# For what value of h the set is linearly dependent?

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
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

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: $v_2=-3v_1$.
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