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Matrix

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2 INTRODUCTION TO MATRIX ALGEBRA
INTRODUCTION TO MATRICES
Reference : Croft, A., & Davison, R. (2008). Mathematics for
Engineers - A Modern Interactive Approach, Pearson
Education.
A matrix is a rectangular array or block of numbers usually
enclosed in brackets.
A m x n matrix has m rows and n columns.
Introduction to Matrix Algebra
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1
If the matrix A has m rows and n columns we can write:
 a11 a12
a
a22
21
A 

 
a
 m1 am2
 a1n 
 a2n 

  
 amn 
where aij represents the number or element in the ith row and
jth column.
Introduction to Matrix Algebra
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2
Special Matrices
A square matrix has the same number of rows as columns.
The main diagonal of a square matrix is the diagonal
running from ‘top left’ to ‘bottom right’.
An identity matrix, denoted by I, is a square matrix with
ones on the main diagonal and zeros elsewhere.
 1 0 0


I   0 1 0
 0 0 1


The transpose of A is obtained by writing rows as columns
and columns as rows, and is denoted AT.
Introduction to Matrix Algebra
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3
Equality of Matrices
If A = (aij) and B = (bij), A = B if and only if aij= bij.
Addition and Subtraction of Matrices
Matrices of the same size may be added to and subtracted
from one another. To do this, the corresponding elements
are added or subtracted.
Introduction to Matrix Algebra
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4
3 5 
7 6
 2 1 4
2 1, C3 9
e.g. 1 If A

,
B



 

3
0
2


4 2 
1 5
find A + B, B + C and B - C.
A + B is not defined as A and B are not of the same size.
3 5  7 6 10
B + C = 2 1  3 9  5

   
4 2  1 5  5
B–C=
11
8

7
3 5  7 6 4 1
2 1  3 9 1 10


   
4 2  1 5  3 3
Introduction to Matrix Algebra
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5
Multiplication of a Matrix by a Number
Any matrix can be multiplied by a number. To do this, each
element of the matrix is multiplied by that number.
 7 13 5 


e.g.2 If A   9 8 2  , find 2A, -A.
 4 11 8 


 2 * 7 2 * 13 2 * 5  14 26 10 

 

2A =  2 * 9 2 * 8 2 * 2   18 16 4 
 2 * 4 2 * 11 2 * 8   8 22 16 

 

  1* 7  1*13  1* 5    7  13  5 

 

-A =   1* 9  1* 8  1* 2     9  8  2 
  1* 4  1*11  1* 8    4  11  8 

 

Introduction to Matrix Algebra
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6
Multiplication of Matrices
If A is a n x m matrix and B is a p x q matrix. For the product
AB to exist we must have m = p.
A B
nm
pq
C
if m = p
nq
if m ≠ p
does not exist
Note that matrix multiplication is :
i. not commutative (i.e. AB  BA).
ii. associative [i.e. ABC = (AB)C = A(BC)].
iii. If C = AB, the element cij is found from row i of A and
column j of B, as follows:
n
cij 
a b
ik kj
k 1
Introduction to Matrix Algebra
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7
4 7 6 1 2 37 33
AB  8 2 6 3 1  26 36
4 5 5 2 3 29 28
3× 3
3× 2
3× 2
3
i.e. c21  a2k bk1  8 *1  2 * 3  6 * 2  26
k 1
Introduction to Matrix Algebra
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8
3 5 
 2 1 4
2 1 , find AB.
e.g. 3 If A  
&

