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I have been trying to establish if certain vectors are linearly dependent and have become confused (in many ways). when inputting the vectors into my augmented matrix should they be done as columns or as rows ?

eg if my vectors are [ 4 -1 2 ], [-4 10 2]

I am looking to solve a[ 4 2 2 ] + b[2 3 9] = [0,0,0]

but am unsure how the matrix should be created

$\begin{bmatrix} 4 & -1 & 2 \\ -4 & 10 & 2 \\ \end{bmatrix}$

vs

$\begin{bmatrix} 4 & -4\\ -1 & 10\\ 2 & -2 \end{bmatrix}$

using the last method I end up with the following in row reduced echelon form (including the 0's as answers on the right)

$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 &0 \end{bmatrix}$

so to me this says a = 0, b = 0 , but I don't get what the last row means ?

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  • $\begingroup$ If I told you that elementary row operations preserve the row space of a matrix and that elementary column operations preserve the column space, does that help? $\endgroup$
    – user137731
    May 25, 2015 at 1:06
  • $\begingroup$ Oh. Now that I've actually finished reading this -- You did it right. The last row of the augmented matrix doesn't tell you anything. Converting back to scalar equations it just says $0=0$ which is always true -- so it doesn't put any conditions on $a$ or $b$. $\endgroup$
    – user137731
    May 25, 2015 at 1:08
  • $\begingroup$ to be honest I'm not 100% sure what that means( that was regarding the initial comment). In terms of the answer to my problem re the row of all 0's , thanks , that helps make things clearer. I thought there were instances where it meant there were many solutions...or does that just apply to linear equations ? $\endgroup$ May 25, 2015 at 1:08
  • $\begingroup$ Well good, if you're $100$%, then we're done ... ;-) $\endgroup$
    – user137731
    May 25, 2015 at 1:11
  • $\begingroup$ was a typo..meant to say not 100% aure $\endgroup$ May 25, 2015 at 1:12

1 Answer 1

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If $ \: \: a \begin{pmatrix} 4 \\ 2 \\ 2 \end{pmatrix} $ + $ b \begin{pmatrix} 2 \\ 3 \\ 9 \end{pmatrix} $ = $ \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \: \:$ then $ \: \: \begin{pmatrix} 4a+2b \\ 2a+3b\\ 2a+9b \end{pmatrix} $ = $ \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$.

So we have three simultaneous equations,

$ 4a+2b=0 \quad (1)\\ 2a+3b=0 \quad (2)\\ 2a+9b=0 \quad (3)$

We can now form a coefficient matrix,

Let $ \: A = \begin{pmatrix} 4 & 2 \\ 2 & 3 \\ 2 & 9 \end{pmatrix} $

Then we put this into reduced row echelon form (RREF) using Gauss-Jordan elimination,

and we get $\; \; \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix} $.

This tells us that there is a unique solution to the simultaneous equation formed above and that is $ a=0 $ and $ b=0 $.

The reason that the bottom row is zero is that the last of the simultaneous equations can be derived from the other two. $ (3) = 4*(2) - \frac{5}{2}(1) $. So it doesn't actually give us any more information.

So if the only solution is $a=0=b$, the vectors $ \begin{pmatrix} 4 \\ 2 \\ 2 \end{pmatrix} $ and $ \begin{pmatrix} 2 \\ 3 \\ 9 \end{pmatrix} $ must be linearly independent.

However, it should be clear from the 'top' elements, 4 and 2, that $b=-2a$, but the the 'middle' elements, 2 and 3, need $b=-\frac{2}{3}a$. So the vectors must be independent.

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  • $\begingroup$ great thanks...I had thought all 0's in the bottom row referred to an instance where there were infinite solutions. $\endgroup$ May 25, 2015 at 1:25

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