Hair stuff, Matrix Multiplication, Complex Numbers and Modal Method

HAIR STUFF! 

This is basically my obligatory non mathematical post. I did this using an inverse french braid (meaning that you cross strands of hair underneath rather than over top) which gives a really distinct platted look which follows the curvature of the head. 

The back bit I did using a technique which is a little hard to explain without a demonstration. But its super duper easy, holds really well and doesn't require as many bobby pins as you think. Intrigued? Ask me next time I see you and I'll give you a demonstration! 

SO! ONTO THE MATH!

Matrix Multiplication (difficulty 2 - Give it a try)

So matrix multiplication is relevant to a lot of things, including today's lesson! Previously we have done a 2x2 matrix multiplied by a 2x1 matrix. Today we will be doing something a little more complicated and multiplying a 2x2 matrix by a 2x2 matrix. Below I am going to show you how to do it by using letters. Don't panic when you see letters being used to explain math! Letters simply represent any number and they are easier to track than numbers. Anyway, here is our matrix multiplication... 

So we can see to make the top left element on the final matrix we used the top row of the first matrix and the left column of the second matrix, to get the top right element of the final matrix we used the top row of the first matrix and the right column of the second matrix etc etc. 

Give it a try on these two matricies:

[2,1; 2,2] and [3,1; 1,1]; did you get [7,3; 8,4]? if you didn't, let me know and I will see if I can sort out your problem. If you did, GOOD JOB YOU, YOU ARE AWESOME!

Modal Method (3)

So as mentioned before the modal method can only be used to find e^At...

• If A is diagonalisable 
• I've talked about this in a previous blog but in a nut shell you can check for diagonalablity (I feel like I'm just making these words up) finding the eigenvectors, putting them into a matrix (called P) and checking that that matrix has two linearly independent columns.

If you go through and find that the matrix is not diagonalisable then too bad how sad you have to use the expansion method.

The rest of this method involves using this formula... e^(At) = Pe^(LAMBDA.t).P^(-1) where

• P is our friendly old matrix made out of our eigenvectors (which we have to find to prove we can use this method anyway)
• LAMBDA (would usually be written as capital lambda and looks like this) is a matrix with our eigenvalues (also already found) on the diagonals
• t is just a scalar representing time

Basically you put all your values into the formula and bob's your uncle. 

Inverting a Matrix (2)

There is one part that I want you to take notice of though... NOTICE how I have coloured the P^(-1) in red above? It means that we need to take the inverse of the matrix. Usually when we see ^(-1) on a scalar number e.g. 5^(-1) it simply means that we need to push the number down to the other side of the fraction line i.e. 5^(-1) == 1/5 but for a matrix EVERYTHING CHANGES. We now have to do an operation to it which might seem weird but it works on paper because algebra. If you don't believe me try converting a matrix into a scalar equation and try inverting that linearly. You'll get the same answer :)

What you need to know for a 2x2 matrix is that we swap the numbers on the diagonal that goes this way \ and we change the sign (e.g. from positive to negative) on the numbers on the diagonal this way /. And then you times the new matrix by one divided by the determinant of the original matrix (which I tell you how to do here in my calculation of eigenvalues). Let's see an example of that...

See if you can cover up my example and give it a go yourself!

Modal Method Example (4 just because there are so many steps!)

We're going to try to find e^(At) using the matrix A = [0,1; 1,0]. To solve this I have to introduce the idea of complex numbers...

COMPLEX NUMBERS!

These are also known as imaginary numbers because in real life they don't exist as a number!! 

Complex numbers are literally the square root of -1. Think about it, a square root sign is the opposite of a squared sign. The square of positive two and the square of negative two both work out to be the same number 2^2 = 4 & (-2)^2 = 4 as a negative times a negative is a positive. Therefore it is impossible to reverse this and take the square root of a negative number. Try it on a calculator! We represent these numbers as the letter "i"  or "j" which literally means the square root of -1. From this we can see...

• j^1 = j
• j^2 = -1
• j^3 = -j
• j^4 = 1 and so on and so forth.  

Although we call them imaginary, this does not go to say that they are useless. They are actually used in a lot of applications and can tell us a lot about a system if they arise. 

BACK TO OUR EXAMPLE...

Doesn't that just come out beautifully!

Euler's formula can be found here for anyone who is interested in seeing for themselves how it works :)