# Express x and y as a linear combination of v's

1. May 13, 2012

### robertjford80

1. The problem statement, all variables and given/known data

3. The attempt at a solution

I can't figure out how to do this problem. If you set the matrix up like this

2 -1 2 1
3 1 1 1

You can't solve it because there are 3 unknowns and two equations.

I still don't really understand quite well when you set the matrix up vertically or horizontally but if you set it up like this

2 3
-1 1
2 1

there's no place to put the values for x and y to solve things.

2. May 13, 2012

### sharks

Reduce the matrix and you'll get two pivots: 2 and 5/2.

Last edited: May 13, 2012
3. May 13, 2012

### robertjford80

But if I reduce it to

x x x x
0 0 x x

where x is any number, I still won't be able to solve it because there will be two unknowns in the first row.

4. May 13, 2012

### sharks

You have only 3 v's. So, in column form, after reducing the matrix:
$$V= \displaystyle\left[ {\begin{array}{*{20}{c}} 2&-1&2 \\ 0&5/2&-2 \\ \end{array}} \right]$$
Hence, you can see that columns 1 and 2 are a basis.

Last edited: May 13, 2012
5. May 13, 2012

### robertjford80

Well, if I reduce it to

$$V= \displaystyle\left[ {\begin{array}{*{20}{c}} 2&-1&2 \\ 0&5/2&-2 \\ \end{array}} \right]$$

Then I don't see what the next step should be.

6. May 13, 2012

### oli4

Hi Robert
You have 3 2D vectors, v1, v2, v3. as you can see, none is perpendicular to any, but they can't form a basis since they are 3 and the dimensionality is 2.
So any two of them will form a basis.
So for instance, you could have a first basis (v1,v2) a second one (v2,v3) and a third one (v1,v3) and that would be the 3 ways in which you can later express x and y
So for instance, in the first base, you would want to express x in (v1,v2)
(and later y too, but first let's look at x)
How do you express x in (v1, V2) ?
x must be a linear combination of v1 and v2, so x=av1+bv2
x being (1,1) that would mean
av1+b2=(1,1), so a(2,3)+b(-1,1)=(1,1)
therefore you would just have to solve (2a-b, 3a+b)=(1,1)
I let you continue at this point

Cheers...

7. May 13, 2012

### robertjford80

by using v1 and v2 and setting up the matrix as follows

2 -1 1
3 1 1

I was able to get .4 -.2 which is the answer listed in 13

Using that technique I was not able to get the other two answers for x which are

-4/5 v + 7/5v + 2v

I don't see how I could even get values for three different v's when I'm only using two equations.