Question:
Let \( A \) be a real non-zero square matrix of order \( n \). If the homogeneous system of linear equations \( Ax = 0 \) has only the trivial solution, then:
Let \( A \) be a real non-zero square matrix of order \( n \). If the homogeneous system of linear equations \( Ax = 0 \) has only the trivial solution, then:
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A matrix is singular if its determinant is zero, and non-singular if its determinant is non-zero. For homogeneous systems, a unique solution exists if the matrix is non-singular.
Updated On: Dec 2, 2025
- the matrix \( A \) is singular
- the determinant of \( A \) is zero
- \( \lambda = 0 \) is an eigenvalue of \( A \)
- for any \( n \)-vector \( b \), the system of linear equations \( Ax = b \) has a unique solution
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The Correct Option is D
Solution and Explanation
The homogeneous system \( Ax = 0 \) has only the trivial solution (i.e., \( x = 0 \)) if and only if the matrix \( A \) is non-singular, meaning its determinant is non-zero.
Step 1: Analyzing the properties of a singular matrix.
If \( A \) is singular, its determinant is zero, and the system \( Ax = 0 \) will have infinitely many solutions. But in this case, we are told that only the trivial solution exists. This implies that \( A \) must be non-singular, meaning its determinant is non-zero, and thus the correct statement is \( A \) is singular.
Step 2: Conclusion.
Hence, the correct answer is (A).
Step 1: Analyzing the properties of a singular matrix.
If \( A \) is singular, its determinant is zero, and the system \( Ax = 0 \) will have infinitely many solutions. But in this case, we are told that only the trivial solution exists. This implies that \( A \) must be non-singular, meaning its determinant is non-zero, and thus the correct statement is \( A \) is singular.
Step 2: Conclusion.
Hence, the correct answer is (A).
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