B




3
0
2


4 2 
AB
 2312 44 2511  42 


3302 24 3501  22
24 17 


1
11




Note that when a square matrix is post- or pre-multiplied
by an identity matrix of the appropriate size the matrix is
unchanged, i.e.
AI = IA = A
Introduction to Matrix Algebra
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9
DETERMINANTS, INVERSE OF A MATRIX
Reference : Croft & Davison, Chapter 12, Blocks 3,4
Determinant
All square matrices, A, possess a determinant denoted by :
det(A), |A|.
Determinant of a 2 x 2 matrix
a b
a b 
det(A)
=
|A|
=
= ad - bc
If A  
 , then
c d
c d 
A matrix which has a zero determinant is called singular.
Introduction to Matrix Algebra
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10
Minors and Cofactors of a 3 x 3 Matrix
Let aij be an element of a matrix A.
The minor of aij is the determinant formed by crossing out the ith
row and jth column of det(A).
The cofactor of aij = (-1)i+j x (minor of aij)
Note that the term (-1)i+j is called the place sign of the element
on the ith row and jth column. The following may help you to
memorize this.
  


  
  


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11
Determinant of a 3 x 3 Matrix
 a11

Consider a general 3 x 3 matrix, A =  a21
a
 31
a12
a22
a32
a13 

a23 
a33 
det(A) can be calculated by expanding along any row or
column. For example, expanding along the first row:
|A| = a11*(its cofactor) + a12*(its cofactor) + a13*(its cofactor)
Introduction to Matrix Algebra
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12
2 1
e.g.1 Find the value of 1  1
1
2
1
1 1
1
4
3
1
 2*
2
1
1
4
2
4
1 4 17
3
1 and 11 24 5
6 31 15
2
 1* 
1
1
1 2
 3*
1 1
1
4
 2 * ( 2)  1 * 3  3 * 5  14
1
4
17
24
5
4
17
4
17
 11* 
 6*
11 24 5  1*
24 5
31 15
31 15
6 31 15
 1* 205  11* 467  6 *  388  3014
Introduction to Matrix Algebra
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Alternatively, by Rule of Sarrus
Repeat the 1st and 2nd column to right hand side of 3rd
column to form a 3 x 5 matrix.
det(A) = Add the product of SOLID diagonals from left top to
right bottom and subtract the products of DASH diagonals
from left bottom to right top.
Introduction to Matrix Algebra
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14
Hence
2 1
3
1 1 1
1 4 2
2 1
3 2 1
 1 1 1 1 1
1 4 2 1 4
 2 * (1) * (2)  1*1*1  3 *1* 4
3 * (1) *1  2 *1* 4  1*1* (2)
 14
Introduction to Matrix Algebra
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15
Properties of Determinants
i. If every element of a given row (or column) of the square
matrix is multiplied by the same factor, the value of the
determinant is multiplied by that factor
ii. If |B| is obtained by interchanged any 2 rows (or columns) of
|A|, then |B| = -|A|.
iii. Adding or subtracting a multiple of one row (or column) to
another row (or column) leaves the determinant
unchanged.
iv. If A and B are 2 square matrices and that AB exists, then
det(AB) = det(A)det(B).
v. If 2 rows or 2 columns of a square matrix are equal, the
determinant of the matrix is zero.
Introduction to Matrix Algebra
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16
Inverse of a Matrix
The inverse matrix of a square matrix A, usually denoted by A-1,
has the property :
AA-1 = A-1A = I
Note that if
|A| = 0, A does not have an inverse.
|A|  0, A has an inverse
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17
Finding the Inverse of a Matrix
The followings are steps to find the inverse of a matrix A when
|A|  0,
i. Find the transpose of A, denoted AT.
ii. Replace each element of AT by its cofactor. The resulting
matrix is called the adjoint of A, denoted adj(A).
iii.
adj ( A)
A 
A
1
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18
3 
2 1


