If A is a matrix such that there exists a square submatrix of order r which is non-singular and eveny square submatrix or order r + 1 or more is singular, then

Options

A. rank (A) = r + 1
B. rank (A) = r
C. rank (A) gt r
D. `"rank "(A)lt r+1`

Answer

To solve the problem, we need to analyze the properties of the matrix \( A \) given the conditions about its submatrices.

### Step-by-Step Solution:

1. **Understanding the Given Conditions**:
- We have a matrix \( A \) such that there exists a square submatrix of order \( r \) which is non-singular. This means that the determinant of this submatrix is not equal to zero.
- Every square submatrix of order \( r + 1 \) or more is singular. This implies that the determinant of any submatrix of order \( r + 1 \) or larger is equal to zero.

2. **Implications of Non-Singularity**:
- Since there is a non-singular submatrix of order \( r \), we can conclude that the rank of this submatrix is \( r \). The rank of a matrix is defined as the maximum number of linearly independent rows or columns, which corresponds to the size of the largest non-singular square submatrix.

3. **Implications of Singularity**:
- The condition that every square submatrix of order \( r + 1 \) is singular indicates that the rank of the entire matrix \( A \) cannot be \( r + 1 \) or higher. If it were, there would exist a non-singular submatrix of order \( r + 1 \).

4. **Conclusion About the Rank of \( A \)**:
- From the above points, we can conclude that the rank of the matrix \( A \) must be exactly \( r \). This is because:
- The existence of a non-singular submatrix of order \( r \) indicates that the rank is at least \( r \).
- The singularity of all submatrices of order \( r + 1 \) indicates that the rank cannot exceed \( r \).

5. **Final Answer**:
- Therefore, the rank of matrix \( A \) is \( r \).