© 2000−2019  P. BogackiFinding a basis of the null space of a matrixv. 1.25 

 PROBLEM

Find the null space of the matrix:

 -3   -2   -2 
 1   0   -2 
 0   0   0 

 SOLUTION

 Step 1: Transform the matrix to the reduced row echelon form  (Hide details)

Row
Operation
1:  
 -3   -2   -2 
 1   0   -2 
 0   0   0 
multiply the 1st row by -1/3
 1   2 
 3 		  2 
 3 
 1   0   -2 
 0   0   0 
Row
Operation
2:  
 1   2 
 3 		  2 
 3 
 1   0   -2 
 0   0   0 
add -1 times the 1st row to the 2nd row
 1   2 
 3 		  2 
 3 
 0   -2 
 3 		  -8 
 3 
 0   0   0 
Row
Operation
3:  
 1   2 
 3 		  2 
 3 
 0   -2 
 3 		  -8 
 3 
 0   0   0 
multiply the 2nd row by -3/2
 1   2 
 3 		  2 
 3 
 0   1   4 
 0   0   0 
Row
Operation
4:  
 1   2 
 3 		  2 
 3 
 0   1   4 
 0   0   0 
add -2/3 times the 2nd row to the 1st row
 1   0   -2 
 0   1   4 
 0   0   0 
 Step 2: Interpret the reduced row echelon form

The reduced row echelon form of the augmented matrix is

 1   0   -2 
 0   1   4 
 0   0   0 

which corresponds to the system

1 x1 -2 x3=0
 1 x2 +4 x3=0
  0=0
The leading entries in the matrix have been highlighted in yellow.

A leading entry on the (i,j) position indicates that the j-th unknown will be determined using the i-th equation.

Those columns in the coefficient part of the matrix that do not contain leading entries, correspond to unknowns that will be arbitrary.

The system has infinitely many solutions:

x1=2 x3
x2=-4 x3
x3=arbitrary

The solution can be written in the vector form:

x1
x2
x3
 =
2
-4
1
x3

Therefore the null space has a basis formed by the set {
2
-4
1
}.
 Comments

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