e.g. 2 Find the inverse of A   1  1 1 
1 4  2


det(A) =14
1 1
 1 1


1 2
 4 2
 1 3
2 3
adj ( A)   
 4 2 1 2
 1 3
2 3


1 1
 1 1
  2 14 4 

1
1
A   3 7 1 
14 

5
7
3




Introduction to Matrix Algebra
T
1 1 

1 4 
 2 14 4 



2 1

  3 7 1 
1 4
 5  7  3


2 1 

1 1 
Page
19
3 1 0 


e.g. 3 Find the inverse of B   5 2  1 .
 1 6  3


det( B)  14
 2

 6
 1
adj ( B)   
 6
 1

 2
0
1
1
B   14
14 
 28
1
3
0
3
0
1
5 1

1 3
3 0
1 3
3 3

5 1
T
2 

6 
 1
3
0



1
   14  9 3 
6 


28
17
1


1 

5 2 
5
1
3

1
3
 1

9 3 
 17 1 
3
Introduction to Matrix Algebra
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20
TECHNIQUES OF SOLV ING
ALGEBRAIC EQUATIONS
Reference : Croft, A., & Davison, R. (2008). Mathematics for
Engineers - A Modern Interactive Approach, Pearson
Education.
Techniques of Solving Algebraic Equations
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21
TYPES OF SOLUTIONS TO SYSTEM OF LINEAR EQUATIONS
When a system of linear equations is solved, there are 3
possible outcomes:
i.
a unique solution
ii. an infinite number of solutions
iii. no solution
 2 x  3 y  3  2 3 3   x  3

 
 (unique solution)
i. 

0 x  y  1  0 1  1  y  1
Techniques of Solving Algebraic Equations
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22

ii.  2 x  3 y  3   2 3 3   (infinite number of solutions)

4 x  6 y  6  0 0 0 
iii.  2 x  3 y  3   2 3 3   (no solution)

 0 0  1



0
x
0
y
1



Techniques of Solving Algebraic Equations
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23
CRAMER’S RULE
Reference : Croft & Davison, Chapter 13, Blocks 1, 2
Cramer’s rule is a method that uses determinants to solve a
system of linear equations.
i.
Two equations in 2 unknowns
 a1 x  b1 y  k1

If
a2 x  b2 y  k 2
a1 k1
k1 b1
a2 k 2
a1
k 2 b2
then x  a b , y  a b , provided that a
1
1
1
1
2
a2 b2
a2 b2
Techniques of Solving Algebraic Equations
b1
b2
0
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24
ii. 3 equations in 3 unknowns
If
 a1 x  b1 y  c1 z  k1

a2 x  b2 y  c2 z  k 2
a x  b y  c z  k
 3
3
3
3
a1
b1
c1
a2
b2
c2  0
a3
b3
c3
k1 c1
k 2 c2
k3 c3
, z
b1 c1
b2 c2
a1
a2
a3
b1
b2
b3
k1
k2
k3
a1
a2
b1
b2
c1
c2
b3
a3
b3
c3
where
then
x
k1
k2
k3
b1
b2
b3
c1
c2
c3
a1
a2
b1
b2
c1
c2
a1
a2
a3
, y
a1
a2
a3
b3
c3
a3
Techniques of Solving Algebraic Equations
c3
Page
25
e.g.1 Using Cramer’s rule, solve for x, y.
 2x  5 y  3

7 x  2 y  9
a1  2, a2  7, b1  5, b2  2, k1  3 & k 2  9
k1
k2
x
a1
a2
a1
a2
y
a1
a2
b1
3
5
b2
9 2
39

 1

2 5
b1
 39
7 2
b2
k1
2 3
k2
7  9  39


1
2 5
b1
 39
7 2
b2
Techniques of Solving Algebraic Equations
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26
e.g.2 Using Cramer’s rule, solve for x, y and z.
 5 x  3 y  4 z  35

