Let be a matrix such that 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 : [2025]

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Solution

Q.
Let $A = [a_{i j}]$ be a $2 \times 2$ matrix such that $a_{i j} \in {0, 1}$ 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 : [2025]

1 $\frac{5}{8}$

2 $\frac{3}{8}$

3 $\frac{1}{4}$

4 $\frac{3}{4}$

Ans.
(2)

We have, $A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]$ , where $a_{i j} \in {0, 1}$

$\Rightarrow | A | = a_{11} a_{22} - a_{12} a_{21} = {- 1, 0, 1}$

Total outcomes = $2^{4}$ = 16, out of which 3 cases will be for value –1, 3 cases for 1, and rest 10 cases for 0.

X –1 0 1

P(X) 3/16 10/16 3/16

Now, $\bar{X} = \sum_{}^{} (X) P (X) = - 1 (\frac{3}{16}) + 0 (\frac{10}{16}) + 1 (\frac{3}{16}) = 0$

$∴ Variance = \sum_{}^{} X_{i}^{2} P (X) - {[\sum_{}^{} X P (X)]}^{2}$

$= 1 \times \frac{3}{16} + 0 \times \frac{10}{16} + 1 \times \frac{3}{16} - 0 = \frac{3}{8}$ .

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