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Economics Masters Refresher Course in Mathematics
• Lecturer: Jonathan Wadsworth
• Teaching Assistant: Tanya Wilson
• Aim: to refresh your maths and statistical skills, to the level needed for success in the Royal Holloway economics masters.
• Why do we need maths/statistics?
- to improve your understanding of Economics.
- As such everything in the course tailored to try to bring out the relevance of the techniques you will (re)learn help the analysis of Economic issues
Economics Masters Refresher Course in Mathematics
• How to do it?
• Lectures 10-1 and classes 2:30-3:30 each day.
• In the afternoons and evening you will be expected to:
– Read the text
– Do the daily problem set.
– Prepare to present your answers to the class the next day.
Vectors and matrices
Learning objectives. By the end of this lecture you should:
– Understand the concept of vectors and matrices
– Understand their relationship to economics
– Understand vector products and the basics of matrix algebra
Introduction
Often interested in analysing the economic relationship between several variables
Use of vectors and matrices can make the analysis of complex linear economic relationships simpler
Lecture 1. Vectors and matrices
1. Definitions
Vectors are a list of numbers or variables – where the order ultimately matters
– E.g. a list of prices, a list of marks in a course test.
– (Pricelabour , Pricecapital )
– (25, 45, 65, 85)
Since this is just a list can store the same information in different ways
Can enter the list either horizontally or vertically
(25, 45, 65, 85) or
85
65
45
25
Dimensions of a vector
If a set of n numbers is presented horizontally it is called a row vector
with dimensions 1 x n
(so the number of rows is always the 1st number in a dimension and the number of columns is always the 2nd number)
Eg
a = (25, 45, 65, 85) is a 1 x 4 row vector
If a set of n numbers is presented vertically it is called a column vector
with dimensions n x 1
Eg
(Note tend to use lower case letters (a b c etc) to name vectors )
Definitions
85
65
45
25
b
Special Types of vector
A null vector or zero vector is a vector consisting entirely of zeros
e.g. (0 0 0) is a 1 x 3 null row vector
A unit vector is a vector consisting entirely of ones denoted by the letter i
e.g. is a 4 x 1 unit column vector
Definitions
1
1
1
1
i
Adding vectors
General rule
If a = (a1, a2 ......an) and b = (b1, b2 ......bn)
then a+b = (a1+ b1, a2 + b2, ......an + bn)
e.g. (0 2 3 ) + ( 1 0 4) = (0+1, 2+0, 3+4) = (1 2 7)
(same rule for addition of column vectors )
Note that the result is a vector with the same dimension 1 x n
Note you can only add two vectors if they have the same dimensions
e.g. you cannot add (0 2 3) and (0 1) (1 x 3 & 1 x 2)
Note also that sometimes adding two vectors may be mathematically ok, but economic nonsense
Eg simply adding factor prices together does not give total input price
More definitions and some rules
Multiplying vectors by a number (a ‘scalar’)
Let x = (x1, …,xn) and a be a scalar, then ax = (ax1, ax2,…,axn)
e.g. a = 2, x = (1 2 3)
ax = (2 4 6)
More definitions and some rules
Multiplying vectors
In general you cannot multiply two row vectors or two column vectors together
Eg a = (1 2) b = (2 3) ?
But there are some special cases where you can
And you can often multiply a column vector by a row vector and vice versa.
We’ll meet them when we do matrices
More on multiplication
Definition: Vector product also known as the dot product or inner product
Let x = (x1, x2, …, xn), y = (y1,…,yn).
The vector product is written x.y and equals x1y1 + x2y2 + …xnyn.
Or
Note that vector products are only possible if the vectors have the same dimensions.
3. Vector products – a special case of vector multiplication
ni
iii yxyx
1
.
Example. Miki buys two apples and three pears from the Spar shop.
Apples cost £0.50 each; pears cost £0.40 each.
How much does she spend in total?
