Reduced Row Echelon Form, Pivot Columns and Special Solutions

Free columns are yours to choose. Pivot columns follow.

Reduce A to R. Columns with a pivot belong to pivot variables. The others are free. Set one free variable to 1 and the rest to 0, then solve for the pivot variables: that is a special solution.

12132438 → R = 12010012

Think: a mixing desk

A mixing desk has four sliders. Two of them are free: you set them however you like. The other two are linked and move by themselves to keep the total at zero.

Free variables are the free sliders. Pivot variables are the linked ones. Each way of setting the free sliders gives exactly one vector in the nullspace.

Reading R

R is the reduced row echelon form: zeros below and above each pivot, and every pivot equal to 1. Here the pivots sit in columns 1 and 3. Columns 2 and 4 are free.

Rx = 0 means: x₁ + 2x₂ + x₄ = 0 and x₃ + 2x₄ = 0

Worked example

Set x₂ = 1, x₄ = 0: x₁ = −2, x₃ = 0 → s₁ = (−2, 1, 0, 0)

Check against A: 1·(−2) + 2·1 = 0 and 2·(−2) + 4·1 = 0 ✓

The other special solution comes from x₂ = 0, x₄ = 1. That is your exercise.

Counting

A has n = 4 columns and r = 2 pivots, so there are n − r = 2 free variables and 2 special solutions. Every vector in N(A) is a combination of them: x = c·s₁ + d·s₂. N(A) is a plane inside 4-dimensional space.

A shortcut: the entries of R in the free columns, with their signs flipped, are the pivot values of the special solutions.