Recipe

Test existence and uniqueness from a matrix in row echelon form

Use when: you have (or can quickly reduce to) an augmented matrix in row echelon form, and need to know how many solutions the system has — not their actual values

Goal: a classification into exactly one of three outcomes: no solution, exactly one solution, or infinitely many

Taught in Existence and uniqueness

1. 1

   If it isn't already, run Gaussian elimination first — leading entries staircasing down and to the right, with zeros below every one of them. Both tests below only read reliably off a matrix already in that shape.

   Example

   \left[\begin{array}{cccc|c} 1 & 2 & -1 & 3 & 5 \\ 0 & 0 & 1 & -2 & 4 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]

   The leading entries staircase from column 1 (row 1) to column 3 (row 2), with only a zero row left over — this is already row echelon form.

   Why: the existence and uniqueness tests below are only well-defined once every row's leading entry sits in its own staircased column — scanning for a pivot or a contradiction assumes that shape.

2. 2

   Check each row: if every coefficient is 0 but the entry after the bar isn't, that row asserts a false equation and no choice of the variables can satisfy it.

   Example

   Row 3 reads 0,0,0,0 \mid 0 — that's the equation 0=0, which is true, not a contradiction. No row here asserts 0=c for a nonzero c, so the system is consistent.

   • a 0=c row with nonzero c exists — the system has no solution; stop here — uniqueness isn't even a meaningful question yet.
   • no such row exists — the system is consistent; continue to Step 3 to test uniqueness.

   Why: a single false row makes the whole system unsatisfiable regardless of what any other row says, since a solution has to satisfy every equation at once.

3. 3

   Find which variable columns (not the augmented column) hold a leading entry. If every variable column has one, the solution is unique; any variable column left without one names a free variable.

   Example

   The pivots sit in column 1 (x_1, from row 1) and column 3 (x_3, from row 2) — 2 pivot columns out of 4 variable columns, so columns 2 and 4 (x_2 and x_4) are free.

   • every variable column has a pivot — exactly one solution.
   • at least one variable column is free — infinitely many solutions — the free variable can be set to any value, and the pivot variables adjust row by row to match.

   Why: a pivot column pins its variable to one exact value once the free variables are fixed; a variable column with no pivot is never pinned down by any row, so it can be set to anything.