Answer: This can be seen as an example of a vector product
– Write the prices as a vector: p=(0.5, 0.4)
– Write the quantities as a vector q=(2, 3)
– Total Expenditure (=Price*Quantity) is the sum of expenditures on each good
– found by multiplying the first element of the first vector by the first element of the second vector and multiplying the second element of the first vector by the second element of the second vector & adding the results
– Spending = 2x0.5 + 3x0.4 = £2.20
3. Vector products – a special case of vector multiplication
ni
iii yxyx
1
.
1. What are the dimensions of the following?
2. Can you add the following (if you can, provide the answer)?
3. Find the dot product
Instant Quiz
22
32
10
.
0
0
0
.10421. iiiiii
21&431.144&
0
0
0
.01&14. iiiiii
013&431.
0
1
2
&
1
0
4
.01&14.
iiiiii
Idea: vectors can also be thought of as co-ordinates in a graph.
A n x 1 vector can be a point in n –dimensional space
E.g. x = (5 3) – which might be a consumption vector C= (x1,x2)
Geometry
5
3
x1
x2
A straight line is drawn out from the origin with definite length and definite direction is called a radius vector
Using this idea can give a geometric interpretation of scalar multiplication of a vector, vector addition or a “linear combination of vectors”
Eg If x = (5 3) then 2x = (10 6) and the resulting radius vector will overlap the original but will be twice as long
Geometry
5
3
x1
x2
Similarly multiplication by a negative scalar will extend a radius vector in the opposite quadrant
10
6
Also can think of a vector addition as generating a new radius vector between the 2 original vectors
Eg If u = (1 4) and v = (3 2) then u+v = (4, 6) and the resulting radius vector will look like this
Geometry
1
2
x1
x2
Note that this forms a parallelogram with the 2 vectors as two of its sides3
6
4
4
u+v
u
v
Given this can depict any linear vector sum (or difference) now called a linear combination geometrically
E.g. x = (5 3); y = (2 2) 2x+3y = (16,12)
Linear combinations depicted
5216
12
y
x3
2
Can also think of a geometric representation of vector inner (dot) product
Remember
A geometric representation of this is that the dot product measures how much the 2 vectors lie in the same direction
Special Cases
a) If x.y=constant then the radius vectors overlap
ni
iii yxyx
1
.
Eg x=(1 1) y =(2 2)
So x.y= (1*2 + 1*2) = 4
Can also think of a geometric representation of vector inner (dot) product
Remember
A geometric representation of this is that the dot product measures how much the 2 vectors lie in the same direction
Special Cases
If x.y= 0 then the radius vectors are perpendicular (orthogonal)
ni
iii yxyx
1
.
Eg x=(5 0) y =(0 4)
So x.y= (5*0 + 0*4) = 0
Y=(0,4)
X=(5 0) x ┴ y
Will return to these issues when deal with the idea of linear programming(optimising subject to an inequality rather than an equality constraint)
Linear dependence
A group of vectors are said to be linearly dependent if (and only if) one of them can be expressed as a linear combination of the other
If not the vectors are said to be linearly independent
Equally linear dependence means that there is a linear combination of them involving non-zero scalars that produces a null vector
E.g. are linearly dependent
Proof. Because x=2y or x – 2y = 0
But are linearly independent
Proof. Suppose x – ay = 0 for some a or other. In other words,
Then 2 – a = 0 and also 1 – 3a = 0. So a = 2 and a = 1/3 – a contradiction
1
1
1
;
2
2
2
yx
3
1;
1
2yx
0
0
31
2
31
2
0
0
3
1
1
2
a
a
a
aora
Linear dependence
Generalising, a group of m vectors are said to be linearly dependent if there is a linear combination of (m-1) of the vectors that yields the mth vector
Formally, the second definition:
Let x1 , x2 , …xm be a set of m nx1 vectors
If for some scalars a1, a2, …am-1
a1x1 + a2x2 + … am-1 xm-1 = xm
then the group of vectors are linearly dependent,
But also
a1x1 + a2x2 + … am-1 xm-1 – xm = 0 which is the first definition
Summary. To prove linear dependence find a linear combination that produces the null vector. If you try to find such a linear combination but instead find a contradiction then the vectors are linearly independent
Quiz II
A group of vectors are said to be linearly independent if there is no linear combination of them that produces the null vector
1. are linearly independent. Prove it.
