Prove that the following two matrices are not row-equivalent:
\[ \left[ \begin{array}{c c c} 2 & 0 & 0 \\ a & - 1 & 0 \\ b & c & 3 \end{array} \right], \qquad \left[ \begin{array}{c c c} 1 & 1 & 2 \\ - 2 & 0 & - 1 \\ 1 & 3 & 5 \end{array} \right]. \]
Solution
Let \[ A=\begin{bmatrix}2&0&0\\a&-1&0\\b&c&3\end{bmatrix}, \qquad B=\begin{bmatrix}1&1&2\\-2&0&-1\\1&3&5\end{bmatrix}. \]
1. The matrix \(A\) has rank \(3\).
\(A\) is lower triangular and \[ \det A=2(-1)3=-6\ne0. \] Hence \(A\) is invertible and has rank \(3\).
2. The matrix \(B\) has rank \(2\).
Its determinant is \[ \det B=0, \] so its rank is at most \(2\). The minor formed by the first two rows and first two columns is \[ \det\begin{bmatrix}1&1\\-2&0\end{bmatrix}=2\ne0, \] so its rank is at least \(2\).
3. Therefore \(A\) and \(B\) are not row-equivalent.
Elementary row operations preserve rank, whereas 1 and 2 give different ranks.