I am emailing a link of this to everyone on the class list every week. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jpmccarthymaths@gmail.com and I will add you to the mailing list.

Five students still owe me E10 for the notes. If this is a problem for you send me a confidential email.

## Week 3

We started our work on linear algebra; specifically linear systems. We showed how to write a linear system in augmented matrix form and showed that the elementary row operations conserved the solutions. We loosed described the Gaussian elimination algorithm that would put the augmented matrix in reduced form. We said that solutions could have none, one or an infinite number of solutions and described how to tell from the reduced form. Depending on the scenario, the solutions are easily found via back-substitution.

## Week 4

Next week we will continue to work on Gaussian elimination. We will discuss Cramer‘s Rule; we will see why computers struggle with Gaussian elimination and learn ways to get around this.

## Tutorials

Deadly serious: ye need to be A-standard on the first two chapters… they are doable for everyone with a bit of work. The later stuff will be challenging and will need a lot of work to reach a good standard.

This week it is

13:30 Groups S2 and S3 (A-L)

14:30 Groups S1 and S3 (M-Z)

Yes the following table should be A-L and M-Z not A-L and K-Z!

### Example on P.33

I never did this in class as I wasn’t happy with how my notes took it up… but actually my notes were right. We won’t have one as complicated as this in an exam anyway so it doesn’t really matter that much.

The augmented matrix is in reduced (row-echelon) form. Find the solutions:

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

Solution: OK the first thing is that there is no row of the form $[0\,\,0\,\,0\,\,0\,\,0\,\,|\,\,k]$ with $k\neq 0$ so there are solutions. We count the number of non-zero rows (morryah independent equations) and there are three. There a five variables, we can call them $x_1,x_2,x_3,x_4,x_5$ and the gap between these ($v-r=5-3=2$) equals the number of parameters (‘variables’ that we will allow take any value), so that we have an infinite number of solutions. In general, it works nice if we pick the ‘last’ variables, i.e. $x_4$ and $x_5$ to be the parameters. We must be careful, however, that they are not already determined. For example, a row of the form $[0\,\,0\,\,0\,\,1\,\,0\,\,|\,\,-3]$ would be equivalent to $1x_4=x_4=-3$ so we couldn’t let $x_4$ be whatever it wants.

Therefore we can let $x_5=t$ and $x_4=s$; i.e. be whatever they want. $t$ and $s$ are parameters of the system and can take on any real number value at all and we might even write $t,\,s\in\mathbb{R}$  This is what I had wrong in my thinking. In fact the proper thing to do is first day that, indeed, $x_5$ is not determined so it can be a parameter $x_5=t$ with $t\in\mathbb{R}$. Now you say, given that $x_5=t$ can we let $x_4$ be a parameter or is it already determined? Well it is determined because the third row corresponds to

$x_4+2x_5=6\Rightarrow x_4=6-2t$,

in other words $x_4$ can’t be anything at all, it depends on $x_5=t$. Similarly the second row implies that $x_3$ is determined:

$x_3+4x_5=-5\Rightarrow x_3=-5-4t$.

Now however, we can say that $x_2$ is not determined; neither is $x_1$. Hence we let the last of these, $x_2=s$ the other, second parameter that we need to cover the gap. Now we have

$x_5=t$

$x_4=6-2t$

$x_3=-5-4t$

$x_2=s$

where $t$ and $s$ are any real numbers. All that remains is to find $x_1$. We can do this from the first row which corresponds to

$x_1+5x_2-2x_5=3\Rightarrow x_1=3-5s+2t$.

### Applications of Linear Algebra: Network Flows

Some applications to traffic flow and  struts/“truss”