# Row reducing a matrix

So my teacher today gave some homework. I need to know how to row reduce the matrix

$\left[ \begin{array}{ccc} 6 & 5 & 6 \\ 1 & 4 & 3 \\ 0 & 2 & 1 \\ \end{array} \right].$

I tried so many times but could not get it. So how do I do it? I don't even know if I have started correctly. Please help me as I am really confused.

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Presumably your teacher also explained something about how to do this, no? What have you tried? Where do you get stuck? – Mariano Suárez-Alvarez Jan 17 '12 at 5:28
@KyleSutherland What is it that you don't know how to do exactly? Besides is the matrix that you want to row reduce this $\left[\begin{array}{ccc} 6 & 5 & 6 \\ 1 & 4 & 3 \\ 0 & 2 & 1 \end{array}\right]$? – user38268 Jan 17 '12 at 5:28
Multiply row 2 by 6 and subtract row 1 from it, that may help? – user23086 Jan 17 '12 at 5:50

We should first get out some definitions for what it means to have a matrix in row-reduced form, and then we can figure out how to get there!

Definition:

An $m \times n$ matrix $R$ is called row-reduced if: (a) the first non-zero entry in each non-zero row of $R$ is equal to 1; (b) each column of $R$ which contains the leading non-zero entry of some row has all its other entries 0.

So, we know that we have tools for exchanging rows, adding equations, and multiplying by a constant. Since, we are doing things in the same manner as "to both sides of the equation" from basic algebra, we are not fundamentally changing the matrix in any way. Formally,

Definition:

The three elementary row operations on an $m \times n$ matrix $A$ over the field $F$ are: (a) multiplication of one row of $A$ by a non-zero scalar $c$; (b) replacement of the $r^{th}$ row of $A$ by row $r$ plus $c$ times row $s$, for any scalar $c$ and $r \neq s$; (c) interchange of two rows of $A$.

While it may be confusing to see this definition, when you see an explicit example in practice, it should be clear. Note that a field is just some set for which the are satisfied. So, for your matrix we can give the following chain of equivalences (I will denote which rule is being used as an (a), (b), or (c), above my arrow.

$$\left[ \begin{array}{ccc} 6 & 5 & 6 \\ 1 & 4 & 3 \\ 0 & 2 & 1 \\ \end{array} \right] \rightarrow^{c} \left[ \begin{array}{ccc} 1 & 4 & 3 \\ 6 & 5 & 6 \\ 0 & 2 & 1 \\ \end{array} \right] \rightarrow^{b} \left[ \begin{array}{ccc} 1 & 4 & 3 \\ 0 & -19 & -12 \\ 0 & 2 & 1 \\ \end{array} \right] \rightarrow^{c} \left[ \begin{array}{ccc} 1 & 4 & 3 \\ 0 & 2 & 1 \\ 0 & -19 & -12 \\ \end{array} \right] \rightarrow^{a} \left[ \begin{array}{ccc} 1 & 4 & 3 \\ 0 & 1 & \frac{1}{2} \\ 0 & -19 & -12 \\ \end{array} \right]$$

At this point I will note that we have satisfied for column #$1$, part $(b)$ of the definition of row-reduced. We will now proceed in reducing the matrix, by removing the $-19$ and $4$ from column #$2$ as to satisfy part $(b)$ of the definition of row-reduced. Continuing on, $$\left[ \begin{array}{ccc} 1 & 4 & 3 \\ 0 & 1 & \frac{1}{2} \\ 0 & -19 & -12 \\ \end{array} \right] \rightarrow^{a} \left[ \begin{array}{ccc} 1 & 4 & 3 \\ 0 & 1 & \frac{1}{2} \\ 0 & 1 & \frac{12}{19} \\ \end{array} \right] \rightarrow^{b} \left[ \begin{array}{ccc} 1 & 4 & 3 \\ 0 & 1 & \frac{1}{2} \\ 0 & 0 & \frac{5}{38} \\ \end{array} \right] \rightarrow^{b} \left[ \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & \frac{1}{2} \\ 0 & 0 & \frac{5}{38} \\ \end{array} \right] \rightarrow^{a} \left[ \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & \frac{1}{2} \\ 0 & 0 & 0 \\ \end{array} \right]$$ At this point we are now done, since part $(a)$ is satisfied, certainly the first non-zero entry in each non-zero row (namely row #$1$ and row #$2$) is 1, and part $(b)$ is satisfied, as for column #$1$ and #$2$ we have that there are no other non-zero entries and column #$3$ does not have a leading non-zero entry.

Note further, that this matrix also happens to be in row-reduced echelon form, which is a stronger condition than just row-reduced and I'm sure you will see that in your course shortly.

Hopefully that helps, if I got some of the arithmetic wrong don't worry about it, the concepts have all been illustrated, and its late for me, so forgive. :)

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I know it's not really important, but should the final matrix have a $1$ in the $3,3$ position, and thus be further reduced to $I_3$? My apologies if I've overlooked something. – yunone Jan 17 '12 at 8:26
Thanks a bunch lot clearer since before. – user23086 Jan 17 '12 at 9:07
@yunone: Well, it is row-reduced as is, if you want to fiddle around with it, it won't achieve anything more solutions-wise. – Samuel Reid Jan 17 '12 at 15:06
@yunone : Yes, you are right. – supremum Dec 18 '14 at 6:14
@Samuel Reid : with which non-zero scalar you multiplied the last row to make it zero?!(I think, yunone was pointing to you this thing) – supremum Dec 18 '14 at 6:17