2. are linearly independent. Prove it.
3. What about ?
;1
0;
1
2
yx
3
0;
2
2yx
0
1;
1
0;
1
2zyx
Rank
The rank of a group of vectors is the maximum number of them that are linearly independent
are linearly independent. So the rank of this group of vectors is 2
are linearly dependent, (2x=y) So the rank is 1
Any two vectors in this group are independent (x=y+2z) so the rank is 2
;1
0;
1
2
yx
4
2;
2
1yx
0
1;
1
0;
1
2zyx
Vectors: Summary
Definitions you should now learn:
– Vector, matrix, null vector, scalar, vector product, linear combination, linear dependence, linear independence, rank
• 4 skills you should be able to do:
– Add two nx1 or 1xn vectors– Multiply a vector by a scalar and do the inner (dot) product– Depict 2 dimensional vectors and their addition in a diagram– Find if a group of vectors are linearly dependent or not.
Matrices
1. Introduction
Recall that matrices are tables where the order of columns and rows matters
– E.g. marks in a course test for each question and each student
Andy Ahmad Anka
Qn. 1 44 74 65
Qn. 2 23 56 48
Qn. 3 8 24 44
1. Storing data
2. Input-output Tables
3. Transition matrices.
(U = unemployed, E = employed, prob. = probability)
Some common types of matrices in applied economics.
US GDP UK GDP PRC GDP
1999 100 80 11
2000 103 82 12
2001 103 84 13
Per kg iron Per kg coal
Kg iron 0.1 0.01
Kg coal 1 0
Hours labour 0.01 0.001
Prob. U in 2011
Prob E in 2011
U in 2010 0.3 0.7
E in 2010 0.1 0.9
Can also derive matrices from economic theory
Eg consider a simple demand and supply system for a good
Supply: Q= - 2 + 3P
Demand: Q=10-2P
Re-arranging
Supply: Q – 3P = - 2
Demand: Q +2P = 10
Or in short-hand Ax = b
where
21
31A
P
Qx
10
2b
In general any system of m equations with n variables (x1, x2, ...xn)
Can be written in matrix form Ax = b
where
mnmnmm
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
.....
:
.....
.....
2211
22222121
11212111
mnmm
n
n
aaa
aaa
aaa
A
..
::
..
..
21
22221
11211
nx
x
x
x:2
1
mb
b
b
b
:
:2
1
Dimensions of a matrix
are always defined by the number of rows followed by the number of columns
So
is a matrix with m rows and n columns.
Example
So A is a 2 x 5 matrix (2 rows , 5 columns)
B is 5x2 matrix
Definitions
106237
443314A
mxnA
144
03
63
231
74
B
Can think of vectors as special cases of matrices
Hence a row vector is a 1 x n matrix
B is a 1 by 3 row vector
Similarly a column vector is an n x 1 matrix
So C is a 3 x 1 column vector
( note the content is the same as B. This means that you can store the same information in different ways)
Definitions
101 B
1
0
1
C
Sometimes we wish to refer to individual elements in a matrix
E.g. the number in the third row, second column.
We use the notation aij (or bij etc.) to indicate the appropriate element
i refers to the row
j refers to the column
Example
So a13 = 3 and a21 = 7
If
What is b21 ?
Definitions
2524232221
1514131211
106237
443314
aaaaa
aaaaaA
144
03
63
231
74
B
The null matrix or zero matrix is a matrix consisting entirely of zeros
A square matrix is one where the number of rows equals the number of columns i.e. nxn
Eg.
