Find a basis for Col A and a basis for Null A A 3 1 7 3 9 2

Find a basis for Col A and a basis for Null A. A [3 -1 7 3 9 -2 2 -2 7 5 -5 9 3 3 4 -2 6 6 3 7]~[3 -1 7 0 6 0 2 4 0 3 0 0 0 1 1 0 0 0 0 0]

Solution

The basis for Col A would be the largest possible set of columns of A such that the columns are linearly independent

Now if columns of A are linearly independent then corresponding columns of row reduced matrix are also linearly independent and vice versa.

We can see that in the row reduced form the linearly independent columns are

First,second and 4th columns

So First ,second and 4th columns of A form basis of Col A

Now for Nul A we need to solve Ax=0

For this instead of A we can use the row reduced form given in the problem

Let, x=[a b c d e]^T

So, Ax=0 using row reduced form gives

d+e=0 ie d=-e

2b+4c+3e=0

b=-3c-3e/2

3a-b+7d+6e=0

3a=b-7d-6e

3a=-3c-3e/2+7e-6e=-3c-e/2

a=-c-e/6

So

x=(-c-e/6,-3c-3e/2,c,-e,e)=c(-1,-3,1,0,0)+e(-1/6,-3/2,0,-1,1)

So basis fo Nul A= {(-1,-3,1,0,0),(-1/6,-3/2,0,-1,1)}