Find Null Space ...
`[[3,-1,-1],[2,-2,1]]`

Solution:

	`3`	`-1`	`-1`
	`2`	`-2`	`1`

Now, reduce the matrix to reduced row echelon form
`R_1 larr R_1-:3`

=	`1` `-1/3` `-1/3` `2` `-2` `1`

`R_2 larr R_2-2xx R_1`

=	`1` `-1/3` `-1/3` `0` `-4/3` `5/3`

`R_2 larr R_2xx(-3/4)`

=	`1` `-1/3` `-1/3` `0` `1` `-5/4`

`R_1 larr R_1+1/3xx R_2`

=	`1` `0` `-3/4` `0` `1` `-5/4`

The rank of a matrix is the number of non all-zeros rows
`:. Rank = 2`

Null Space :
Now, solve the matrix equation

`1` `0` `-3/4` `0` `1` `-5/4`

`x_1` `x_2` `x_3`

=

`0` `0`

`x_1-3/4x_3=0`

`x_2-5/4x_3=0`

Add equation for each free variable
`x_1-3/4x_3=0`

`x_2-5/4x_3=0`

`x_3=x_3`

Solve for each variable in terms of the free variables
`x_1=3/4x_3`

`x_2=5/4x_3`

`x_3=x_3`

Convert this into vectors

`x_1` `x_2` `x_3`

=

`3/4x_3` `5/4x_3` `x_3`

=

`[[3/4],[5/4],[1]]`

`x_3`

Thus, the basis for the null space is
`[[3/4],[5/4],[1]]`

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