Parametric Form

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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.

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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 .

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