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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 word: 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

E2x+y+12z=1x+2y+9z=−1.

We solve it using row reduction:

C21121129−1DR1←→R2−−−−→C129−121121D(Optional)R2=R2−2R1−−−−−−→C129−10−3−63D(Step1c)R2=R2÷−3−−−−−→C129−1012−1D(Step2b)R1=R1−2R2−−−−−−→C1051012−1D(Step2c)

This row reduced matrix corresponds to the linear system

Ex+5z=1y+2z=−1.

In what sense is the system solved? We rewrite as

Ex=1−5zy=−1−2z.

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

Fx=1−5zy=−1−2zz=zzanyrealnumber.

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 x1,x2,...,xn. Let A be a row echelon form of the augmented matrix for this system.

We say that xi 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

C1051012−1D.

In the matrix

C1A0AA001AAD,

the free variables are x2 and x4. (The augmented 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

E2x+y+12z=1x+2y+9z=−1

is a line in R3, 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

Ex=1−5zy=−1−2z.

can be written as follows:

(x,y,z)=(1−5z,−1−2z,z)zanyrealnumber.

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 parameteric form is much more explicit: it gives a concrete recipe for producing all solutions.

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

Free variables come from the columns without pivots 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
    A100010001B
    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
    A100a010b001cB
    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
    C1−20310014−1D,
    any values for x2 and x4 yield a solution to the system of equations.