8.5. LINEAR INDEPENDENCE AND BASES 171

You need to solve the equation Ax = 0. To do this you write the augmented matrix andthen obtain the row reduced echelon form and the solution. The augmented matrix is 1 2 1 | 0

0 −1 1 | 02 3 3 | 0

Next place this matrix in row reduced echelon form, 1 0 3 | 0

0 1 −1 | 00 0 0 | 0

Note that x1 and x2 are basic variables while x3 is a free variable. Therefore, the solution tothis system of equations, Ax = 0 is given by 3t

tt

 : t ∈ R.

Example 8.5.25 Let

A =

1 2 1 0 12 −1 1 3 03 1 2 3 14 −2 2 6 0

Find the null space of A.

You need to solve the equation, Ax = 0. The augmented matrix is1 2 1 0 1 | 02 −1 1 3 0 | 03 1 2 3 1 | 04 −2 2 6 0 | 0

Its row reduced echelon form is

1 0 35

65

15 | 0

0 1 15 − 3

525 | 0

0 0 0 0 0 | 00 0 0 0 0 | 0

It follows x1 and x2 are basic variables and x3,x4,x5 are free variables. Therefore, ker(A)is given by 

(− 3

5

)s1 +

(−65

)s2 +

( 15

)s3(

− 15

)s1 +

( 35

)s2 +

(− 2

5

)s3

s1

s2

s3

 : s1,s2,s3 ∈ R.