For a square matrix, the main (or leading) diagonal is all the elements aij
where i = j
So in the above the main diagonal is (2 1)
Definitions
13
32
000
000
000
A
The identity matrix is a square matrix consisting of zeros except for the leading diagonal which consists of 1s:
We write I for the identity matrix
If we wish to identify its size (number of rows or columns) we write In
Definitions
10000
01000
00100
00010
00001
I
A diagonal matrix is a square matrix with aik = 0 whenever i ≠ k
And so consists of zeros everywhere except for the main diagonal which consists of non-zero numbers
(so the identity matrix is a special case of a diagonal matrix since it has just ones along the main diagonal)
Definitions
100000
01000
00400
00010
00002
A
Trace of a matrix
is the sum of the elements on the main (leading) diagonal of any square matrix
So tr(A) = 2+1+4-1-10 = -4
100000
01000
00400
00010
00002
A
1. What are the dimensions of A?
2. What is a21?
3. Is B a square matrix?
4. What is the largest element on the main diagonal of B?
5. What is the value of the largest element on the main diagonal of B?
Mini quiz
030
712
601
B
0037
3314A
The transpose of a matrix A is obtained by by turning rows into columns and vice versa
swapping aij for aji for all i and j
We write the transpose as A’ or AT ( A “prime”)
A symmetric matrix is one where A’ = A
A positive matrix is one where all of the elements are strictly positive
A non-negative matrix is one where all of the elements are either positive or zero
More Definitions
'13
32AA
076
310
021
030
712
601
AsoA
Useful Properties of Transposes
1. (A’)’ = A
- The transpose of a transpose is the original matrix
2. (A + B)’ = A’ + B’
- The transpose of a sum is the sum of the transposes
3. (AB)’ = B’A’
- The transpose of a product is the product of the transposes in reverse order
Eg. Given
Find (AB)’ and B’A’
1 2
3 4A
0 1
6 7B
A negative matrix is one where none of the elements are positive
A strictly negative matrix is one where all of the elements are strictly negative
C is strictly positive and symmetric; B is negative; A is neither positive nor negative.
More Definitions
13
32C
030
712
601
A
0
1
1
332
B
Adding matrices
- add each element from the corresponding place in the matrices.
i.e. if A and B are m x n matrices, then A+B is the m x n matrix where
cij= aij+bij for i = 1,..,m and j = 1,…,n.
You can only add two matrices if they have the same dimensions.
e.g. you cannot add A and B
Some rules
0037
3314A
030
712
601
B
040
724
601
;
030
712
601
;
010
012
002
BABA
Multiplying by a scalar
When you multiply by a scalar (e.g. 3, 23.1 or -2), then you multiply each element of the matrix by that scalar
Example 1: what is 4A if
Example 2: what is xB if
Matrix Multiplication
0037
3314A
001228
12124164A
030
712
601
B
030
72
60
x
xxx
xx
xB
Matrix MultiplicationIn general multiplication of 2 or more matrices has some special rules
1. The first rule is that the order of multiplication matters.
In general AxB (or AB) is not the same as BA
(so this is very different to multiplying numbers where the order doesn’t matter – e.g. 3x4 = 4x3 = 12 )
Asides:
• Addition, multiplication, matrix multiplication etc. are examples of operators
• An operator is said to be commutative if x operator y = y operator x
for any x and y (the order of multiplication does not matter)
• Addition is commutative: x+y = y+x; multiplication is commutative; subtraction is not commutative (2-1 ≠ 1-2)
• Matrix multiplication is commutative but matrix multiplication is not
2. The second rule is that you can only multiply two matrices if they are conformable
• Two matrices are conformable if the number of columns for the first matrix is the same as the number of rows for the second matrix
• If the matrices are not conformable they cannot be multiplied.
• Example 1: does AB exist?
– Answer: A is a 2x4 matrix. B is a 3x3.
– So A has 4 columns and B has 3 rows.
– Therefore AB does not exist. A and B are not conformable.
Multiplying two matrices
0037
3314A
030
712
601
B
Example 2: does AB exist?
–Answer: A is a 2x5 matrix. B is a 5x2. So A has 5 columns and B has 5 rows. A and B are conformable.
–Therefore AB exists.
–How to find it?
