Matrices problem, K != ? so that its a linear transformation R^3->R^2?

1. Nov 26, 2005

mr_coffee

Hello everyone, i'm confused on this problem:
It says:
A linear transformation T:R^3->R^2 whose matrix is
2 -4 -3
-3 6 0+k

is onto if k != ?
!= meaning, not equal.

So my thinking was, For it to be a transformation into R^2, doesn't that mean k isn't suppose mean that the column
-3
0+k
has to some how come out to be
0
0
2 -4
-3 6

so that would be in R^2 right?
i'm confused, am i thinking wrong? or is there some trick to easily find what k isn't suppose to be? Thanks!

2. Nov 26, 2005

HallsofIvy

Staff Emeritus
T will be "onto" R2 as long as, after row reduction, the second row is not [0 0 0]. What value of k would make the second row [0 0 0]?

3. Nov 26, 2005

mr_coffee

THanks Halls, but I'm slighly confused,
I row reduced and got:
2 -4 -3
0 0 2k

so 2k = 0?
so no value of k will equal 0, unless its 0, which isn't right.
So i thought maybe i have to write it out like:
2x - 4y -3k = 0
0 + 0 +2k = 0
k = 0;
2x -4y = 0;
so if x = 2, and y = 1 then u'll get 0 = 0, but thats for the top row....hm..what am i not getting? THanks!

4. Nov 26, 2005

HallsofIvy

Staff Emeritus
You started with
[2 -4 -3]
[-3 6 0+k]
which I take to mean
[2 -4 -3]
[-3 6 k]

To "row-reduce" add 3/2 the first row to the second row.
3/2(2)= 3 and adding that to -3 gives 0
3/2(-4)= -6 and adding that to 6 gives 0
3/2(-3)= -9/2 and adding that to k give k- 9/2. that will equal 0 when
k= 9/2.

5. Nov 27, 2005

mr_coffee

ahhh i c were I messed up! thanks Ivy it worked great!!