Let A 1 4 2 5 0 1 0 1 2 8 4 6 x 1 3 2 3 r10 2 13 a Is x Null

Let A= [1 4 2 5 0 1 0 1 -2 8 -4 6] x= [1 -3 -2 3] r=[10 2 13] a) Is x Null(A)? b) Find a basis for the Nullspace of A as span of a set of vectors.

Solution

Take the given matrix and find the reduced row echelon form.

Add 2 times the 1st row to the 3rd row
1   4   2   5
0   1   0   1
0   16   0   16

Add -16 times the 2nd row to the 3rd row

1   4   2   5
0   1   0   1
0   0   0   0

Add -4 times the 2nd row to the 1st row
Matrix
1   0   2   1
0   1   0   1
0   0   0   0

This is the reduced row echelon form of the augmented matrix.

The corresponding system is,
x1 +2 x3+ x4 = 0
   x2 + x4 = 0
      0 = 0
Clearly x3,x4 are free variables.

x1=-2x3-x4+0x1+0x2
x2=-x4+0x3+0x2+0x1
x3=x3+0x4+0x2+0x1
x4=x4+0x3+0x2+0x1

Then, x3[-2,0,1,0]^T+x4[-1,-1,0,1]^T
Therefore, Null(A)={[-2,0,1,0]^T,[-1,-1,0,1]^T}
Clearly, x=[-2,0,1,0]^T-3[-1,-1,0,1}^T
Therefore, x is in Null(A).