Multiplying two matrices
10627
03314A
11
03
63
31
04
B
Finding AB
– A and B are conformable. So C = AB exists and will be a 2 x 2 matrix
(no. of rows of A by no. Columns of B)
To calculate it:
i. To get the first element on the first row of C take the first row of A and multiply each element in turn against its corresponding element in the first column of B. Add the result.
– Example: c11 = (4x4) + (-1x-1) + (3x3) + (3x3) + (0x1) = 35
ii. To get the remaining elements in the first row: repeat this procedure with the first row of A multiplying each column of B in turn.
Multiplying two matrices
10627
03314A
11
03
63
31
04
B
2221
1211
cc
ccABC
So top left hand element is
And top right hand element is
Multiplying two matrices
..
.)1)(0()3)(3()3)(3()1)(1()4)(4(
11
03
63
31
04
10627
03314AB
..
)1)(0()0)(3()6)(3()3)(1()0)(4(35
11
03
63
31
04
10627
03314AB
Suppose A is an mxn matrix and and B is an nxr matrix with typical elements aik and bkj respectively
then AB =C where element cij is :
Note that the result is an mxr matrix
Multiplying two matrices - formally
nk
kkjikij bac
1
Example 2: calculate AB
– First we note that A is 1x4 and B is 4x2
– so C=AB exists and is a 1x2 matrix
– The first element c11 = (1)(1)+(0)(2)+(2)(1)+(0)(0)= 3
– The second element c12 = (1)(0)+(0)(0)+(2)(3)+(0)(1)= 6
– So C = (3 6)
Another example
0201A
10
31
02
01
B
Can you multiply the following matrices? If so, what is the dimension of the result?
1. BA
2. BC
3. AA’
4. A’A
5. CC
Quiz.
0201A
10
31
02
01
B
11
22C
Answers
Finding CC (sometimes written C2).
33
55
1.12.11.12.1
1.22.21.22.2
11
22
11
22CC
0201A
Recall:
1. A square matrix has the same number of rows and columns – it’s nxn
2. The identity matrix is a square matrix with 1s in the leading diagonal and 0s everywhere else.
E.g.
Multiplying square matrices by the identity matrix
10
01I
The usefulness of the identity matrix is similar to that of the number 1 in number algebra
Since IA = AI = A
- if multiply a matrix by the identity matrix the product is the original matrix
Eg
(leave it to you to show AI=A)
A 1 2
3 4
IA 1 0
0 1
1 2
3 4
(1*1) (0* 3) (1*2) (0* 4)
(0*1) (1* 3) (0*2) (1* 4)
1 2
3 4
A
1. If A is a square matrix then IA = AI = A
NB. This only applies to square matrices
A general result for square matrices
nnn
n
nnn
n
aa
aaa
IA
aa
aaa
AIf
1
11211
1
11211
10
001
A
aa
aa
aaaaaa
aaaaaa
nnn
n
nnnn
nnnn
1
111
12112111
12112111
100100
001001
This result can be useful sometimes to help solve matrix algebra
Since if AI = A
Then
AIB = (AI) B = A (BI) = AB
-The inclusion of the identity matrix does not affect the matrix product result (since like multiplying by “1”
(will see example of this in econometrics EC5040)
Also note that
-An identity matrix squared is equal itself
Any matrix with this property AA = A
Is said to be idempotent
2n nI I
Idea:
In standard multiplication every number has an inverse
(except maybe zero unless you count infinity)
The inverse of 3 is 1/3; the inverse of 27 is 1/27, the inverse of -1.1 = -1/1.1
Also a number times its inverse equals 1: x(1/x) = 1
and the inverse times the number equals 1: (1/x)x = 1
and the inverse of the inverse is the original number 1/(1/x) = x
Inverses for (square) matrices
The rules for the inverse of a matrix are similar (but not identical)
If A is an nxn matrix then the inverse of A, written A-1 , is an nxn matrix such that:
1. AA-1 = I
2. A-1A = I
Notes:
