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:
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:
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.
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
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A matrix is a rectangular array of numbers, variables, symbols, or expressions that are defined for the operations like subtraction, addition, and multiplications. The size of a matrix is determined by the number of rows and columns in the matrix.
