Example: \[ A = \begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \] After \( R_1 \leftrightarrow R_3 \): \[ \begin{bmatrix}7 & 8 & 9 \\ 4 & 5 & 6 \\ 1 & 2 & 3 \end{bmatrix} \]

Example: \[ A = \begin{bmatrix}2 & 3 \\ 4 & 5 \end{bmatrix}, \quad R_1 \to 3R_1 \] Result: \[ \begin{bmatrix}6 & 9 \\ 4 & 5 \end{bmatrix} \]

Example: \[ A = \begin{bmatrix}1 & 2 \\ 3 & 4 \end{bmatrix}, \quad R_2 \to R_2 - 3R_1 \] Result: \[ \begin{bmatrix}1 & 2 \\ 0 & -2 \end{bmatrix} \]

\[ \begin{bmatrix}1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1\end{bmatrix} \]

Explanation: The pivots are at positions (1,1), (2,2), (3,3). Each pivot is to the right of the pivot above, and all entries below pivots are zero. ✔

Example 2

❌ Not in echelon form because the third row has its pivot in column 2, which comes below a row with a pivot in column 3. Pivots must move strictly to the right as we go down rows.

Example 3

❌ Not in echelon form because all-zero rows must be at the bottom, but here a nonzero row appears after a zero row.

Example 4

❌ Not in echelon form because the second row’s leading entry is in the same column as the first row’s pivot. Pivots must move strictly to the right.

\[ \begin{bmatrix}1 & 5 & 2 \\ 0 & 1 & 7 \\ 0 & 0 & 0\end{bmatrix} \]

Explanation: Nonzero rows are above the zero row. Pivots at (1,1) and (2,2) are in proper order. ✔

\[ \begin{bmatrix}2 & 3 & 4 & 5 \\ 0 & 1 & 6 & 7 \\ 0 & 0 & 0 & 0\end{bmatrix} \]

Explanation: Pivot in row 1 is at column 1, pivot in row 2 is at column 2. All-zero row is at the bottom. ✔

\[ \begin{bmatrix}1 & -1 & 2 \\ 0 & 3 & 5 \\ 0 & 0 & 7\end{bmatrix} \]

Explanation: Pivots appear at (1,1), (2,2), (3,3). Each pivot moves strictly right as we go down rows. ✔

\[ \begin{bmatrix}1 & 4 & 6 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix} \]

Explanation: Pivot of row 1 is at column 1, pivot of row 2 is at column 3. Row 3 is zero. Strictly increasing pivot positions are maintained. ✔

\[ \begin{bmatrix}3 & 2 & 1 \\ 0 & 5 & 4 \\ 0 & 0 & 6\end{bmatrix} \]

Explanation: Pivots are at diagonal positions (1,1), (2,2), (3,3). Matrix is upper triangular, which always satisfies REF conditions. ✔

\[ \begin{bmatrix}1 & 7 & 3 & 2 \\ 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0\end{bmatrix} \]

Explanation: Row 1 pivot is at column 1, row 2 pivot is at column 3. The zero row is at the bottom. This is valid REF. ✔

\[ \begin{bmatrix}1 & 2 \\ 0 & 1 \\ 0 & 0\end{bmatrix} \]

Explanation: Pivot at (1,1), next pivot at (2,2), and row 3 is zero. Classic triangular REF form. ✔

\[ \begin{bmatrix}2 & 4 & 6 \\ 0 & 3 & 5 \\ 0 & 0 & 8\end{bmatrix} \]

Explanation: Upper triangular with pivots at (1,1), (2,2), (3,3). Automatically satisfies REF rules. ✔

\[ \begin{bmatrix}1 & 0 & 2 & 3 \\ 0 & 1 & 4 & 5 \\ 0 & 0 & 0 & 1\end{bmatrix} \]

Explanation: Pivot in row 1 is column 1, row 2 pivot in column 2, row 3 pivot in column 4. Each pivot is to the right of the one above. ✔