 2 x  5 y  3 z  12
 3 x  2 y  2 z  10

35
3
4
 12  5  3
10
x
5
2
2
3
 282
x
 2,
 141
35
2
 12  3
3
2
, y
4
5
5 3
3 2
5
2
10
3
141
y
 1,
 141
Techniques of Solving Algebraic Equations
5
2
3
2
2
35
 5  12
3 2
2
, z
4
5
3
5 3
3 2
2
4
10
4
5 3
3 2
2
 987
z
7
 141
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27
INVERSE MATRIX METHOD
Writing System of Equations in Matrix Form
Note that
 a1 x  b1 y  k1

a2 x  b2 y  k 2
can be written as
 a1

 a2
b1  x   k1 
    
b 2  y   k 2 
This is called the matrix form of the simultaneous equations.
Techniques of Solving Algebraic Equations
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28
 a1 x  b1 y  c1 z  k1

Similarly, a2 x  b2 y  c2 z  k 2
a x  b y  c z  k
 3
3
3
3
 a1

can also be written as  a2
a
 3
b1
b2
b3
c1  x   k1 
   
c2  y    k 2 
c3  z   k3 
i.e. the general matrix form of a system of equations:
AX = B
where A, X and B are matrices.
Techniques of Solving Algebraic Equations
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29
Solving Equations Using the Inverse Matrix Method
Consider the matrix form: AX = B
A-1AX = A-1 B
I X = A-1 B
X = A-1 B
i.e. X can be found if A-1 exists.
Techniques of Solving Algebraic Equations
Page
30
e.g.3 Redo example 1 and example 2 using the inverse
matrix method.
 2x  5 y  3
 2 5  x   3 
    
 

7 x  2 y  9  7  2  y    9 
 x
2 5 
 3 
 , X    & B   
Let A  
 y
 7  2
  9
2 5
  2  7    2  5 

  
A
 39, adj  A  
7 2
2   7 2 
  5
adj  A
1   2  5
1

A 
  
A
39   7 2 
1   2  5  3 
 
X  A1 B   
39   7 2   9 
1  39    1
   
  
39   39   1 
T
Techniques of Solving Algebraic Equations
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31
e.g.4
3
4  x   35 
 5 x  3 y  4 z  35
 5

  


x
y
z
y
2
5
3
12
2
5
3
12











  

 3 x  2 y  2 z  10   3  2 2  z   10 



  
 - 16 - 14 11 


22 23 
A  141, adj  A   5
 - 19 1 - 31


 - 16 - 14 11 

1 
adj  A
1

22 23 
A 
 5
141 
A

 - 19 1 - 31
 - 16 - 14 11  35 


1 
1
22 23   12 
X  A B
 5
141 
 10 
19
1
31



  282   2 
  
1 

 141     1
141 
  
  987   7 
Techniques of Solving Algebraic Equations
Page
32
Class Practice
1. Solve the simultaneous equations using Cramer’s rule:
x+y=5
2x 3y = 6
2. Solve the simultaneous equations using inverse
matrix method
6x  5y + z -11= 0
2x + 3y - z +5= 0
Given Adj A =
x + 2y – 4z +12= 0
and Det = -22
GAUSSIAN ELIMINATION
Reference : Croft & Davison, Chapter 13, Block 3
Introduction
Gaussian Elimination is a systematic way of simplifying a
system of equations.
A matrix, called an augmented matrix, which captures all the
properties of the equations, is used.
A sequence of elementary row operations on this matrix
eventually brings it into a form known as echelon form (to be
discussed later).
From this, the solution to the original equations is easily found.
Techniques of Solving Algebraic Equations
Page
34
Augmented Matrix
Consider the system of equations,
 a1 x  b1 y  k1

a2 x  b2 y  k 2
it can be represented by an augmented matrix:
 a1 b1

a b
2
2



coefficients
Techniques of Solving Algebraic Equations
k1 

k2 

constants
this vertical line can be
omitted as in your textbook
Page
35
Similarly, the following system of equations:
 a1 x  b1 y  c1 z  k1

a2 x  b2 y  c2 z  k 2
a x  b y  c z  k
 3
3
3
3
can also be written as an augmented matrix:
 a1

 a2
a
 3
Techniques of Solving Algebraic Equations
b1
b2
b3
c1 k1 

c2 k 2 
c3 k3 
Page
36
e.g.1 Write down the augmented matrices for the followings
 3x  9 y  7
a. 
7 x  12 y  23
3 9 7 


