a_{11}a_{22} - a_{12}a_{21}$$

So the possible values of $X$ are $-1, 0, 1$.

From all possible matrices, the probability distribution is:

$$P(X=-1)=\frac{3}{16}, \quad P(X=0)=\frac{10}{16}=\frac{5}{8}, \quad P(X=1)=\frac{3}{16}$$

Now,

$$\text{Var}(X)=E(X^2)-\left(E(X)\right)^2$$

Also,

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

And,

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

Therefore,

$$\text{Var}(X)=\frac{3}{8}-0=\frac{3}{8}$$

Hence, the correct option is B.

Given: $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ where each of $a,b,c,d \in \{0,1\}$.

Find: The variance of $X=\det(A)$.

Since each of the four entries can be chosen independently as $0$ or $1$, the total number of possible matrices is

$$2^{2 \times 2}=16$$

For a $2 \times 2$ matrix,

$$\det(A)=ad-bc$$

Now $ad$ and $bc$ can each be either $0$ or $1$, so the possible values of $ad-bc$ are

$$-1, 0, 1$$

To get $\det(A)=1$, we need $ad=1$ and $bc=0$.

• $ad=1$ happens only when $a=1$ and $d=1$, which is $1$ way.
• $bc=0$ happens in $3$ ways: $(b,c)=(0,0),(0,1),(1,0)$. So,

$$P(X=1)=\frac{3}{16}$$

To get $\det(A)=-1$, we need $ad=0$ and $bc=1$.

• $bc=1$ happens only when $b=1$ and $c=1$, which is $1$ way.
• $ad=0$ happens in $3$ ways. So,

$$P(X=-1)=\frac{3}{16}$$

The remaining matrices have determinant $0$:

$$16-(3+3)=10$$

Hence,

$$P(X=0)=\frac{10}{16}=\frac{5}{8}$$

Now compute expectation:

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

Next,

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

Therefore,

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

Thus, the variance is $\frac{3}{8}$ and the correct option is B.