Let `S={{:[(a,b),(c,d)]:a,b,c,d in {0,1}:}}` A matrix A is picked up at random from the set S and is found to be invertible. The probability that det(A) `gt` 0 , is ____

Answer

To solve the problem, we need to determine the probability that the determinant of a randomly selected invertible matrix \( A \) from the set \( S \) is greater than 0. Here’s a step-by-step breakdown of the solution:

### Step 1: Understanding the Set \( S \)
The set \( S \) consists of matrices of the form:
\[ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]
where \( a, b, c, d \in \{0, 1\} \). This means each entry of the matrix can either be 0 or 1.

### Step 2: Conditions for Invertibility
A matrix is invertible if its determinant is non-zero. The determinant of matrix \( A \) is given by:
\[ \text{det}(A) = ad - bc \]
For the matrix to be invertible, we need:
\[ ad - bc \neq 0 \]

### Step 3: Analyzing the Determinant
The determinant can be either positive or negative. We need to find cases where:
1. \( ad - bc > 0 \)
2. \( ad - bc < 0 \)

### Step 4: Finding Favorable Cases
To satisfy \( ad - bc > 0 \), we need:
- \( ad = 1 \) and \( bc = 0 \)

This means:
- \( a = 1 \) and \( d = 1 \) (to make \( ad = 1 \))
- Either \( b = 0 \) or \( c = 0 \) (to make \( bc = 0 \))

The possible combinations for \( b \) and \( c \) that satisfy \( bc = 0 \) are:
1. \( b = 0, c = 0 \)
2. \( b = 1, c = 0 \)
3. \( b = 0, c = 1 \)

Thus, we have 3 favorable cases where \( ad - bc > 0 \).

### Step 5: Total Invertible Matrices
Next, we need to count the total number of invertible matrices. The determinant \( ad - bc \) must be non-zero, which gives us two cases:
1. \( ad = 1 \) and \( bc = 0 \) (as discussed above)
2. \( ad = 0 \) and \( bc = 1 \)

For \( ad = 0 \):
- Possible combinations are \( (a, d) = (0, 0), (0, 1), (1, 0) \) which gives us 3 cases.
- For \( bc = 1 \), we need \( b = 1, c = 1 \) which gives us 1 case.

Thus, the total number of invertible matrices is:
- From \( ad = 1 \) and \( bc = 0 \): 3 cases
- From \( ad = 0 \) and \( bc = 1 \): 3 cases (since \( a \) and \( d \) can be either 0 or 1)

Total invertible matrices = 3 (from the first case) + 3 (from the second case) = 6.

### Step 6: Calculating the Probability
The probability that the determinant of the matrix is greater than 0, given that it is invertible, is given by:
\[ P(\text{det}(A) > 0 | A \text{ is invertible}) = \frac{\text{Number of favorable cases}}{\text{Total invertible cases}} = \frac{3}{6} = \frac{1}{2} \]

### Final Answer
The probability that \( \text{det}(A) > 0 \) given that \( A \) is invertible is \( \frac{1}{2} \).

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