1. Find Column Space ...
`[[1,-2,0,3,-4],[3,2,8,1,4],[2,3,7,2,3],[-1,2,0,4,-3]]`

Solution:

	`1`	`-2`	`0`	`3`	`-4`
	`3`	`2`	`8`	`1`	`4`
	`2`	`3`	`7`	`2`	`3`
	`-1`	`2`	`0`	`4`	`-3`

Now, reduce the matrix to row echelon form
interchanging rows `R_1 harr R_2`

=	`3` `2` `8` `1` `4` `1` `-2` `0` `3` `-4` `2` `3` `7` `2` `3` `-1` `2` `0` `4` `-3`

`R_2 larr R_2-1/3xx R_1`

=	`3` `2` `8` `1` `4` `0` `-8/3` `-8/3` `8/3` `-16/3` `2` `3` `7` `2` `3` `-1` `2` `0` `4` `-3`

`R_3 larr R_3-2/3xx R_1`

=	`3` `2` `8` `1` `4` `0` `-8/3` `-8/3` `8/3` `-16/3` `0` `5/3` `5/3` `4/3` `1/3` `-1` `2` `0` `4` `-3`

`R_4 larr R_4+1/3xx R_1`

=	`3` `2` `8` `1` `4` `0` `-8/3` `-8/3` `8/3` `-16/3` `0` `5/3` `5/3` `4/3` `1/3` `0` `8/3` `8/3` `13/3` `-5/3`

`R_3 larr R_3+5/8xx R_2`

=	`3` `2` `8` `1` `4` `0` `-8/3` `-8/3` `8/3` `-16/3` `0` `0` `0` `3` `-3` `0` `8/3` `8/3` `13/3` `-5/3`

`R_4 larr R_4+ R_2`

=	`3` `2` `8` `1` `4` `0` `-8/3` `-8/3` `8/3` `-16/3` `0` `0` `0` `3` `-3` `0` `0` `0` `7` `-7`

interchanging rows `R_3 harr R_4`

=	`3` `2` `8` `1` `4` `0` `-8/3` `-8/3` `8/3` `-16/3` `0` `0` `0` `7` `-7` `0` `0` `0` `3` `-3`

`R_4 larr R_4-3/7xx R_3`

=	`3` `2` `8` `1` `4` `0` `-8/3` `-8/3` `8/3` `-16/3` `0` `0` `0` `7` `-7` `0` `0` `0` `0` `0`

The rank of a matrix is the number of non all-zeros rows
`:. Rank = 3`

Column Space :
The matrix has 3 pivots and Pivots are in the columns 1,2 and 4.
We know that these columns in the original matrix define the column space of the matrix.
`:.` The Column Space is

`[[1],[3],[2],[-1]],[[-2],[2],[3],[2]],[[3],[1],[2],[4]]`

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