Question ID: #568
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 possible values of the determinant of the matrix $A$. Then, the variance of $X$ is:
- (1) $\frac{1}{4}$
- (2) $\frac{3}{8}$
- (3) $\frac{5}{8}$
- (4) $\frac{3}{4}$
Solution:
Total number of matrices with entries $\{0, 1\}$ is $2^4 = 16$.
Let $D = \det(A) = a_{11}a_{22} – a_{12}a_{21}$.
Since entries are 0 or 1, the products $a_{11}a_{22}$ and $a_{12}a_{21}$ can only be 0 or 1.
Possible values for $X$ (determinant) are:
1. $1 – 0 = 1$
2. $0 – 1 = -1$
3. $0 – 0 = 0$ or $1 – 1 = 0$
**Case $X = 1$:**
Requires $a_{11}a_{22}=1$ (1 choice: 1,1) and $a_{12}a_{21}=0$ (3 choices: 0,0; 0,1; 1,0).
Total matrices $= 1 \times 3 = 3$.
$$ P(X=1) = \frac{3}{16} $$
**Case $X = -1$:**
Requires $a_{11}a_{22}=0$ (3 choices) and $a_{12}a_{21}=1$ (1 choice).
Total matrices $= 3 \times 1 = 3$.
$$ P(X=-1) = \frac{3}{16} $$
**Case $X = 0$:**
Remaining matrices $= 16 – (3 + 3) = 10$.
$$ P(X=0) = \frac{10}{16} $$
**Calculation of Variance:**
$$ E[X] = \sum x_i p_i = 1\left(\frac{3}{16}\right) + (-1)\left(\frac{3}{16}\right) + 0\left(\frac{10}{16}\right) = 0 $$
$$ E[X^2] = \sum x_i^2 p_i = (1)^2\left(\frac{3}{16}\right) + (-1)^2\left(\frac{3}{16}\right) + (0)^2\left(\frac{10}{16}\right) $$
$$ E[X^2] = \frac{3}{16} + \frac{3}{16} = \frac{6}{16} = \frac{3}{8} $$
$$ \text{Variance} = E[X^2] – (E[X])^2 = \frac{3}{8} – 0^2 = \frac{3}{8} $$
Ans. (2)
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