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My lecturer has obviously missed something crucial because this isn't the only question I've been having trouble with.

The question: Show that {$u_1,u_2,u_3$} is linearly independent, where:

$u_1= \left[\begin{array}{cccc} 3\\ 1\\ -2\\ \end{array}\right] $ $u_2= \left[\begin{array}{cccc} 4\\ 3\\ 6\\ \end{array}\right] $ $u_3= \left[\begin{array}{cccc} -3\\ 4\\ 7\\ \end{array}\right] $

I understand the process so I'm assuming I'm doing something incorrect in regards to the transformations.

What I did: $c_1$$u_1$+$c_2$$u_2$+$c_3$$u_3$=$0$

$[A|0]= \left[\begin{array}{cccc} 3&4&-3&|&0\\ 1&3&4&|&0\\ -2&6&7&|&0\\ \end{array}\right] $

Row 2 + (1/2)*Row 3

Row 3 +2*Row 2

$[A|0]= \left[\begin{array}{cccc} 3&4&-3&|&0\\ 0&6&\frac{15}{2}&|&0\\ 0&12&15&|&0\\ \end{array}\right] $

Row 3 - 2*Row 2

$[A|0]= \left[\begin{array}{cccc} 3&4&-3&|&0\\ 0&6&\frac{15}{2}&|&0\\ 0&0&0&|&0\\ \end{array}\right] $

$\therefore r(A)=2<n=3$ so the system has non-trivial solutions and the set of vectors are linearly dependent - infinite solutions.

According to the worked solution the final transformed matrix (which I was also able to produce via transformations) is:

$[A|0]= \left[\begin{array}{cccc} 1&3&4&|&0\\ 0&1&3&|&0\\ 0&0&-21&|&0\\ \end{array}\right] $

$\therefore r(A)=n=3 $ which would imply that the system has only a trivial solution and the set of vectors must be linearly independent.

So my question is, why am I able to produce a row of zeroes? What am I doing wrong? Earlier when I was doing problems where I had to find the determinant, more often than not I was getting the answer wrong because nearly every single time I was able to produce a row of zeroes. Obviously there is something wrong with my method but nothing in my lecture notes says why and a few online searches have not provided me with a solution.

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You cannot perform type 3 ERO for 2 rows where the pivot row are the other one at the same time. The EROs should be performed one by one, so when you add $0.5$ times row 3 to row 2, row 2 has already changed. \begin{equation} \begin{pmatrix} 3 & 4 & -3 & \vert & 0\\1 & 3 & 4 & \vert & 0\\ -2 & 6 & 7 & \vert & 0 \end{pmatrix} \rightarrow \begin{pmatrix} 3 & 4 & -3 & \vert & 0\\0 & 6 & 7.5 & \vert & 0\\ -2 & 6 & 7 & \vert & 0 \end{pmatrix} \end{equation} Then you try to use row 2 as a pivot row and perform type 3 ERO to row 3, but the multiple would not be $2$ then. In fact $a_{21} = 0$, therefore you cannot use row 2 as a pivot row anymore. You should use row 1 instead.

Sometimes instructors would like to write 2 or more EROs at the same line, but this is legit because usually they (usually) use the same pivot row throughout these EROs. Therefore, performing them one by one is equivalent to performing them at once.

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  • $\begingroup$ Thankyou! I knew it would have to be something as simple as that. $\endgroup$ Jun 10, 2015 at 10:08

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