# For what values of a are these vectors linearly dependent?

1. Apr 10, 2013

### 1question

1. The problem statement, all variables and given/known data
For what value(s) of a are the following vectors v1=[1,2,-1], v2=[0,1,3], and v3=[a,4,5] linearly dependent?

2. Relevant equations

Since linear dependence means that any one of the vectors can be expressed as a linear combination of the other vectors:
sv1v1+s2v2=v3

where s1and s2 are some coefficients.

3. The attempt at a solution

Looking at the vectors, I tried v1+2v2=v3

Solving, I get a=1

I'm fairly certain I did the question correctly. The only thing that really bothers me is the plural of "value" in the question - is there ever a case where there can be more than one solution? How can I tell and solve those cases?

2. Apr 10, 2013

### clamtrox

The vectors [a,4,5] define a plane. So do [1,2,-1] and [0,1,3]. The linearly dependent cases are the intersections of these two planes.

3. Apr 10, 2013

### ehild

It is the only solution, but instead of trial and error, you can find the solution(s) by solving the equation valid for the components of the vectors.

ehild

4. Apr 10, 2013

### 1question

Doesn't that imply that any one of the vectors can be expressed as a linear combination of the other vectors? Basically equate the two planes and solve (which I did)?

Could you elaborate, please? What equation?

I did Gaussian Elimination and wound up with a free variable. How am I supposed to solve for a specific case other than to plug in random values (which is essentially what I did in the first place with far less work)? How does it help me determine all of the possible solutions in other cases?

5. Apr 10, 2013

### ehild

s1v1+s2v2=v3. That means three equations for the x,y,z components

s1=a
2s1+s2=4
-s1+3s2=5.

Three equations with three unknowns. No free parameters.

ehild

6. Apr 10, 2013

### clamtrox

Yes. Two planes intersect either on only one line, or everywhere.

7. Apr 10, 2013

### 1question

Ah. Just solve the matrix/system of equations for a.

Thank you.