# Simple Matrices Help

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.

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. $$\begin{pmatrix} 3 & 4 & -3 & \vert & 0\\1 & 3 & 4 & \vert & 0\\ -2 & 6 & 7 & \vert & 0 \end{pmatrix} \rightarrow \begin{pmatrix} 3 & 4 & -3 & \vert & 0\\0 & 6 & 7.5 & \vert & 0\\ -2 & 6 & 7 & \vert & 0 \end{pmatrix}$$ 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.