Consistency of a linear system

A system AX = B is consistent if rank A = rank [A|B]. If this common rank equals the number of unknowns, the solution is unique.

The augmented matrix is

[A|B] = [1 −2 2 | 4]
        [3  1 4 | 6]
        [1  1 1 | 1]

Row reduction

R₂ → R₂ − 3R₁, R₃ → R₃ − R₁:

[1 −2  2 |  4]
[0  7 −2 | −6]
[0  3 −1 | −3]

R₂ → R₂ − 2R₃:

[1 −2  2 |  4]
[0  1  0 |  0]
[0  3 −1 | −3]

R₃ → R₃ − 3R₂:

[1 −2  2 |  4]
[0  1  0 |  0]
[0  0 −1 | −3]

Consistency

rank A = 3 and rank [A|B] = 3 (three non-zero rows). Since the ranks are equal, the system is consistent. The rank equals the number of unknowns (3), so the solution is unique.

Solution (back substitution)

• −z = −3 ⇒ z = 3
• y = 0
• x − 2(0) + 2(3) = 4 ⇒ x = −2

Check: 3(−2) + 0 + 4(3) = 6 ✓; −2 + 0 + 3 = 1 ✓.

Answer: x = −2, y = 0, z = 3.