Question

In a matrix \( A \), if all the sub matrices of \( k^{th} \) order are singular and there is one non-singular sub matrix of order \( r \) \( (r<k) \), then the rank \( (\rho) \) of the matrix \( A \)

• A. satisfies \( r \le \rho<k \)
• B. is equal to \( r \)
• C. is equal to \( (k - 1) \)
• D. is equal to \( (k + 1) \)

Hint:

The rank of a matrix is the size of the largest non-zero minor. If all minors of order \( k \) are zero, then the rank is less than \( k \). If there exists a non-zero minor of order \( r \), then the rank is at least \( r \). Combining these gives the range for the rank of the matrix.

Answer 1

The Correct Option is A

The rank of a matrix \( A \) is the order of the largest non-singular square sub-matrix of \( A \).
We are given that there exists at least one non-singular sub-matrix of order \( r \).
This implies that the rank of the matrix \( A \) is at least \( r \), so \( \rho \ge r \).
We are also given that all the sub-matrices of \( k^{th} \) order are singular.
This means that there is no non-singular square sub-matrix of order \( k \) or greater.
Therefore, the rank of the matrix \( A \) must be less than \( k \), so \( \rho<k \).
Combining these two conditions, we have \( r \le \rho<k \).
Let's analyze the other options:
Option (B) is equal to \( r \): We only know that the rank is at least \( r \).
There might be a non-singular sub-matrix of an order greater than \( r \) but less than \( k \).

Option (C) is equal to \( (k - 1) \): We know that all \( k^{th} \) order sub-matrices are singular, which implies the rank is at most \( k - 1 \).
However, we are only guaranteed a non-singular sub-matrix of order \( r \), where \( r \) could be less than \( k - 1 \).

Option (D) is equal to \( (k + 1) \): This contradicts the given information that all \( k^{th} \) order sub-matrices are singular.
The rank cannot be greater than or equal to \( k \).
Therefore, the only correct statement is that the rank \( \rho \) satisfies \( r \le \rho<k \).