1. This means that A is the inverse of A-1
2. But...
3. A-1 may not always exist
Inverses for (square) matrices
1. Suppose and
Then
So given AA-1=I it must be that in this case B=A-1
An Inverse matrix example
11
03A
13/1
03/1B
10
01
)1)(1()0)(1()3/1)(1()3/1)(1(
)1)(0()0)(3()3/1)(0()3/1)(3(
13/1
03/1
11
03AB
In general need to introduce some more terminology before can invert a matrix
1. Only square matrices can be inverted.
Not all square matrices can be inverted however
2. A matrix that can be inverted is said to be nonsingular
(so squareness is a necessary but not sufficient condition to invert)
3. The sufficient condition is that the columns (or rows since it is square) be linearly dependent
- think of this as being separate equations so the equations must be independent (n equations and n unknowns) if a solution is to be found
Eg
So that the 1st row of A is twice that of the 2nd row and there is linear dependence
One equation is redundant (no extra information) and the system reduces to a single equation with 2 unknowns
So no unique solution for x1 and x2 exists
1 1
2 2
1 2 1
1 2 2
10 4
5 2
10 4
5 2
x dAx d
x d
x x d
x x d
Rank of a matrix
The idea of vector rank can be easily extended to a matrix
The rank of a matrix is the maximum number of linearly independent rows or columns
If the matrix is square the maximum number of independent rows must be the same as the maximum number of independent columns
If the matrix is not square then the rank is equal to the smaller of the maximum number of rows or columns, ρ<=min(rows, cols)
If a matrix of order n is also of rank n, the matrix is said to be of full rank
Important: Only full rank matrices can be inverted
Matrix ranks are closely linked to the concept of determinants
Let A be an nxn matrix then the determinant of A is a unique number (scalar), defined as:
(1)
Notes: In each term there are three components:
1. (-1)1+j
2. a1j
3. Det(A1j)
4. What does this mean?
Start with a 2 x 2 matrix
which gives a single number (scalar) as the answer – as do all determinants
Can you see how this relates to equation (1) ?
Determinants
1 11
1
det( ) ( 1) det( )j n
j jj
j
A A a A
11 12
21 22
11 22 12 21
a aA
a a
A a a a a
Eg
What is the determinant of
So matrices that are not full rank – have linear dependent rows/columns - have zero determinants (will come back to this) and are singular
Determinants
10 4
8 5A
(10*5) (4*8) 18A
3 5
0 1B
2 6
8 24C
The determinant of a matrix is defined iteratively
1. An nxn is calculated as the sum of terms involving the determinants of nx(n-1)x(n-1) (ie n!) matrices
2. Each (n-1)x(n-1) matrix determinant is the sum of terms involving n-1 determinants of (n-2)x(n-2) matrices and so on
3. Since we know how to calculate the determinant of a 2x2 matrix we can always use this definition to find the determinant of an nxn matrix
4. In practice we shall not go above 3x3 matrices (unless using a computer program) but we need to know the general formula for an inverse
Determinants.
General properties of determinants
1) If B = A’, then det. B = det. A
2) If B is constructed from A by swapping two rows, then det. B = -det. A
3) If B is constructed from A by swapping two columns, then det. B = -det. A
4) If B is constructed from A by multiplying one row (or column) by a constant, c, then det. B = c det. A
5) If B is constructed from A by adding a multiple of one row to another, then det. B = det. B
1 2
3 4A
1 3
2 4B
Determinants of triangular matrices
1 2 3
0 4 5
0 0 6
A
6 0 0
5 4 0
1 2 3
B
Are examples of – respectively – an upper triangular and a lower triangular matrix
(zeros below or above the main diagonal)
The determinant of either an upper or lower triangular matrix is equal to the product of the elements on the main diagonal
Eg det.A = 1(24-0) - 2(0) + 3(0) = 24 = 1*4*6
For a 3 x 3 matrix, using
Question: What is the determinant of
Determinants
0 2 1
0 0 1
1 0 2
A
11 12 13
21 22 23
31 32 33
22 23 21 23 21 2211 12 13
32 33 31 33 31 32
a a a
A a a a
a a a
a a a a a aA a a a
a a a a a a
1 11
1
det( ) ( 1) det( )j n
j jj
j
A A a A
Method 1: Laplace expansion of an n x n matrix.