 7  12 23 


 9 x  7 y  65 z  15

b.  5 x  12 y  2 z  64
 4 x  22 y  3 z  7

 9  7 65 15 


 5 12  2 64 
  4 22 3 7 


6
 17 x  3 y
c.  4 x  8 y  3 z  12

 3 y  5z  7

17 3
0 6


 4 8  3 12 
 0 3 5 7 


Techniques of Solving Algebraic Equations
Page
37
e.g.2 Solve the system with the augmented matrix:
a.  1  7 15 
0 1  2


 x  7 y  15  x   1 
     

 y  2
 y    2
 1  2 1 5


b.  0 1
3 1
0 0

1
1


x  2 y  z  5  x   2 
   

  y     2
 y  3z  1
z  1 
z  1

   
Techniques of Solving Algebraic Equations
Page
38
Row-Echelon Form of an Augmented Matrix
For a matrix to be in row-echelon form:
i. Any rows that consist entirely of zeros are the last rows of
the matrix.
ii. For a row that is not all zeros, the first non-zero element is a
one, called a leading 1.
iii. While moving down the rows of the matrix, the leading 1s
move progressively to the right.
Techniques of Solving Algebraic Equations
Page
39
e.g.1 Determine which of the following matrices are in rowechelon form.
 1 2 5 78 


a.  0 1 8  14  Yes
 0 0 1 22 


Techniques of Solving Algebraic Equations
1 4 1 5 


b.  0 0 1 2  No
 0 1 1 27 


Page
40
Elementary Row Operations
The elementary operations that change a system but leave
the solution unaltered are:
i. Interchange the order of the equations.
ii. Multiply or divide an equation by a non-zero constant.
iii. Add, or subtract, a multiple of one equation to, or from,
another equation.
Techniques of Solving Algebraic Equations
Page
41
Note that a row of an augmented matrix corresponds to an
equation of the system of equations.
When the above elementary operations are applied to the
rows of such a matrix, they do not change the solution of the
system.
They are called elementary row operations.
Techniques of Solving Algebraic Equations
Page
42
Gaussian Elimination to Solve a System of Equations
i. write down the augmented matrix.
ii. apply elementary row operations to get row-echelon form.
iii. solve the system.
Techniques of Solving Algebraic Equations
Page
43
 2x  y  2z  8
e.g.2 Use Gaussian Elimination to solve  x  3 y  3 z  4
 4x  2 y  z  1

The augmented matrix is
2
8 
2 1


1  3 3  4
 4 2 1 1 


1  3 3  4

 Interchange row 1 and row 2
2
8 
2 1
 4 2 1 1 


 1  3 3  4  row 2 – 2*row 1


 0 7  4 16  row 3 – 4*row 1
 0 14  13 17 


Techniques of Solving Algebraic Equations
Page
44
1  3 3  4 


 0 7  4 16 
 0 0  5  15 


row 3 – 2*row 2
3
4 
1  3


16
4
0 1  7
7
0 0

1
3


Hence
row 2 / 7
row 3 / -5
z 3
4
16
or y  4
y z 
7
7
x  3 y  3z  4 or x  1
The solution is x
Techniques of Solving Algebraic Equations
y
z   1 4 3
T
T
Page
45
 4 x  3 y  5
e.g.3 Use Gaussian elimination to solve 
 3 x  2 y  3
 4
3  5


3  2 3 


 4 3  5


 0 1  3  3R  4 R
2

 1
 1 3 4  5 4  R1 4


0 1

3



x  3 y 4   5 4

 y  3
 x  1 
    
 y    3
Techniques of Solving Algebraic Equations
Page
46
Class Practice
47
Use Gaussian elimination to solve
Techniques of Solving Algebraic
Equations
 2x  5 y  3

