For what values of a are these vectors linearly dependent?

[1question]

Homework Statement

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?

Homework Equations

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

where s₁and s₂ are some coefficients.

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?

 

[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.

 

[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

 

[1question]

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.

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

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

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?

 

[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

 

[clamtrox]

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

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

 

[1question]

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

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

Thank you.