## Example: Solving a 3-by-3 Homogeneous Matrix Equation

Solve `Ax = 0` where:
```
A = [ 1  2  3]
    [ 2  5  3]
    [ 1  0  8]
```

**Step 1:** Form the augmented matrix `[A | 0]`
```
[ 1  2  3 | 0]
[ 2  5  3 | 0]
[ 1  0  8 | 0]
```

**Step 2:** R₂ → R₂ - 2R₁, R₃ → R₃ - R₁
```
[ 1  2  3 | 0]
[ 0  1 -3 | 0]
[ 0 -2  5 | 0]
```

**Step 3:** R₃ → R₃ + 2R₂
```
[ 1  2  3 | 0]
[ 0  1 -3 | 0]
[ 0  0 -1 | 0]
```

**Step 4:** R₃ → -R₃
```
[ 1  2  3 | 0]
[ 0  1 -3 | 0]
[ 0  0  1 | 0]
```

**Step 5:** Back-substitute:
- z = 0
- y - 3(0) = 0, so y = 0
- x + 2(0) + 3(0) = 0, so x = 0

**Solution:** x = [0, 0, 0]ᵀ (the trivial solution)

The homogeneous system has only the trivial solution because A has 3 pivots (full rank). If A had fewer pivots, free variables would yield nontrivial solutions.