If the elements of a matrix `A` are real positive and distinct such that `det(A+A^(T))^(T)=0` then

Options

A. `detA gt 0`
B. `det A ge 0`
C. `det (A-A^(T)) gt 0`
D. `det (A.A^(T)) gt 0`

Explanation

To solve the problem, we need to analyze the given conditions and derive the necessary conclusions step by step.

### Given:
- The elements of matrix \( A \) are real, positive, and distinct.
- \( \text{det}(A + A^T) = 0 \).

### Step 1: Define the Matrix
Let’s define the matrix \( A \) as follows:
\[ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]
where \( a, b, c, d \) are positive and distinct real numbers.

### Step 2: Calculate the Transpose
The transpose of matrix \( A \) is:
\[ A^T = \begin{pmatrix} a & c \\ b & d \end{pmatrix} \]

### Step 3: Calculate \( A + A^T \)
Now, we compute \( A + A^T \):
\[ A + A^T = \begin{pmatrix} a & b \\ c & d \end{pmatrix} + \begin{pmatrix} a & c \\ b & d \end{pmatrix} = \begin{pmatrix} 2a & b+c \\ c+b & 2d \end{pmatrix} \]

### Step 4: Calculate the Determinant
Next, we calculate the determinant of \( A + A^T \):
\[ \text{det}(A + A^T) = \text{det}\left(\begin{pmatrix} 2a & b+c \\ b+c & 2d \end{pmatrix}\right) \]
Using the determinant formula for a 2x2 matrix, we have:
\[ \text{det}(A + A^T) = (2a)(2d) - (b+c)(b+c) = 4ad - (b+c)^2 \]

### Step 5: Set the Determinant to Zero
Given that \( \text{det}(A + A^T) = 0 \), we set up the equation:
\[ 4ad - (b+c)^2 = 0 \]
This implies:
\[ 4ad = (b+c)^2 \]

### Step 6: Apply the AM-GM Inequality
From the AM-GM inequality, we know:
\[ \frac{b+c}{2} \geq \sqrt{bc} \]
Thus:
\[ b+c \geq 2\sqrt{bc} \]
Squaring both sides gives:
\[ (b+c)^2 \geq 4bc \]

### Step 7: Relate \( 4ad \) and \( 4bc \)
From our earlier equation \( 4ad = (b+c)^2 \), we can substitute:
\[ 4ad \geq 4bc \]
This simplifies to:
\[ ad \geq bc \]

### Step 8: Conclusion about Determinants
Now, we can relate this back to the determinant of \( A \):
\[ \text{det}(A) = ad - bc \]
Since \( ad \geq bc \), we conclude:
\[ ad - bc \geq 0 \implies \text{det}(A) \geq 0 \]
However, since \( a, b, c, d \) are distinct and positive, \( ad - bc > 0 \). Therefore:
\[ \text{det}(A) > 0 \]

### Final Options Analysis
1. **Option 1**: \( \text{det}(A) > 0 \) - **True**
2. **Option 2**: \( \text{det}(A) \geq 0 \) - **False** (since it is strictly greater than 0)
3. **Option 3**: \( \text{det}(A - A^T) > 0 \) - **True** (as shown in the video)
4. **Option 4**: \( \text{det}(A \cdot A^T) > 0 \) - **True** (as shown in the video)

### Summary of Results
- The correct options are: 1, 3, and 4.