Consistency of a linear system

The system AX = B is consistent when rank A = rank [A|B]; we reduce the augmented matrix to echelon form to compare the two ranks.

[A|B] = [1 1 1 | 3]
        [1 2 3 | 4]
        [2 3 4 | 7]

Step 1. R₂ → R₂ − R₁, R₃ → R₃ − 2R₁:

[1 1 1 | 3]
[0 1 2 | 1]
[0 1 2 | 1]

Step 2. R₃ → R₃ − R₂:

[1 1 1 | 3]
[0 1 2 | 1]
[0 0 0 | 0]

Consistency

rank A = 2 and rank [A|B] = 2. The ranks are equal, so the system is consistent. Since the rank (2) is less than the number of unknowns (3), there are infinitely many solutions, with 3 − 2 = 1 free variable.

Solution

Put z = k.

• y + 2k = 1 ⇒ y = 1 − 2k
• x + y + z = 3 ⇒ x = 3 − (1 − 2k) − k = 2 + k

Answer: x = 2 + k, y = 1 − 2k, z = k, for any real k.

Check with k = 1: (3, −1, 1): 3 − 1 + 1 = 3 ✓; 3 − 2 + 3 = 4 ✓; 6 − 3 + 4 = 7 ✓.