K cannot be what, if you want this matrix to form a basis, confused

[mr_coffee]

4 -4 4
-5 6 -3
0 4 k
form a basis for R^3 if and only if k != ?
since the last col has a k in it, i wanted to see if the first 2 colmuns are linear indepdant, adn they are, becuase i row reduced w/calculator:
4 -4 0
-5 6 0
0 4 0
and got
1 0 0
0 1 0
0 0 0
is there a quick way to figure out the k, without mantually row reducing with k? Also if there isn't a quick way, what am i going to be looking for? it will span R^3 if i get
0
0
1
for the last col with k in it. Thanks.

 

[matt grime]

You don't *have* to get down to 1's, merely linearly independent rows/cols, 1s are just nice. And it's better if you say "when do the rows/cols form a basis", rather than "when is the matrix a basis of R^3": the matirx is never a basis of R^3; it isn't even an element of R^3.

Remember, you're just looking for a solution of Mx=0 to imply x=0, so that's an easy criterion to check when there is only 1 variable (k) to check