Let \(A = \begin{bmatrix} 1 & 1 & 1\\0 & 1 & 0\\1 & 0 & -1\end{bmatrix}\).
Is \(A\) singular?
Yes
No
The answer is “No”.
Row-reducing \(A\) gives \(\begin{bmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\). Hence, every column is a pivot column. Since there is no free variable, the only solution to \(Ax = 0\) is the trivial solution.