Recipe

Detect that a system is inconsistent while eliminating

Use when: you're running Gaussian elimination on an augmented matrix and want to catch (or rule out) a contradiction as early as possible, rather than waiting until the whole matrix reaches echelon form

Goal: either a confirmed 0=c row — permission to stop immediately with no solution — or confidence that no such row exists anywhere, even after combining every row

Taught in Inconsistent systems, and what the picture looks like

1. 1

   After every replace operation, check the row you just changed: if every coefficient on the left has become 0, look at what's left on the right before doing anything else.

   Example

   For \left[\begin{array}{ccc|c} 1 & 1 & 0 & 3 \\ 0 & 1 & 1 & 4 \\ 1 & 0 & -1 & -2 \end{array}\right], eliminate column 1 from row 3 with R_3\to R_3-R_1: (1,0,-1,-2)-(1,1,0,3)=(0,-1,-1,-5). The coefficients aren't all zero yet, so keep going.

   Why: catching a contradiction the moment it appears saves redoing work on rows that no longer matter — it's the same elimination already underway, just watched closely as it happens.

2. 2

   The moment a row's coefficients have all vanished but its constant hasn't, no further row operation can repair it — swapping, scaling, or replacing an all-zero coefficient row only ever produces another all-zero coefficient row.

   Example

   Eliminate column 2 from the new row 3 using row 2, R_3\to R_3+R_2: (0,-1,-1,-5)+(0,1,1,4)=(0,0,0,-1). Every coefficient is 0, but the constant is -1\neq0 — the system is inconsistent, full stop.

   • a 0=c row with nonzero c appears — stop immediately; no solution, regardless of any row not yet touched.
   • elimination finishes with no such row ever appearing — the system is consistent; the uniqueness question (a separate recipe) is the one worth asking next.

   Why: this is the same 0=c signal the existence test watches for — a row no completion of the reduction can rescue, so finishing the elimination on the remaining rows can't change the verdict.

3. 3

   Checking whether any two rows' coefficient parts are scalar multiples of each other catches an obvious contradiction early, but it's a shortcut, not the whole test — three or more equations can be pairwise fine and still contradict once genuinely combined.

   Example

   None of (1,1,0), (0,1,1), (1,0,-1) is a scalar multiple of either of the others — no single pair looks parallel — yet the full elimination above still produced 0=-1. The contradiction was hiding in the three-way combination, not in any one pair.

   Why: parallel rows are only one way a contradiction can arise; a genuine contradiction can also sit in a hidden combination of three or more rows that no pairwise glance ever reveals.