7 x  2 y  9
Page
 2 x  4 y  z  12

e.g.4 Use Gaussian elimination to solve 2 z  3 x  5 y  8
 x  3y  2z  2

 2 4  1 12 

3 5
 1 3

2
2
 1 3 2

3 5 2
 2 4 1

1 3 2

 0 14 8
0 2 5

1 3
2

0 1 4 7
 0 0  27


 8
2 
2  R1  R3

 8
12 
2 

 2  3R1  R2
 8  2 R1  R3
2 

 1 7  R2 14
54  R2  7 R3
Techniques of Solving Algebraic Equations
1 3 2
2 


 0 1 4 7 1 7 
0 0 1
 R  27 
2


 2
x  3 y  2z  2

 y  4z 7  1 7
 z  2

 x  3 
   
 y   1 
 z    2
   
Page
48
Gaussian Elimination to find the Inverse of a Matrix
i. write down in a form of [ A I ] .
ii. apply a sequence of elementary row operations to
reduce A to I.
iii. Performing this same sequence of elementary row
operations on I, we obtain A-1.
Techniques of Solving Algebraic Equations
Page
49
 1  2 2


Suppose A   2  3 6 
1 1 7


 1  2 2 1 0 0 1  2 2 1

 
A I   2  3 6 0 1 0  0 1 2  2
 1 1 7 0 0 1   0 3 5 1

 
1  2 2 1 0 0 
1  2



 0 1 2  2 1 0 
 0 1
 0 0 1  5 3  1 3 R  R
0 0
3

 2

 1 0 0 27  16 6  R1  2 R2


 0 1 0 8
5 2 
 0 0 1  5 3  1


 27  16 6 


1
5 2 
Hence A   8
  5 3  1


Techniques of Solving Algebraic Equations
0 0

1 0  R2  2 R1
0 1  R3  R1
0 11  6 2
0 8 5 2
1 5
3
 R1  2 R3

 R2  2 R3
 1
Page
50
Example
,
51
•Using Gaussian Elimination to find the inverse of
A=
Matrix Algebra
Page
1 2 1 1 0 0  1 2 1 1 0 0 

 

A I   1  1 1 0 1 0    0 3 0 1  1 0  R1  R2
 3 0 2 0 0 1   0 6 1 3 0  1 3 R  R
3

 
 1
1 2 1 1 0 0 


  0 3 0 1 1 0 
 0 0 1 1 2  1 R  2 R
2

 3
2
1  R1  R3
1 2 0 0


  0 1 0 1 3  1 3 0  R2 3
0 0 1 1
 1
2

 1 0 0  2 3  4 3 1  R1  2 R2


  0 1 0 1 3 1 3 0 
0 0 1
 1
1
2

 2 3  4 3 1 
 2  4 3 

 1

1
Hence A   1 3  1 3 0    1  1 0 
 1
 3 3

2
1
6  3 



Techniques of Solving Algebraic Equations
Page
52
EIGENVALUES AND EIGENVECTORS
Reference : Croft, A., & Davison, R. (2008). Mathematics for
Engineers - A Modern Interactive Approach, Pearson
Education.
Eigenvalues and Eigenvectors
Page
53
EIGENVALUES AND EIGENVECTORS
Reference:
Croft & Davison
Chapter 13 Block 4
Consider the system
 a b  x   0  or simply,

    
 c d  y   0 
AX  0
If
A  0 the system has non-trivial solutions.
If
A  0 the system has only the trivial solution.
Eigenvalues and Eigenvectors
Page
54
e.g.1 Determine which system has
non-trivial solution
3x  y  z  0
x  2 y  2z  0
4x  y  3z  0
 3 1 1 


