Question

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:

• A. \(\frac{1}{4}\)
• B. \(\frac{3}{8}\)
• C. \(\frac{5}{8}\)
• D. \(\frac{3}{4}\)

Hint:

For variance problems involving matrices, ensure all possible determinant values are calculated correctly with corresponding probabilities.

Answer 1

The Correct Option is B

Step 1: Identifying possible determinant values. The determinant is calculated as: \[ |A| = a_{11}a_{22} - a_{12}a_{21} \] Using all possible combinations of 0 and 1, the possible determinant values are: \(\{-1, 0, 1\}\)

Step 2: Probability distribution. - Probability for \(X = -1\) = \(\frac{3}{16}\)
- Probability for \(X = 0\) = \(\frac{10}{16} = \frac{5}{8}\)
- Probability for \(X = 1\) = \(\frac{3}{16}\)

Step 3: Calculating Variance. \[ \text{Variance} = E(X^2) - (E(X))^2 \] \[ = \frac{3}{16}(-1)^2 + \frac{5}{8}(0)^2 + \frac{3}{16}(1)^2 - (0)^2 \] \[ = \frac{3}{16} + 0 + \frac{3}{16} = \frac{6}{16} = \frac{3}{8} \]

Answer 2

Step 1: Define the matrix A and possible entries.
We are told that \( A = [a_{ij}] \) is a \( 2 \times 2 \) matrix, and each entry \( a_{ij} \in \{0, 1\} \).
Thus, the total number of possible matrices is \( 2^{2 \times 2} = 16 \).

Step 2: Write the determinant formula for a 2×2 matrix.
If \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), then the determinant is:
\[ \text{det}(A) = ad - bc \] Each of \( a, b, c, d \) can be either 0 or 1.

Step 3: Find all possible values of the determinant.
We consider all combinations:
- If \( a = 0, d = 0 \) → \( ad = 0 \).
- If \( a = 1, d = 1 \) → \( ad = 1 \).
- If \( b = 0, c = 0 \) → \( bc = 0 \).
- If \( b = 1, c = 1 \) → \( bc = 1 \).

Thus, possible determinant values are \( ad - bc \in \{-1, 0, 1\} \).

Step 4: Compute probabilities of each value.
We enumerate cases for all \( a, b, c, d \in \{0, 1\} \):
- Number of matrices with det = 1 → occurs when \( ad = 1 \) and \( bc = 0 \).
- \( ad = 1 \Rightarrow a = 1, d = 1 \) (1 way).
- \( bc = 0 \Rightarrow \) either \( b = 0 \) or \( c = 0 \) (3 combinations).
So, 3 cases → Probability \( = \frac{3}{16} \).

- Number of matrices with det = -1 → occurs when \( ad = 0 \) and \( bc = 1 \).
- \( bc = 1 \Rightarrow b = 1, c = 1 \) (1 way).
- \( ad = 0 \Rightarrow \) either \( a = 0 \) or \( d = 0 \) (3 combinations).
So, 3 cases → Probability \( = \frac{3}{16} \).

- Remaining 16 − (3 + 3) = 10 matrices have determinant 0 → Probability \( = \frac{10}{16} = \frac{5}{8} \).

Step 5: Define the random variable X.
Possible values of X: −1, 0, 1.
Respective probabilities:
\[ P(X = -1) = \frac{3}{16}, \quad P(X = 0) = \frac{10}{16}, \quad P(X = 1) = \frac{3}{16}. \]

Step 6: Compute the mean \( E(X) \).
\[ E(X) = (-1)\left(\frac{3}{16}\right) + 0\left(\frac{10}{16}\right) + (1)\left(\frac{3}{16}\right) = 0. \]

Step 7: Compute \( E(X^2) \).
\[ E(X^2) = (-1)^2\left(\frac{3}{16}\right) + 0^2\left(\frac{10}{16}\right) + (1)^2\left(\frac{3}{16}\right) = \frac{6}{16} = \frac{3}{8}. \]

Step 8: Compute the variance.
\[ \text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{3}{8} - 0 = \frac{3}{8}. \]

Final Answer:
\[ \boxed{\frac{3}{8}} \]