Question

Consider the following statements. (I) Let \( A \) and \( B \) be two \( n \times n \) real matrices. If \( B \) is invertible, then \( \text{rank}(BA) = \text{rank}(A) \). (II) Let \( A \) be an \( n \times n \) real matrix. If \( A^2x = b \) has a solution for every \( b \in \mathbb{R}^n \), then \( Ax = b \) also has a solution for every \( b \in \mathbb{R}^n \). Which of the above statements is/are true?

• A. Only (I)
• B. Only (II)
• C. Both (I) and (II)
• D. Neither (I) nor (II)

Hint:

- {Statement (I)} holds true because the rank of a matrix is unchanged by multiplication with an invertible matrix.
- {Statement (II)} is true because surjectivity of \( A^2 \) guarantees that \( A \) is surjective, ensuring a solution for every \( b \in \mathbb{R}^n \).

Answer 1

The Correct Option is C

1) Analysis of Statement (I):
This statement is true. If \( B \) is invertible, we know that multiplying by \( B \) does not change the rank of the matrix. Therefore, \( \text{rank}(BA) = \text{rank}(A) \), making statement (I) correct.
2) Analysis of Statement (II):
This statement is also true. If \( A^2x = b \) has a solution for every \( b \in \mathbb{R}^n \), it implies that the matrix \( A^2 \) is surjective (onto). If \( A^2 \) is surjective, it guarantees that \( A \) is also surjective, meaning \( Ax = b \) will have a solution for every \( b \in \mathbb{R}^n \). Hence, statement (II) is correct as well.
Thus, the correct answer is (C) Both (I) and (II).