Animated Solution for Mathematics - Probability: Let $A=[a_{ij}]$ be a $2\times 2$ matrix such that $a_{ij}\in \{0,1\}$ 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 :

Imagine you are standing before a $2\times 2$ grid. It is not just a collection of numbers; it is a canvas of potential. Each cell, $a_{11},a_{12},a_{21},a_{22}$, is a binary switch—it can be either $0$ or $1$.

When we ask about the 'variance' of the determinant of this matrix, we are essentially asking: "How much does the determinant fluctuate across all possible configurations of this grid?" This is the heart of probability in linear algebra.

First, we must define our universe. We have four positions in our matrix:

Since each entry has two choices, the total number of distinct matrices we can construct is $2^4=16$. This is our sample space. It is small enough to handle, yet rich enough to reveal deep patterns.

The determinant, $X=\text{det}(A)=a_{11}{a}_{22}-a_{12}{a}_{21}$, is our random variable. Because the entries are restricted to the set $\{0,1\}$, the products $a_{11}{a}_{22}$ and $a_{12}{a}_{21}$ can only take values $0$ or $1$.

Consequently, the difference $X$ can only take three values: $1$, $-1$, or $0$. This is the beauty of discrete probability—we have collapsed an infinite sea of possibilities into three distinct outcomes.

To find $P(X=1)$, we need $a_{11}{a}_{22}=1$ and $a_{12}{a}_{21}=0$. The first condition, $a_{11}{a}_{22}=1$, happens only when both are $1$ (1 case). The second condition, $a_{12}{a}_{21}=0$, happens in three cases (0-0, 0-1, 1-0).

Thus, we have $1\times 3=3$ favorable cases out of 16. So, $P(X=1)=\frac{3}{16}$.

By symmetry, the logic for $P(X=-1)$ is identical. We need $a_{11}{a}_{22}=0$ and $a_{12}{a}_{21}=1$. This yields $3\times 1=3$ cases. Thus, $P(X=-1)=\frac{3}{16}$.

Finally, for $P(X=0)$, we simply subtract the others from the total probability:

Now, we apply the heavy machinery of statistics. The variance is defined as $Var(X)=E(X^2)-[E(X){]}^2$.

We calculate the expected value $E(X)$ first:

The terms $1$ and $-1$ cancel out perfectly, resulting in $E(X)=0$. This is a profound result—it tells us that on average, the determinant of these random matrices is zero.

Now, for the variance, we need $E(X^2)$. We square the values of $X$: $1^2=1$, $(-1{)}^2=1$, and $0^2=0$. The expectation becomes:

Plugging these into our variance formula, we get:

We have arrived at our destination. The variance of the determinant is $\frac{3}{8}$.

It is a clean, elegant fraction that emerges from the chaos of 16 possible matrices. Remember, in JEE Advanced, it is rarely about the calculation itself; it is about the structure you see before you calculate. You have mastered the structure today.