Gaussian Elimination

Infinite Solutions

Gaussian elimination on an underdetermined 3x3 system where one equation is a multiple of another. Elimination produces a zero row [0 0 0 | 0], revealing a free variable. The remaining pivot variables are expressed in terms of the free variable t.

free-variable A variable is free when its column has no pivot after RREF. It can take any value t; the pivot variables are then determined as functions of t, giving infinitely many solutions.
parametric-solution The solution set is expressed with the free variable as a parameter: each pivot variable equals a constant plus a multiple of t. Different values of t give different solutions, all valid.

Example

Row-reduce to expose a free variable and write a parametric solution.

highlighted = computed this step

Step 1 — Set up

Start with the augmented matrix.

[1126224121238]\left[\begin{array}{ccc|c}1&1&2&6\\2&2&4&12\\1&2&3&8\end{array}\right]

Step 2 — Update row 2

Update row 2 using row 1: r2 <- r2 - 2*r1.

[112600001238]\left[\begin{array}{ccc|c}1&1&2&6\\\hl{0}&\hl{0}&\hl{0}&\hl{0}\\1&2&3&8\end{array}\right]

Step 3 — Update row 3

Update row 3 using row 1: r3 <- r3 - r1.

[112600000112]\left[\begin{array}{ccc|c}1&1&2&6\\0&0&0&0\\\hl{0}&\hl{1}&\hl{1}&\hl{2}\end{array}\right]

Step 4 — Swap rows

Swap rows 2 and 3: r2 <-> r3.

[112601120000]\left[\begin{array}{ccc|c}1&1&2&6\\\hl{0}&\hl{1}&\hl{1}&\hl{2}\\\hl{0}&\hl{0}&\hl{0}&\hl{0}\end{array}\right]

Step 5 — Back-sub row 1

Update row 1 using row 2: r1 <- r1 - r2.

[101401120000]\left[\begin{array}{ccc|c}\hl{1}&\hl{0}&\hl{1}&\hl{4}\\0&1&1&2\\0&0&0&0\end{array}\right]

Step 6 — Result

State the parametric solution: x1 = 4 - t, x2 = 2 - t, x3 = t.

[101401120000],parametric: x1=4−t, x2=2−t, x3=t\left[\begin{array}{ccc|c}1&0&1&4\\0&1&1&2\\0&0&0&0\end{array}\right],\quad \text{parametric: }x1=4-t,\ x2=2-t,\ x3=t