2×2 Matrix With Entries 0 or 1 — Variance of the Determinant

JEE Main 2025 — 29 January, Evening Shift. Previous Year Question.

Problem

Let $A=[a_{ij}]$ be a $2\times2$ matrix such that $a_{ij}\in\{0,1\}$ for all $i$ and $j$. Let the random variable $X$ denote the determinant of the matrix $A$. Then the variance of $X$ is:

Key insight. With only 4 entries, each independently $0$ or $1$, there are just $16$ equally likely matrices in total — few enough to reason about directly. The determinant $a_{11}a_{22}-a_{12}a_{21}$ only takes the values $-1$, $0$, or $1$, since each product term is itself just $0$ or $1$ — this makes $X$’s probability distribution straightforward to work out from first principles.

Watch this explained step by step →  ·  More on the Shiwam’s Classes channel

Approach

Note that $\det(A)=a_{11}a_{22}-a_{12}a_{21}$, where $a_{11}a_{22}$ and $a_{12}a_{21}$ are each independently $1$ with probability $\frac{1}{4}$ (only when both relevant entries are $1$) and $0$ otherwise. Work out the probability of each possible value of $X$ ($-1$, $0$, $1$), then compute $E(X)$ and $E(X^2)$ to find the variance.

Solution

Step 1 — Find the probability that a₁₁a₂₂ = 1

Since $a_{11}$ and $a_{22}$ are each independently $0$ or $1$ with equal probability, $a_{11}a_{22}=1$ only when both equal $1$:

$$P(a_{11}a_{22}=1) = \frac{1}{4}, \qquad P(a_{11}a_{22}=0) = \frac{3}{4}$$

By the same reasoning, $P(a_{12}a_{21}=1)=\frac{1}{4}$ and $P(a_{12}a_{21}=0)=\frac{3}{4}$, independently of the first pair.

Step 2 — Find P(X = 1)

$X=1$ requires $a_{11}a_{22}=1$ and $a_{12}a_{21}=0$:

$$P(X=1) = \frac{1}{4}\times\frac{3}{4} = \frac{3}{16}$$

Step 3 — Find P(X = −1)

$X=-1$ requires $a_{11}a_{22}=0$ and $a_{12}a_{21}=1$:

$$P(X=-1) = \frac{3}{4}\times\frac{1}{4} = \frac{3}{16}$$

Step 4 — Find P(X = 0)

$X=0$ when both products are equal — either both $0$ or both $1$:

$$P(X=0) = \frac{3}{4}\times\frac{3}{4} + \frac{1}{4}\times\frac{1}{4} = \frac{9}{16}+\frac{1}{16} = \frac{10}{16}$$

Checking: $\frac{3}{16}+\frac{3}{16}+\frac{10}{16}=\frac{16}{16}=1$ ✓

Step 5 — Compute E(X)

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

Step 6 — Compute E(X²)

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

Step 7 — Compute the variance

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

Answer

Common mistakes

• Assuming $X$ is uniformly distributed over $\{-1,0,1\}$. The value $0$ is actually more likely than $\pm1$ (probability $\frac{10}{16}$ versus $\frac{3}{16}$ each), since it results from two different underlying cases (both products $0$, or both products $1$), not just one.
• Forgetting to square $E(X)$ when it isn’t obviously zero. Here $E(X)=0$ by symmetry, so $\text{Var}(X)=E(X^2)$ directly — but this shortcut only applies because of the symmetric probabilities; it shouldn’t be assumed without checking.

Practise next

• Let $B=[b_{ij}]$ be a $2\times2$ matrix with entries from $\{-1,1\}$ instead of $\{0,1\}$. Let $Y=\det(B)$. Find the variance of $Y$, using the same direct-enumeration approach.