Linear algebraConcept reference

Free variables

A free variable corresponds to a nonpivot variable column; a parameterization expresses every variable using independent real parameters for the free variables.

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On this page 7 sections

Overview
In the reduced equation x+2y=6, the x-column has a pivot and the y-column does not
Free does not mean zero
Use a separate parameter for each free variable
Key takeaway
Sources & further reading
Concept connections

01In the reduced equation x+2y=6, the x-column has a pivot and the y-column does not#

In the reduced equation $x+2y=6$ , the x-column has a pivot and the y-column does not. Choose any real value $y=t$ . The equation then forces $x=6-2t$ .

Figure 1Every point on the plotted line satisfies x+2y=6. Parameter t=y gives x=6−2t. The examples t=0 and t=2 produce (6,0) and (2,2). The line represents infinitely many real choices, not just these points.

For $x+2y=6$ , the formulas $x=6-2t$ and $y=t$ describe every solution. Substitution cancels the parameter. Choosing $t=2$ gives $(x,y)=(2,2)$ .

(6-2t)+2t=6

Check your reasoning

Columns: $x,y$ . Give all solutions; $t$ is real. $\left[\begin{array}{rr|r}1&2&6\\0&0&0\end{array}\right]$

A $x=6-2t;\ y=t$
B $x=6+2t;\ y=t$
C $x=6;\ y=0$

Show answer and explanation

$x=6-2t;\ y=t$

$y=t$ is free.

02Free does not mean zero#

Free does not mean zero. In $x-y=3$ , choosing $y=0$ gives one solution, $(3,0)$ . Choosing $y=2$ gives another, $(5,2)$ . Keeping $y=t$ preserves them all.

Check your reasoning

For $x-y=3$ , a learner claims only $(3,0)$ . Repair all solutions; $t$ is real.

A $x=3;\ y=0$
B $x=3-t;\ y=t$
C $x=3+t;\ y=t$

Show answer and explanation

$x=3+t;\ y=t$

Every $y=t$ is allowed; $x=3+t$ .

03Use a separate parameter for each free variable#

Use a separate parameter for each free variable. In $x+y-z=4$ , only x has a pivot. Set $y=s$ and $z=t$ independently, so $x=4-s+t$ . The augmented column is a right-hand side, not a free variable.

Check your reasoning

Settings obey $x+2y-z=5$ . Give all $(x,y,z)$ ; real $s,t$ vary independently.

A $(5,0,0)$
B $(5-2s+t,s,t)$
C $(5+2s-t,s,t)$

Show answer and explanation

$(5-2s+t,s,t)$

Let $y=s,z=t$ ; solve for $x$ .

Key takeaway

Assign one real parameter per free variable. Express pivot variables in terms of those choices.

Parameterize the free variables of a consistent reduced system.

Sources & further reading

[1]
Margalit and Rabinoff: Interactive Linear Algebra, 1.3 Parametric Form ↗Interactive Linear Algebra · Book

01In the reduced equation x+2y=6, the x-column has a pivot and the y-column does not#

02Free does not mean zero#

03Use a separate parameter for each free variable#

Assign one real parameter per free variable. Express pivot variables in terms of those choices.

Sources & further reading

Reference this concept