The Determinant of a Matrix

I may end up putting various bits and pieces from lectures 4 and 5 into this class. The determinant is a useful quantity associated to a matrix. We will look at why it is important, and it's geometrical properties.
We saw that the inverse of a matrix
/ a b \
\ c d /
1/(ad-bc) / d -b \
\ -c a /

The number ad-bc, which appears in this formula is pretty important, since it has several applications:
Anything with several good uses gets given an name. "ad-bc" is called the Determinant of a matrix. It can also be defined for matrices of larger size, and has just the same uses. We can define it to be the factor of enlargement of the area or volume, or higher dimensional equivalent of volume. This material is covered in chapter 4 of the text book.
We will look at the algebra, but probably concentrate on the geometric meaning of the determinant in the class.

Notes on text book chapters

4.1: Introduction to determinants
This section shows how the determinant comes up when you row reduce a matrix - it's what you get on the diagonals, so that's how it's related to finding if the system is consistant. A "recursive" formula is given. We will look at this in class.
4.2: Properties of Determinants:
This shows how row operations change the determinant:
  • If you add a multiple of a row to another row, the determinant is not changed - this is a shearing effect.
  • If you multiply a row by a number k, this multiplies the determinant by k - this is a stretching effect.
  • If you swap two rows, this multiplies the determinant by -1. This is a reflection effect.
Note that any such row opperation can be given by a matrix multiplication. We also have similar results for column operations.
Also this section gives some important properties:
  • A is invertible if and only if det(A) is not zero
  • det(AT)=det(A)
  • det(AB)=det(A)det(B)
The book gives these from a more algebraic point of view; I will present a more geometric point of view.
4.3: Cramers Rule and linear transformations.
This is a formula for how to find solutions of linear systems by using determinant. This section also gives the geometric interpretation of the determiant (pages 180-184), and this is the part of this chapter which I shall be emphasising most.


This is to be handed in on Thursday 12th June. (Note this is not the last homework)


There will be a quiz on Thursday with a question similar to the following:

If I understand the first lecture, what is the probability I'll understand the 12th lecture?
Hint: Think of everyone as having the same chances, and then the class falls into two "populations", the population that understands, and the population that doesn't, so this is like the population movement problem.
un=probability I'll understand the nth lecture
wn=probability I won't understand the nth lecture
Alterntatively, we can think of the two states "understanding" and "not understanding" as two populations, and let
un=proportion of people in class who understand the nth lecture
wn=proportion of people in class who don't understand the nth lecture

However you think about it, you'll get the same answer.
In terms of popluations, each lecture, 1/3 of the people who understand move to not understanding, and 1/4 of the people who do not understand move to understanding.
Now we can write the linear relations between un and wn, and un+1 and wn+1:
un+1 = 2/3un + 1/4wn
wn+1 = 1/3un + 3/4wn

Next, we want to fix some matrix and vector notation.
Let xn=(un,wn)
A = / 2/3 1/4 \
\ 1/3 3/4 /

So we have

We start with x1=(1,0).
To find what happens in the long run, we want to find the eigen vectors of A, and express x1 in terms of them.
So, first we find the eigen values, then the eigen vectors, then we see how x1 is in the span of them, and finally we put it all together to find x12
finding eigen values
We have to find L with a solution to Ax=Lx, ie (A-LI)x=0, which means finding the characteristic equation, det(A-LI).
det(A-LI)= | 2/3-L 1/4 | =(2/3-L)(3/4-L)-1/3*1/4
| 1/3 3/4-L |

So the characteristic equation is 6/12-3/4L-2/3L+L2-1/12=5/12-17/12L+L2
The roots are 1 and 5/12.
Finding the eigen vectors
L=1: we want a solution of Ax=x, ie, a solution of (A-1I)x=0 We write an augemented matrix and reduce to echelon form:
/ 2/3-1 1/4 | 0 \
\ 1/3 3/4-1 | 0 /

We end up with top row (-1/3 1/4 | 0), and zero bottom row, so -1/3x+1/4y=0, so x=3/4y, and the eigen vector is any multiple of (3/4,1), eg (3,4).
L=5/12: we want a solution of Ax=5/12x, ie, a solution of (A-5/12I)x=0 We write an augemented matrix and reduce to echelon form:
/ 2/3-5/12 1/4 | 0 \
\ 1/3 3/4-5/12 | 0 /
We end up with top row (3/12 1/4 | 0), and zero bottom row, so x+y=0, so x=-y, and the eigen vector is any multiple of (1,-1).
express x1 in terms of the eigen vectors:
We want to find a and b so that
/ 1 \ = a / 3 \ +b / 1 \
\ 0 / \ 4 / \ -1 /

We could solve this by writing an augmented matrix, and reduce to echelon form, etc, but since this is quite a simple example, we can solve it by inspection. Since we need a zero second component, b=4a, then we see that we can take a=1/7 and b=4/7, and we have:
/ 1 \ = 1/7 / 3 \ +4/7 / 1 \
\ 0 / \ 4 / \ -1 /

Solution at lecture 12:
let v1 be the eigen vector with eigen value 1, (3,4)
let v2 be the eigen vector with eigen value 5/12, (1,-1)
x12=A11x1= A11(1/7v1+4/7v2) =A11(1/7v1) +A11(4/7v2) =1/7v1+4/7(5/12)11v2

So the chance I'll understand the 12th lecture is approximately 3/7.
(Or in "population" terms, I might expect 3/7 of the class to understand all the last lecture)