Example of using the echelon form to determine if a linear system is consistent.

x + 2y - z =  1
2x + 4y - 2z =  3
-x - 2y + z = -1

[ 1  2 -1 |  1]
[ 2  4 -2 |  3]
[-1 -2  1 | -1]

[ 1  2 -1 |  1]
[ 0  0  0 |  1]
[ 0  0  0 |  0]

x + 2y - z = 1
2x + 4y - 2z = 2
-x - 2y + z = -1

[ 1  2 -1 |  1]
[ 0  0  0 |  0]
[ 0  0  0 |  0]

## Example: Using Echelon Form to Determine Consistency

Determine whether the following system is consistent:
```
x + 2y - z =  1
2x + 4y - 2z =  3
-x - 2y + z = -1
```

**Step 1:** Write augmented matrix
```
[ 1  2 -1 |  1]
[ 2  4 -2 |  3]
[-1 -2  1 | -1]
```

**Step 2:** R₂ → R₂ - 2R₁, R₃ → R₃ + R₁
```
[ 1  2 -1 |  1]
[ 0  0  0 |  1]
[ 0  0  0 |  0]
```

**Step 3:** Analyze echelon form

Row 2 reads `[0 0 0 | 1]`, which corresponds to the equation `0 = 1`. This is a contradiction.

**Conclusion:** The system is **inconsistent** -- it has no solution.

---

**Contrast with a consistent system:**
```
x + 2y - z = 1
2x + 4y - 2z = 2
-x - 2y + z = -1
```

Augmented matrix after reduction:
```
[ 1  2 -1 |  1]
[ 0  0  0 |  0]
[ 0  0  0 |  0]
```

No contradictory row. The system is consistent with free variables y and z, giving infinitely many solutions: `x = 1 - 2y + z` for any y, z in R.Copy