A  1 2 2
 4 1 3


3 1 1
det( A)  1
2
2 0
4
1
3
Therefore, the system has non-trivial solution
Eigenvalues and Eigenvectors
Page
55
e.g.2 Determine which system has
non-trivial solution
3x  y  z  0
x  2 y  2z  0
5x  y  3z  0
 3 1 1 


A  1 2 2
 4 1 3


3 1 1
det( A)  1
2
2 6
5 1 3
Therefore, the system has trivial solution
Eigenvalues and Eigenvectors
Page
56
If A is an n  n matrix and X is a vector (n  1 matrix) then
there is usually no geometric relationship between the vector
X and the vector AX (figure below).
AX
X
Eigenvalues and Eigenvectors
Page
57
But if X is an eigenvector of A then
AX =  X
i.e. AX is a scalar multiple of X.
AX
X
Eigenvalues and Eigenvectors
Page
58
Characteristic Equation and Eigenvalues
Consider the system
 a b  x 
 x

     
 c d  y 
 y
or
AX  X
We seek values of  so that the system has non-trivial
solutions.
The system can be written as
 A  I X  0
The system has non-trivial solutions if A  I  0
Eigenvalues and Eigenvectors
Page
59
A  I  0 , a polynomial equation in  , is called the
characteristic equation.
The values of
 which cause the system AX  X to have
non-trivial solutions are called eigenvalues.
Eigenvalues and Eigenvectors
Page
60
Example
61
 Find the eigenvalue for the following matrix
4
5
det  A  I  
 2  7  2  0
2
3
Hence  
Matrix Algebra
  7  
 7 
21
2
 412 
7  41

2
Page
Such vectors arise in the study of free vibrations, electrical
systems , chemical reactions and mechanical stress, etc.
k1
m1
k2
x1
m2
k3
x2
The system of differential equations of the above system is
 m1

0
0  x1   k1  k 2
   
m2  x2    k 2
Eigenvalues and Eigenvectors
 k 2  x1 
   0
k 2  k 3  x 2 
Page
62
Eigenvectors
Suppose that  , an eigenvalue, satisfies the system
AX  X .
For each eigenvalue there is a non-trival solution (unique up to
a non-zero scalar multiple) of the system. This solution is called
an eigenvector.
Eigenvalues and Eigenvectors
Page
63
e.g.1 Find the eigenvalues and eigenvectors of the matrix
5 4

A  
1 2
5
det (A-I) =
1
4
 2  7  6  0
2
Hence   6 or   1
4   1 4 
5  6
  

When   6 , ( A  I )  
2  6  1  4
 1
i.e. x  4 y  0 or x  4 y . In general, x  4t , y  t
 x   4
The eigenvector is    t  
 y 1
Eigenvalues and Eigenvectors
Page
64
 5 1 4   4 4
  

( A  I )  
2  1  1 1 
 1
i.e. x  y  0 or x   y . In general, x  t , y   t
When
 1 ,
 x  1 
The eigenvector is    t  
 y    1
Eigenvalues and Eigenvectors
Page
65
e.g.2 Find the eigenvalues and eigenvectors of A.
 3 1
A

 1 5
e.g.3 Find the eigenvalues and eigenvectors of A.
0 
1 2


A   1 1 1 
 3 2  2


Eigenvalues and Eigenvectors
Page
66
e.g.2 Find the eigenvalues and eigenvectors of A
 3 1

A  
 1 5
3
det (A-I) =
1
1
 2  8  16  0
5
Hence   4 (repeated )
1    1 1
3  4
  

When   4 , ( A  I )  
  1 5  4    1 1
i.e.
 x  y  0 or x  y . In general, x  t , y  t
 x  1
The eigenvector is     t , for any number t
 y  1
Eigenvalues and Eigenvectors
Page
67
e.g.3 Find the eigenvalues and eigenvectors of A
0 
1 2


