for the null space you are going to have one vector since there is 1 zero row,

first you should reduce the matrix as much as you can because this will make the calculation easier,

1 0 0 0

0 1 0 -1

0 0 1 -1

0 0 0 0

you will have to let x_4 = k, some real number k

then all the x's will be either 0 or k

x1 = 0, x2 = k = x3

so you will get a nullspace of {0,1,1,1}

try following

http://www.math.jhu.edu/~jmb/note/rowcol.pdf

the row and column space are easy,

the row space are just the remaining non-zero rows after reduction to row-ech form (so you should continue the reduction until you have number 1 down the diagonal)

it will be {1 0 0 0}, {0 1 0 -1}, { 0 0 1 -1 }

then for the column space you take each each column of the original matrix corresponding to a pivot (a leading 1 on the diagonal)

so the column space for this matrix will be the first 3 columns of the original matrix,

{1,-1,-2,0}, {4,-2,-4,-4}, {5,-1,0,-8}