Can generalise this rule for the determinant of any n by n matrix
As part of this method, you need to know the following:
1. Minor M
2. Co-factor C
(which are also essential to invert a matrix)
11 1 1
1 1
( 1)j n j n
jj j j ij
j j
A a M a C
There is a minor Mij for each element aij in the square matrix.
1. To find it construct a new matrix by deleting the row i and deleting the column j.
2. Then find the determinant of what’s left
3. E.g. M11
4. Example M12
nxn Matrix inversion - minors
nnn
n
nnn
n
aa
aa
Mso
aa
aa
A
1
222
11
1
111
;
.6423697
64;
987
654
321
.. 12
MsoAei
1 2 3
A= 4 5 6
7 8 9
1 2 3
Delete row and column
4 5 6
7 8 9
The cofactor is Cij is a minor with a pre-assigned algebraic sign given to it
1.For each element aij, work out the minor
2.Then multiply it by (-1)i+j
3.In simpler language: if i+j is even then Cij = Mij
4.If i+j is odd, then Cij = -Mij
5.The co-factor matrix is just
6.The adjoint matrix is C’ – i.e. the transpose of C.
N xn matrix inversion: Co-factor and adjoint matrices
nnn
n
cc
cc
C
1
111
1. The inverse matrix, A-1 is just
So
i) find the determinant
– if it is non-zero, the matrix is non-singular so its inverse exists
ii) Find the cofactors of all the elements of A and arrange them in the cofactor matrix
iii) Transpose this matrix to get the adjoint matrix
iv) Divide the adjoint matrix by the determinant to get the inverse
Co-factor, adjoint matrices and the inverse matrix
1 1 1.A adj A C
A A
1. If find A-1
Use the formula
First find the determinant
which is non-zero so can continue
Now find matrix of cofactors, which in the 2 x 2 case is a set of 1 X 1 determinants
Example 1 (2 x 2 matrix)
22 211
12 1111 22 12 21
1 a aA
a aa a a a
10
12A
11 22 12 21 (2)(1) (1)(0) 2A a a a a
11 12
21 22
1 0
1 2
C CC
C C
1 1 1.A adj A C
A A
Now transpose the matrix of cofactors to get the adjoint matrix
Now using the formula above
which is non-zero so can continue
then
NB. Always check that the answer is right by looking if AA-1 = I
Example 1 (2 x 2 matrix)
11 21
12 22
1 1. '
0 2
C Cadj A C
C C
11 22 12 21 (2)(1) (1)(0) 2A a a a a
1 1 1 1/ 2 1/ 21
0 2 0 12A
10
01
2000
2202
2
1
20
11
10
12
2
11AA
1 1 1.A adj A C
A A
While
Example 2: 3 x 3 matrix
54
20
64
30
65
3287
20
97
30
98
3287
54
97
64
98
65
;
987
654
320
MsoA
;
8123
14216
363
801201512
1402102418
353264845
M
;
8143
12216
363
;
8123
14216
363
CC
11 11 12 12 1 1 (0)( 3) (2)(6) (3)( 3) 3n nA a c a c a c
Example continued
8143
12216
363
3
1;
987
654
3201AsoA
1. In each case find the matrix of minors
2. Find the determinant
3. Find the inverse and check it.
Quiz
0 2 1
0 0 1
1 0 2
B
3 2
1 0A
The term is called the determinant of A, often written det.(A). Vertical lines surrounding the original matrix entries also means ‘determinant of A’.
