Question

Which of the following systems has non-trivial solution?

• A. $AX=0$, $|A|=4$
• B. $AX=0$, $|A|=-4$
• C. $AX=0$, $|A|=0$
• D. $AX=B$, $|B|=5$

Hint:

No Determinant ($|A|=0$) = No Unique Solution = Non-Trivial possibilities!

Answer 1

The Correct Option is C

To determine which of the given systems has a non-trivial solution, let's analyze each option individually:

\(AX=0\), \(|A|=4\)

• If \(|A| \neq 0\), matrix \(A\) is invertible, indicating that the only solution to \(AX=0\) is the trivial solution \(X=0\). Therefore, this system has only the trivial solution.

\(AX=0\), \(|A|=-4\)

• Similar to the first case, since \(|A| \neq 0\) (even though it is negative), matrix \(A\) is invertible, and the system has only the trivial solution \(X=0\).

\(AX=0\), \(|A|=0\)

• In this case, since \(|A|=0\), matrix \(A\) is singular. This implies there are infinitely many solutions, meaning the system has a non-trivial solution (a solution other than the zero vector).

\(AX=B\), \(|B|=5\)

• The determinant of \(B\) being non-zero is irrelevant when determining if \(AX=B\) has a non-trivial solution. The solution depends on whether \(A\) is invertible (i.e., \(|A|\neq 0\)).

Based on the analysis above, the system that has a non-trivial solution is option 3: \(AX=0\), \(|A|=0\). This occurs because a singular matrix (\(|A|=0\)) implies an infinite number of solutions.