What Values of x Make These Vectors Linearly Dependent?

[mkir]

Homework Statement

For what values of x are the vectors,

[1]
[x]
[2x]

[1]
[-1]
[-2]

[2]
[1]
[x]

linearly dependent?

Homework Equations

The Attempt at a Solution

I made a matrix,

[1 ; 1 ; 2 ; 0]
[x ;-1 ; 1 ; 0]
[2x;-2 ; x ; 0]

but I'm having trouble figuring out how to get rid of the x's in the first row, if I get rid of one of the x's then I'm left with nothing to get rid of the other.

 

[gabbagabbahey]

Why not just compute the following determinant:
\begin{vmatrix} 1 & 1 & 2 \\ x & -1 & 1 \\ 2x & -2 & x \end{vmatrix}
The determinant will be zero iff the vectors are linearly dependent.

 

[mkir]

Well, I haven't learned about determinants in class yet so I don't know how to do that. What we learned in class was to find pivots and if there are 1 or more free variables then it is linearly dependent.

 

[mkir]

Ok, trying out the determinant idea, I got x = 2 and x = -1. Did I do it right?

 

[Defennder]

[1 ; 1 ; 2 ; 0]
[x ;-1 ; 1 ; 0]
[2x;-2 ; x ; 0]

but I'm having trouble figuring out how to get rid of the x's in the first row, if I get rid of one of the x's then I'm left with nothing to get rid of the other.

There isn't any x in the first row. So what you have to do is to use row-reduction to reduce it to reduced row echelon form. Along the way, note all those cases whereby you multiply each row with a factor involving x which may not be defined for certain values of x. For example, if you multiplied row 1 by 1/x, you should note x=0 is undefined.

So collect all those special cases and perform from the start assuming x is given those forbidden values. So as you can see, this makes it rather tedious.

The other way as you have done it yields -2 and -1. That's correct.

 

[HallsofIvy]

How about just using the definition of "linearly dependent"?

A set of vectors, {u, v, w} is "linearly dependent" if and only if there exist numbers, a, b, c, not all 0, such that au+ bv+ cw= 0.

Here, u= <1, x, 2x>, v= <1, -1, -2> and w= <2, 1, x> so our equation is
a<1, x, 2x>+ b<1, -1, -2>+ c<2, 1, x>= 0 or
<a+ b+ 2c, ax-b+ c, 2ax-2b+ cx>= <0, 0, 0> which gives the three equations
a+ b+ 2c= 0, ax- b+ c= 0, 2ax- 2b+ cx= 0. "x" is a parameter so this is a set of three equations in 3 variables a, b, and c. In general, we can solve such a set for single values of a, b, and c and, since a=b=c= 0 obviously satisfy the equations, that would be the "usual" solution. So we are really asking "for what values of x" can we NOT find a single solution. To answer that, TRY to solve the equations.

Obviously, adding the first two equations eliminates b: (1+x)a+ 3c= 0. Almost as obviously, adding twice the first equation to the third also eliminates b: 2(1+x)a+ (4+ x)c. Multiply the first of those equations by 2 and subtract from the other:(4+ x)c- 6c= (-2+ x)c= 0. If we just divide by -2+ x, we get c= 0 and then we have (1+x)a= 0. Dividing that by -1 gives a= 0. Of course, if a and b are both 0, any of the first three equations gives b= 0. That is the condition that the three vectors be independent. For what values of x can we NOT solve for a or c so the three vectors are dependent?