The matrix part of the solution is called the ‘adjoint of A’ written adj. A. The elements of the adj.A are called co-factors. So,
Some more jargon
1121
1222
21122211
1
aa
aa
aaaaB
21122211 aaaa
11 12
21 22
det .a a
A Aa a
AadjA
B ..det
1
1121
1222.aa
aaAadj
2x2 Matrix Inversion Quiz
Suppose
1. What is det. A?
2. What is A-1
3. What is det. B?
4. Can you find B-1
11
13A
11
22B
Summary
11 Definitions you should now memorise:
Matrix dimensions, null matrix, identity matrix, transpose, symmetric matrix, square matrix, leading diagonal, nonnegative, positive, nonpositive, and negative matrices.
5 skills you should be able to do:
– Add two nxm matrices
– Multiply a matrix by a scalar
– Transpose a matrix.
– Identify the element aij in any matrix
– Understand and have practised matrix multiplication
For home study: Method - the Gaussian approach.
• In a row operation multiples of one row are added or subtracted from multiples of another row to produce a new matrix.
• Example: transform A by replacing row 1 by the sum of row 1 and row 2:
• A row operation can be represented by matrix multiplication. A’ = BA where
;
200
120
121
A
;
200
120
001
200
120
112201
'
A
;
100
010
011
B
Method 2: the Gaussian method.
• Suppose, by a series of m row operations we transform A into I, the identity matrix.
• Let Bi indicate the series of matrices in this sequence of m row operations:
• BmBm-1…B1A = I.
• Let C = BmBm-1…B1 so that CA = I.
• It follows that, by the definition of the inverse that C = A-1.
• Since C = CI we can find C by taking the row operations conducted on A and conducting them in parallel on I.
• Method. Begin with the extended matrix [A I]:
• Carry out row operations on the extended matrix until it has the form [I C]
• C = A-1
;
100
010
001
200
120
121
''
A
1. Use the Gaussian method to find the inverse and check it works.
Quiz
;
202
100
124
A
5. An odd example.
• In Cafeland there are only two goods: x1 = latte, x2 = muffin.
• Peculiarly, it is not possible to buy and sell the goods separately.
• Instead the following combinations are available
– x: Latte lover : x1 = 2, x2 = 1.
– y: Muffintopia: x1 = 0, x2 = 3.
Any fraction of these combinations can be bought and sold
Joan wishes to own and consume exactly one latte and one muffin. Can she buy to achieve her goal?
Answer:
She buys 1/2 of latte lover combo and 1/6 of muffintopia combo.
This mix of vectors is called a linear combination.
More formally, if x and y are n x 1 vectors and a and b are scalars,
ax + by is a linear combination.
Joan’s purchase is (1/2)x + (1/6)y
The example again
• In Cafeland there are only two goods: x1 = latte, x2 = muffin.
• Peculiarly, it is not possible to buy and sell the goods separately.
• Instead the following combinations are available
– x: Latte lover : x1 = 2, x2 = 1.
– y: Muffintopia: x1 = 0, x2 = 3.
Any fraction of these combinations can be bought and sold
Can Joan construct any combination of latte and muffin out of x and y?
Any combination: So formally, is there an a and a b such that:
Or
;3
0;
1
2
yx
2
1
x
x
3
0
1
2
2
1 babyaxx
x
ba
a
x
x
3
2
2
1
The example again
I.e. two equations in two unknowns:
x1 = 2a and
x2 = a+3b
Or
0.5x1 = a
and so
x2 = a + 3b = 0.5x1 + 3b or
x2 –0.5x1= 3b or
b = (x2 -0.5x1)/3
For instance if x1 = 1 and x2 = 1, then a = 0.5 and b = 1/6
ba
a
x
x
3
2
2
1
10. Vector Space
• The set of vectors generated by the various linear combinations of 2 vectors is called a 2-dimensional vector space
• Consider a space with n dimensions
• It follows that a single nx1 vector is a point in that space
• A group of nx1 vectors is said to form a basis for the space if any point in that space can be represented as a linear combination of the vectors in the group.
• In the example, formed a basis for 2 dimensional space.
• These vectors are also said to span 2 dimensional space• In other words if a group of vectors form a basis for an n-dimensional
space that’s the same as saying that they span the n-dimensional space.
;3
0;
1
2
yx