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For what value of $h$ set $(\vec v_1 \ \vec v_2 \ \vec v_3)$ is linearly dependent? $$\vec v_1=\left[ \begin{array}{c} 1 \\ -3 \\ 2 \end{array} \right];\ \vec v_2=\left[ \begin{array}{c} -3 \\ 9 \\ -6 \end{array} \right] ;\ \vec v_3=\left[ \begin{array}{c} 5 \\ -7 \\ h \end{array} \right]$$

Attempt: After row reducing the augmented matrix of $A\vec x=\vec 0$ where $A=(\vec v_1 \ \vec v_2 \ \vec v_3)$:

$$\begin{bmatrix} 1 & -3 & 5 & 0 \\ -3 & 9 & -7 & 0 \\ 2 & -6 & h & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & -3 & 5 & 0 \\ 0 & 0 & 8 & 0 \\ 0 & 0 & h-10 & 0 \end{bmatrix} $$

I am not sure whether the set is linearly dependent when $h=10$ or for any $h$. Help please.

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The set is always linearly dependent since $v_2 = -3v_1.$ –  user2468 Jul 19 '12 at 2:27
    
@J.D. so it is enough for the set of three vectors to have two vectors that are collinear to be a linearly dependent set, right? –  Koba Jul 19 '12 at 3:39
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Indeed. A quick geometric reminder for yourself: the basis in $\Bbb{R}^3.$ If you pick two vectors collinear in the direction of the $x$-axis & a vector in the $z$ direction, would you be able to describe every vector in $\Bbb{R}^3$? Of course not. –  user2468 Jul 19 '12 at 3:45
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1 Answer 1

up vote 4 down vote accepted

That reduced matrix shows you that the set of vectors is linearly dependent for every value of $h$. If $h\ne 10$, the system has no solution, and if $h=10$, it has infinitely many, so there is no value of $h$ that gives it exactly one solution.

Indeed, you can see this directly from the vectors themselves: $v_2=-3v_1$.

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I think you may have confused the data, @Brian: if $\,h=10\,$ then the third row becomes all zero and, thus, the original set of three vectors is linearly dependent, as asked. True, if $\,h\neq 10\,$ then the homogeneous system is inconsistent, but we don't really care about that as nothing was asked about solutions of linear systems, homogeneous or non-homog. –  DonAntonio Jul 19 '12 at 2:17
    
In fact, I think the OP confused himself by writing down an augmented matrix as if he wanted to solve some linear system, whereas a $\,3\times 3\,$ matrix with the vectors' components is enough to find out whether they're l.i. or not. And then yes, as you wrote: for any value of $\,h\,$ the three vectors are l.d. This is also easy to check calculating the easy determinant of that square matrix, which is zero no matter what $\,h\,$ is. –  DonAntonio Jul 19 '12 at 2:21
    
@DonAntonio: I suspect that you’re right about the confusion, but you missed the point of my answer. If the three vectors were linearly independent for some $h$, then for that $h$ the homogeneous system would have only the trivial solution. But there is no value of $h$ for which this is the case, so for every $h$ the vectors must be linearly dependent. –  Brian M. Scott Jul 19 '12 at 2:24
    
I think I got it, @Brian. I just meant the phrase "...so there is no value of h that gives it exactly one solution." seemed to imply we were looking for solutions of some (homogeneous) system oflinear eq's and this seemed to me going astray from the question's point, but I see you followed the OP's work he himself showed. +1, anyway. –  DonAntonio Jul 19 '12 at 2:52
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