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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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    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.2012-07-19

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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 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.2012-07-19