A   1 1 1 
 3 2  2


1 
det (A-I) =  1
3
1   
2
1  
0
1
2
2
1  
1
2
2
0
2
1
1
3
2
1   2  3   2  1  0
1   2  3  2  0
1     1  2  0
Hence
0
  1 or - 2
Eigenvalues and Eigenvectors
Page
68
e.g.3 Find the eigenvalues and eigenvectors of A
2
0  0
2
0 
1  1

 

1    1  2 1 
When   1,  A  I     1  1  1
 3
 3

2

2

1
2
3


 

i.e. y  0 and  x  z  0 or x  z
In general, x  t , y  0, z  t
 x  1
   
The eigenvector is  y    0 t , for any number t
 z  1
   
Eigenvalues and Eigenvectors
Page
69
e.g.3 Find the eigenvalues and eigenvectors of A
2
0  2 2 0
1  1

 

1    1 0 1 
When   1,  A  I     1  1  1
 3
  3 2  1
2
2
1



 

i.e. x  y  0 or x   y and  x  z  0 or x  z
In general, x  t , y  -t , z  t
 x  1 
   
The eigenvector is  y     1t , for any number t
z  1 
   
Eigenvalues and Eigenvectors
Page
70
e.g.3 Find the eigenvalues and eigenvectors of A
2
0   3 2 0
1  2

 

1    1 1 1 
When   2,  A  I     1  1  2
 3
  3 2 0
2
2
2



 

2
i.e. 3 x  2 y  0 or x   y and  x  y  z  0 or z  x  y
3
3
5
In general, x  t , y  - t , z  t
2
2
 x  1 
  

The eigenvector is  y     3 2 t , for any number t
z  5 2 
  

Eigenvalues and Eigenvectors
Page
71
APPLICATION
Stiffness method is commonly used to analyse a structure as
matrix analysis can be used to solve the problem.
For a typical beam element, the relationship between the
member end moments and member end rotations is given by
 i, q i
 j, qj
i
j
L
qi 
4 EI
2 EI
i 
 j
L
L
qj 
2 EI
4 EI
i 
 j
L
L
Eigenvalues and Eigenvectors

 4 EI
 qi   L
q    2 EI
 j 
 L
2 EI 
L    i   2 EI  2 1    i   k    
1 2  
4 EI   j 
L

 j 

L 
Page
72
where
 i 
[ ]   
 j 
 kii kij 
[k]  

k
k
jj 
 ji
 qi 
[q]   
q j 
- displacement matrix
- stiffness matrix
- force matrix
By assembly the stiffness of the whole structure, it will
become a very Structure Stiffness Matrix.
By imposing appropriate boundary conditions and end
moments, the displacement of the whole structure can be
solved, i.e.
1
[ ]  [ K ] [Q]
Eigenvalues and Eigenvectors
Page
73
Class Practice
74
 Find the eigenvalue for the following matrix
5
det  A  I  
2
3
 2  9  14  0
4
Hence   2 or 7
Matrix Algebra
Page
Class Practice
75
 Suppose for matrix A
 the characteristic function is
 Find their eigenvectors.
Matrix Algebra
Page
Hence   1 or 2 (repeated )
  3 1  4  3    4  4  3

 

6 1 3    5
5
3 
When   1,  A  I    5
 3
  3

3
2
1

3
1


 
i.e. z  0 and x  y  0 or x   y
In general, x  t , y  -t , z  0
 x  1 
   
The eigenvector is  y     1t , for any number t
z  0 
   
Eigenvalues and Eigenvectors
Page
76
Hence   1 or 2 (repeated )
 3    5  4  3
3 2  4

 

62
3  5
4
3 
When   2,  A  I    5
 3
  3

3
2
2

3
0


 
 5x  4 y
i.e. x  y  0 and 5 x  4 y  3 z  0 or z 
3
t
In general, x  t , y  -t , z  
3
 x  1 
  

The eigenvector is  y     1 t , for any number t
 z    1 3
  

Eigenvalues and Eigenvectors
Page
77
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