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Show that the vectors $(-1,1,2)$, $(1,1,0)$ and $(3,-3,6)$ are linearly dependent.

I keep trying to solve it but get values for the constant coefficients all zeros which shows they are independent. I get 3 equations: \begin{align} -x + y + 3z &= 0 \\ x + y - 3z &= 0 \\ 2x +0y + 6z &= 0 \end{align} When these equations are solved simultaneously they yield $x = 0; y= 0; z = 0$. Which shows they are independent since the constants are all zeros.

I even tried solving the system of linear equations using row echelon matrix method but still got the same results. None of the vectors is a scalar multiple of the other as that would obviously make them dependent. Tried googling this question exactly as it is but didn't get any useful results to solve this riddle. Please help as it is an exam question.

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    $\begingroup$ The answer is simple: in fact, they are not linearly dependent! $\endgroup$ – Daniel Robert-Nicoud Jun 2 '17 at 23:41
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    $\begingroup$ Maybe there's a misprint, or a typo in your transcription? Any of changing the first vector to $(-1,1,-2)$, changing the second vector to $(1, -1, 0)$ or $(-1, 1, 0)$, or changing the third vector to $(3,-3,-6)$ would make them linearly dependent. $\endgroup$ – Daniel Schepler Jun 2 '17 at 23:41
  • $\begingroup$ They are linearly independent. They would linearly dependent if for example the last coordinat of the last vector will change sign $\endgroup$ – Minz Jun 2 '17 at 23:43
  • $\begingroup$ Thanks. I thought as much. Perhaps my lecturer wants me to think out of the box and say they are linearly independent even when we were asked to show they were dependent. This exam question keeps repeating itself in many years. I think it is a trick. Trust me, it is not a typo as I have checked over and over and the same exact question keeps repeating itself in subsequent years with little changes in some but the results always show they are independent though we were asked to show they were dependent. Thanks Minz $\endgroup$ – numyeo Jun 3 '17 at 0:23
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  1. We want to find the values of the constants $a,b,c$ such that $$a\begin{pmatrix}-1\\1\\ 2\end{pmatrix}+b\begin{pmatrix}1\\1\\ 0\end{pmatrix}+c\begin{pmatrix}3\\-3\\ 6\end{pmatrix} = 0$$ This gives the system $$-a+b+3c=0 $$ $$a+b-3c=0$$$$2a+0b+6c=0$$ Now if solve this using your favourite method, i.e. standard elimination, row reduction, you find that $a=b=c=0$. So they are in fact linearly independent! Why? Either you copied out the question wrong or the question has an error!

  2. However, as a side note, I couldn't resist to make this post and show that this can be done by evaluating the determinant of a $3\times 3$ matrix.

$$\begin{vmatrix} -1 & 1 & 3 \\ 1 & 1 & -3 \\ 2 & 0 & 6 \\ \end{vmatrix} = 2 \begin{vmatrix} 1 & 3 \\ 1 & -3 \\ \end{vmatrix}+6\begin{vmatrix} -1 & 1 \\ 1 & 1 \\ \end{vmatrix} = -12-12=-24 \neq 0$$ so the vectors are linearly independent. (I have used the result that the determinant of a matrix is 0 iff the rows/columns are linearly independent.) But remember, this method only works if you have an $n \times n$ system. I usually find it easier in exams because the vectors tend to have 0s somewhere, which makes expanding quite easy.

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