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Section2.3Parametric Form

Objectives
  1. Learn to express the solution set of a system of linear equations in parametric form.
  2. Understand the three possibilities for the number of solutions of a system of linear equations.
  3. Recipe: parametric form.
  4. Vocabulary: free variable.

Subsection2.3.1Free Variables

There is one possibility for the row reduced form of a matrix that we did not see in Section 2.2.

Example(A System with a Free Variable)

Consider the linear system

E 2 x + y + 12 z = 1 x + 2 y + 9 z = − 1.

We solve it using row reduction:

C 2112 1129 − 1 D R 1 ←→ R 2 −−−−→ C 129 − 12112 1 D (Optional) R 2 = R 2 − 2 R 1 −−−−−−→ C 129 − 10 − 3 − 6 3 D (Step1c) R 2 = R 2 ÷− 3 −−−−−→ C 129 − 1012 − 1 D (Step2b) R 1 = R 1 − 2 R 2 −−−−−−→ C 105 1012 − 1 D (Step2c)

This row reduced matrix corresponds to the linear system

E x + 5 z = 1 y + 2 z = − 1.

In what sense is the system solved? We rewrite as

E x = 1 − 5 zy = − 1 − 2 z .

For any value of z , there is exactly one value of x and y that make the equations true. But we are free to choose any value of z .

We have found all solutions: it is the set of all values x , y , z , where

F x = 1 − 5 zy = − 1 − 2 zz = zz anyrealnumber.

This is called the parametric form for the solution to the linear system. The variable z is called a free variable.

Figure2A picture of the solution set (the yellow line) of the linear system in this example. There is a unique solution for every value of z ; move the slider to change z .

Given the parametric form for the solution to a linear system, we can obtain specific solutions by replacing the free variables with any specific real numbers. For instance, setting z = 0 in the last example gives the solution ( x , y , z )=( 1, − 1,0 ) , and setting z = 1 gives the solution ( x , y , z )=( − 4, − 3,1 ) .

Definition

Consider a consistent system of equations in the variables x 1 , x 2 ,..., x n . Let A be a row echelon form of the augmented matrix for this system.

We say that x i is a free variable if its corresponding column in A is not a pivot column.

In the above example, the variable z was free because the reduced row echelon form matrix was

C 105 1012 − 1 D .

In the matrix

C 1 A 0 A A 001 A A D ,

the free variables are x 2 and x 4 . (The augmentation column is not free because it does not correspond to a variable.)

Recipe: Parametric form

The parametric form of the solution set of a consistent system of linear equations is obtained as follows.

  1. Write the system as an augmented matrix.
  2. Row reduce to reduced row echelon form.
  3. Write the corresponding (solved) system of linear equations.
  4. Move all free variables to the right hand side of the equations.

Moving the free variables to the right hand side of the equations amounts to solving for the non-free variables (the ones that come pivot columns) in terms of the free variables. One can think of the free variables as being independent variables, and the non-free variables being dependent.

Implicit Versus Parameterized Equations

The solution set of the system of linear equations

E 2 x + y + 12 z = 1 x + 2 y + 9 z = − 1

is a line in R 3 , as we saw in this example. These equations are called the implicit equations for the line: the line is defined implicitly as the simultaneous solutions to those two equations.

The parametric form

E x = 1 − 5 zy = − 1 − 2 z .

can be written as follows:

( x , y , z )=( 1 − 5 z , − 1 − 2 z , z ) z anyrealnumber. (2.3.1)

This called a parameterized equation for the same line. It is an expression that produces all points of the line in terms of one parameter, z .

One should think of a system of equations as being an implicit equation for its solution set, and of the parametric form as being the parameterized equation for the same set. The parametric form is much more explicit: it gives a concrete recipe for producing all solutions.

Parametric Forms in Vector Notation

While you can certainly write parametric solutions in point notation, it turns out that vector notation is ideally suited to writing down parametric forms of solutions.

Example

It is sometimes useful to introduce new letters for the parameters. For instance, instead of writing

( x , y , z )=( 1 − 5 z , − 1 − 2 z , z ) z anyrealnumber.

you can write

( x , y , z )=( 1 − 5 s , − 1 − 2 s , s ) s anyrealnumber.

or in vector notation

A xyz B = A 1 − 10 B + s A − 5 − 21 B .

Of course, since this implies z = s , you might think that we haven't gained anything by the extra complexity.

Note

This note describes a subtle point, which you can disregard on first reading.

What we gain from the extra complexity is flexibility to change the parameter. For instance, we could start with

( x , y , z )=( 1 − 5 s , − 1 − 2 s , s ) s anyrealnumber.

and decide we would prefer to parametrize usingt = − s . Then we can write the solution set as

( x , y , z )=( 1 + 5 t , − 1 + 2 t , − t ) t anyrealnumber.

We could go even further, and change the parameter to u = 1 − s . Now we can write the solution set as

( x , y , z )=( 5 u − 4, u − 3,1 − u ) u anyrealnumber.

The three parameterizations above all describe the same line in R 3 . For instance, if you plug in s = 1, you get ( − 4, − 3,1 ) , which you can also get by setting t = − 1 or u = 0.

Note

This note describes a subtle point, which you can disregard on first reading.

What we gain from the extra complexity is flexibility to change the parameter. For instance, we could start with

( x , y , z )=( 1 − 5 s , − 1 − 2 s , s ) s anyrealnumber.

and decide we would prefer to parametrize usingt = − s . Then we can write the solution set as

( x , y , z )=( 1 + 5 t , − 1 + 2 t , − t ) t anyrealnumber.

We could go even further, and change the parameter to u = 1 − s . Now we can write the solution set as

( x , y , z )=( 5 u − 4, u − 3,1 − u ) u anyrealnumber.

The three parameterizations above all describe the same line in R 3 . For instance, if you plug in s = 1, you get ( − 4, − 3,1 ) , which you can also get by setting t = − 1 or u = 0.

You can choose any value for the free variables in a (consistent) linear system.

Free variables come from the columns without pivots (excluding the augmentation column) in a matrix in row echelon form.

Example
Example(A Parameterized Plane)

Subsection2.3.2Number of Solutions

There are three possibilities for the reduced row echelon form of the augmented matrix of a linear system.

  1. The last column is a pivot column. In this case, the system is inconsistent. There are zero solutions, i.e., the solution set is empty. For example, the matrix
    A 10 001 000 1 B
    comes from a linear system with no solutions.
  2. Every column except the last column is a pivot column. In this case, the system has a unique solution. For example, the matrix
    A 100 a 010 b 001 c B
    tells us that the unique solution is ( x , y , z )=( a , b , c ) .
  3. The last column is not a pivot column, and some other column is not a pivot column either. In this case, the system has infinitely many solutions, corresponding to the infinitely many possible values of the free variable(s). For example, in the system corresponding to the matrix
    C 1 − 203 10014 − 1 D ,
    any values for x 2 and x 4 yield a solution to the system of equations.