9: 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

The number ad-bc, which appears in this formula is pretty important, since it has several applications:

• It tells us whether the matrix has an inverse or not
• It tells us about the size change of the transformation
• It can tell us whether a linear system can be solved or not.
• It's useful for finding the eigen values of a transformation.

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(A^T)=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.

Assignment

This is to be handed in on Thursday 12th June.

• Section 4.1: 8,12,39
• Section 4.2: 2,22,24,36
• Section 4.3: 20 (What matrix is the answer a determinant of?)
• Section 5.1: 4, 16, 18
• Section 5.2: 2, 4
• Section 5.3: 10, 12, 22
• Section 5.4: 2, 4, 8, 10
• Section 5.5: 2, 6
• Section 5.6: 20
• Section 5.7: 2

Quiz

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

• If I understand a lecture, the probability I'll understand the next is 2/3
• If I don't understand a lecture, the probability I'll understand the next is 1/4.

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.

Solution

Let

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 u_n and w_n, and u_n+1 and w_n+1:

un+1	=	2/3un	+	1/4wn
wn+1	=	1/3un	+	3/4wn

Next, we want to fix some matrix and vector notation.

Let x_n=(u_n,w_n)

Let

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)

So

1/7v₁=1/7(3,4)=(3/7,4/7)

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)

Please send comments, questions, suggestions, corrections, ideas, to me.