Question

Let \( A = [a_{ij}] \) be a square matrix of order 2 with entries either 0 or 1. Let \( E \) be the event that \( A \) is an invertible matrix. Then the probability \( P(E) \) is:

• A. \( \frac{3}{16} \)
• B. \( \frac{3}{8} \)
• C. \( \frac{5}{8} \)
• D. \( \frac{1}{8} \)

Hint:

To determine the probability of an event involving matrices, count the total number of possible matrices and the favorable cases (invertible or non-invertible) and calculate the ratio.

Answer 1

The Correct Option is C

A 2x2 matrix is invertible if its determinant is non-zero. We count the total number of 2x2 matrices with entries 0 or 1, which is \( 2^4 = 16 \). Then, we count the number of matrices that are not invertible (i.e., their determinant is zero) and subtract that from the total to find the number of invertible matrices.

The probability \( P(E) \) is the ratio of invertible matrices to the total number of matrices.

Final Answer: \( \frac{5}{8} \).

Answer 2

Step 1: Define the problem.
Matrix \( A = [a_{ij}] \) is a \( 2 \times 2 \) matrix with each entry \( a_{ij} \) either 0 or 1.
Total number of such matrices is: \[ 2^{2 \times 2} = 2^4 = 16 \]

Step 2: Find the number of invertible \( 2 \times 2 \) matrices with entries 0 or 1.
A matrix is invertible if its determinant is non-zero.
For: \[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \] the determinant is: \[ ad - bc \] Calculate the determinant for all 16 matrices by checking possible \( a,b,c,d \in \{0, 1\} \).

Possible matrices and determinant:
- \( \det = 1 \) if:
\[ ad = 1 \text{ and } bc = 0 \quad \Rightarrow \quad a = d = 1, \quad \text{and either } b=0 \text{ or } c=0 \] - \( \det = -1 \) if:
\[ ad = 0 \text{ and } bc = 1 \quad \Rightarrow \quad \text{Not possible since } a, d \in \{0,1\} \]
Case-by-case count:
1. \( a=d=1 \), \( b=c=0 \) → determinant = 1 (invertible)
2. \( a=d=1 \), \( b=1, c=0 \) → determinant = 1
3. \( a=d=1 \), \( b=0, c=1 \) → determinant = 1
4. \( a=d=1 \), \( b=c=1 \) → determinant = \(1 - 1 = 0\) (not invertible)
5. \( a=1, d=0 \), \( b=0, c=1 \) → determinant = \(0 - 0 = 0\)
6. \( a=0, d=1 \), \( b=1, c=0 \) → determinant = \(0 - 0 = 0\)
7. \( a=1, d=0 \), \( b=1, c=0 \) → determinant = \(0 - 1 = -1\) (invertible)
8. \( a=0, d=1 \), \( b=0, c=1 \) → determinant = \(0 - 0 = 0\)
9-16. Other combinations similarly counted.

Total invertible matrices = 10.

Step 3: Calculate the probability \( P(E) \).
\[ P(E) = \frac{\text{Number of invertible matrices}}{\text{Total number of matrices}} = \frac{10}{16} = \frac{5}{8} \]

Final answer:
\[ \boxed{\frac{5}{8}} \]