Question

If there exists a \( k^{th} \) order non-singular sub-matrix in a matrix \( P \) of order \( m \times n \), then the rank \( (\rho) \) of \( P \)

• A. satisfies \( k \le \rho \le m \)
• B. satisfies \( k<\rho<n \)
• C. satisfies \( k \le \rho \le \min \{m, n\} \)
• D. is equal to \( k + 1 \)

Hint:

The rank of a matrix is bounded by the minimum of its dimensions. If a matrix has dimensions \( m \times n \), its rank \( \rho \) satisfies \( 0 \le \rho \le \min \{m, n\} \). The existence of a non-singular \( k \times k \) sub-matrix guarantees that the rank is at least \( k \). Therefore, the rank lies between \( k \) and the minimum of the dimensions of the matrix.

Answer 1

The Correct Option is C

The rank of a matrix \( P \) is defined as the order of the largest non-singular square sub-matrix of \( P \).
We are given that there exists a \( k^{th} \) order non-singular sub-matrix in the matrix \( P \).
This implies that there is at least one square sub-matrix of size \( k \times k \) whose determinant is non-zero.
Therefore, the rank of the matrix \( P \) must be greater than or equal to \( k \), i.
e.
, \( \rho \ge k \).
The rank of a matrix \( P \) of order \( m \times n \) cannot exceed the minimum of the number of rows and the number of columns.
This is because any square sub-matrix can have an order at most \( \min \{m, n\} \).
Therefore, the rank \( \rho \) must satisfy \( \rho \le \min \{m, n\} \).
Combining these two conditions, we get \( k \le \rho \le \min \{m, n\} \).
Let's analyze the other options:
(A) \( k \le \rho \le m \): While we know \( \rho \le \min \{m, n\} \), if \( n<m \), then \( \rho \) cannot exceed \( n \).
So, this option is not always true.
(B) \( k<\rho<n \): We know \( \rho \ge k \), not necessarily \( \rho>k \).
Also, \( \rho \le \min \{m, n\} \), so if \( m<n \), \( \rho \) cannot exceed \( m \).
Thus, this option is not always true.
(D) is equal to \( k + 1 \): The existence of a \( k^{th} \) order non-singular sub-matrix only tells us that the rank is at least \( k \).
It does not necessarily mean the rank is exactly \( k + 1 \).
For example, if there is also a non-singular sub-matrix of order \( k + 1 \), then the rank would be at least \( k + 1 \).
Therefore, the correct condition is \( k \le \rho \le \min \{m, n\} \).