Let A = [a_{ij}] be a 2 times 2 matrix such t | vedclass

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(D) The matrix $A$ is given by $\begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix}$ where each $a_{ij} \in \{0, 1\}$. There are $2^4 = 1

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$\frac{3}{8}$

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(D) The matrix $A$ is given by $\begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix}$ where each $a_{ij} \in \{0, 1\}$. There are $2^4 = 16$ possible matrices.The determinant is $|A| = a_{11}a_{22} - a_{12}a_{21}$.The possible values for $|A|$ are $\{-1, 0, 1\}$.- $|A| = -1$ if $(a_{11}, a_{22}, a_{12}, a_{21}) = (0, 0, 1, 1)$,so $P(X = -1) = \frac{1}{16}$. Wait,let's re-evaluate: $a_{11}a_{22} - a_{12}a_{21} = -1 \implies a_{11}a_{22} = 0$ and $a_{12}a_{21} = 1$. This happens when $(a_{12}, a_{21}) = (1, 1)$ and $a_{11}a_{22} = 0$. The pairs $(a_{11}, a_{22})$ can be $(0,0), (0,1), (1,0)$,which are $3$ cases. So $P(X = -1) = \frac{3}{16}$.- $|A| = 1$ if $a_{11}a_{22} = 1$ and $a_{12}a_{21} = 0$. This happens when $(a_{11}, a_{22}) = (1, 1)$ and $(a_{12}, a_{21}) \in \{(0,0), (0,1), (1,0)\}$,which are $3$ cases. So $P(X = 1) = \frac{3}{16}$.- $|A| = 0$ in the remaining $16 - 3 - 3 = 10$ cases. So $P(X = 0) = \frac{10}{16}$.The variance is $Var(X) = E[X^2] - (E[X])^2$.$E[X] = (-1)(\frac{3}{16}) + (0)(\frac{10}{16}) + (1)(\frac{3}{16}) = 0$.$E[X^2] = (-1)^2(\frac{3}{16}) + (0)^2(\frac{10}{16}) + (1)^2(\frac{3}{16}) = \frac{3}{16} + \frac{3}{16} = \frac{6}{16} = \frac{3}{8}$.Thus,$Var(X) = \frac{3}{8} - 0^2 = \frac{3}{8}$.

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:

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