Finding eigenvalues

To find eigenvalues of a matrix, subtract the unknown value λ (lambda, a letter of the greek alphabet) from all values on the main diagonal.

Then multiply the values on the main diagonal, the secondary diagonal, and then subtract them. Then put that equation to zero. Then solve for λ.

This value or values is/are the eigenvalue(s).
(Click on the images on the right to view pictorial representation in full size).

DEFINITIONS (in basic terms)
Main Diagonal - the diagonal that runs from the top-left to the bottom-right of the matrix.
Secondary Diagonal - the diagonal that runs from the top-right to the bottom-left of the matrix.

Finding the determinant

To find the determinant of a matrix, there are 2 different methods that can be used, depending on the size of the matrix.

In a 2x2 matrix (two columns, two rows) the determinent is found by:

1. multiplying the two terms on the main diagonal
2. multiplying the two terms on the other diagonal
3. subtracting the result in (2) from (1), ie. (1) - (2) = determinant
   

In other sized matricies, the determinent is found in a few more steps:

1. write down the term in the top-left corner. Now, draw a line through all terms below it and in line with it. This leaves a smaller matrix. Write this matrix in brackets, next to the term you wrote earlier. You will multiply this later.
2. Repeat step 1 for all the terms across the top row. Note you must follow the 'checkerboard of signs' when doing this.
3. Now with all the terms you've got written down, find the determinents of the smaller matricies and multiply them by the term before it.
4. Once you have done this, you have the determinant (click on the image to enlarge and see a graphic representation of this)
   

DEFINITIONS (in basic terms)
Checkerboard of signs - when you need to use this, it means that every second term must be multiplied by (-1) in the first row.
iff - used commonly in mathematics. Means if and only if.
Main Diagonal - the diagonal that runs from the top-left to the bottom-right of the matrix

EXAMPLE

An example can be found in the image on the right. If you cannot read it, click on it to enlarge.


IS A MATRIX SINGULAR?
A matrix is singular if and only if (iff) the determinant of that matrix is zero.

Finding the inverse of a matrix

To find the inverse of a matrix:

1. Put your matrix next to an identity matrix of the same size, on the right hand side, drawing a vertical line to clearly distinguish between the two
   
2. Perform row operations until the identity matrix is on the left hand side
3. The matrix now on the right is the inverse of your original matrix

'LAYMANS TERMS' DEFINITIONS
Identity Matrix - a matrix that contains all 1s on the main diagonal, and zeros elsewhere
Inverse Matrix - a matrix that when multiplied by another matrix results in the identity matrix
Row Operation - performing calculations row by row on a whole matrix. This could be multiplying, adding, subtracting or dividing a row by another row or a number.


EXAMPLE
(See image on right for example -
Click on image to make it larger if needed)

Homogenous linear differential equations

To find the 'general solution' of equations that look like these:

y'' + 2y' + 3y = 0
2y'' - y' - 6y = 0
y'' - 3y' - y = 0

Simply follow these steps:

1) Rewrite the equation in terms of 'm', and for each (') count as a power (ie. ' -> x¹)

2) Now find the roots (or find a value for m). You can do this by either:
a) splitting the middle term then factorising, or
b) use the quadratic equation


3) Now you must 'sub' those roots into one of the three 'general solutions.' Which equation you must use is dependant on what your roots (m) are.

If roots are REAL and DISTINCT: use y = c1 e^mt + c2 e^mt
If roots are REAL but THE SAME: use y = c1 e^mt + c2 (t) e^mt
If roots have a REAL part and an IMAGINARY part: use y = e^rt(c1cosqt+c2sinqt)

Then you have the general solution.

EXAMPLE

Find the general solution of y'' + 2y' - 3y = 0

1) m² + 2m - 3 = 0

2) You can use either method (a) or (b) to find the roots. Both are shown:

a) a=1, b=2, c=-3
The two numbers 3 and -1 will add to b and multiply to c
m² - m + 3m - 3 = 0
m(m-1)+3(m-1) = 0
(m+3)(m-1) = 0
m=-3, 1

b) Using the quadratic equation, the outcome is m=-3 or m=1

3) Roots (m) are REAL and DISTINCT.
So... use y=c1 e^mt + c2 e^mt (see graphic above)

Sub in m=-3 and 1: y = c1 e^-3t + c2 e<sup>t

That is the general solution!</sup>