Let A be `3xx3` symmetric invertible matrix with real positive elements. Then the number of zero elements in `A^(-1)` are less than or equal to :

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to determine the maximum number of zero elements in the inverse of a \(3 \times 3\) symmetric invertible matrix \(A\) with real positive elements.

### Step-by-Step Solution:

1. **Understanding the Matrix Properties**:
- A symmetric matrix is one that is equal to its transpose, i.e., \(A = A^T\).
- An invertible matrix is one that has a non-zero determinant and thus has an inverse \(A^{-1}\).

2. **Using the Formula for Zero Elements**:
- The maximum number of zero elements in the inverse of a matrix can be estimated using the formula:
\[ \text{Number of zero elements in } A^{-1} \leq n^2 - 2n \]
where \(n\) is the dimension of the matrix.

3. **Substituting the Value of \(n\)**:
- For our \(3 \times 3\) matrix, we have \(n = 3\).
- Plugging this value into the formula:
\[ \text{Number of zero elements in } A^{-1} \leq 3^2 - 2 \times 3 \]

4. **Calculating the Expression**:
- Calculate \(3^2\):
\[ 3^2 = 9 \]
- Calculate \(2 \times 3\):
\[ 2 \times 3 = 6 \]
- Now, substitute these values back into the inequality:
\[ \text{Number of zero elements in } A^{-1} \leq 9 - 6 = 3 \]

5. **Conclusion**:
- Therefore, the maximum number of zero elements in \(A^{-1}\) is \(3\).

### Final Answer:
The number of zero elements in \(A^{-1}\) is less than or equal to \(3\).