Question

Let A and B be two \(3 × 3\) non-zero real matrices such that AB is a zero matrix. Then

• A. the system of linear equations AX = 0 has a unique solution
• B. the system of linear equations AX = 0 has infinitely many solutions
• C. B is an invertible matrix
• D. adj(A) is an invertible matrix

Answer 1

The Correct Option is B

To understand why the given statement "the system of linear equations AX = 0 has infinitely many solutions" is correct, let's analyze the scenario where \( A \) and \( B \) are non-zero \( 3 \times 3 \) matrices such that their product is a zero matrix, i.e., \( AB = 0 \). 

1. Given that \( AB = 0 \) (the zero matrix), this implies that the rank of the matrix \( A \) must be less than 3. Otherwise, if \( A \) were invertible (rank 3), the product would not result in a zero matrix unless \( B \) was also a zero matrix, which contradicts the given non-zero condition.

2. If the rank of \( A \) is less than 3 (say less than full rank), it means that \( A \) does not have full row or column rank. Therefore, the homogeneous system of equations \( AX = 0 \) must have infinitely many solutions.

3. By linear algebra, a homogeneous system \( AX = 0 \) has:

• A unique solution (only the trivial solution) if and only if \( A \) is full rank (invertible).
• Infinitely many solutions if \( A \) is not full rank (i.e., rank is less than the number of variables).

Since \( AB = 0 \) and both \( A \) and \( B \) are non-zero, \( A \) cannot be of full rank, hence the system \( AX = 0 \) has infinitely many solutions.

4. Analyzing the other options:

• The system of linear equations AX = 0 has a unique solution: As explained, this is incorrect because \( A \) is not full rank.
• B is an invertible matrix: If \( B \) were invertible, multiplying both sides of \( AB = 0 \) by \( B^{-1} \) would give \( A = 0 \), contradicting the non-zero condition of \( A \).
• adj(A) is an invertible matrix: The adjugate matrix adj(A) is invertible if and only if \( A \) is not singular, which would imply \( A \) has full rank, which we've determined is not the case.

Thus, the correct conclusion is that the system of linear equations \( AX = 0 \) has infinitely many solutions, due to \( A \) being a non-full rank matrix.

Answer 2

AB is zero matrix
\(⇒ |A| = |B| = 0\)
Hence, neither A nor B is invertible
If \(|A| = 0\)
\(⇒ |adj A| = 0\) so adj A is not invertible
\(AX = 0\) is homogeneous system and \(|A| = 0\)
Therefore, it is having infinitely many solutions.
So, the correct option is (B): the system of linear equations \(AX = 0\) has infinitely many solutions