Introduction to Matrix

What is a matrix?
A rectangular array of numbers (we will concentrate on
real numbers). A nxm matrix has ‘n’ rows and ‘m’
columns
 M 11 M 12 M 13 M 14  First row
 
M 3x4   M 21 M 22 M 23 M 24  Second row
 M 31 M 32 M 33 M 34  Third row

First Second Third Fourth

column column column column

M12 Row number

Column number
What is a vector?

A vector is an array of ‘n’ numbers

A row vector of length ‘n’ is a 1xn matrix

a 1 a2 a3 a4
A column vector of length ‘m’ is a mx1 matrix

 a1
a 
 2
 a3 
Special matrices

Zero matrix: A matrix all of whose entries are zero

 0 0 0 0
03x 4   0 0 0 0 
 0 0 0 0 
Identity matrix: A square matrix which has ‘1’ s on the
diagonal and zeros everywhere else.

 1 0 0
I3x 3 
  0 1 0 
 0 0 1 
Matrix operations
Equality of matrices

If A and B are two matrices of the same size,

then they are “equal” if each and every entry of one
matrix equals the corresponding entry of the other.

 1 2 4 a b c

A   3 0 7 
B  d e f  
 9 1 5   g h i 
a 1, b  2, c  4,
A  B  d   3, e 0, f  7,
g 9, h 1, i 5 .
Matrix operations Addition of two
matrices

If A and B are two matrices of the same size,

then the sum of the matrices is a matrix C=A+B whose
entries are the sums of the corresponding entries of A
and B
 1 2 4   1 3 10 
  
A    3 0 7 B    3 1 0  
 9 1 5  1 0 6 
 0 5 14 

C A  B   6 1 7  
 10 1 11 
 Addition of of matrices
Matrix operations
Properties

Properties of matrix addition:

1. Matrix addition is commutative (order of
addition does not matter)
A  B B  A
2. Matrix addition is associative
A  B  C   A  B   C
3. Addition of the zero matrix
A 0 0  A A
Matrix operations Multiplication by a
scalar

If A is a matrix and c is a scalar, then the product cA is a

matrix whose entries are obtained by multiplying each of
the entries of A by c

 1 2 4

A   3 
0 7  c 3
 9 1 5
 3 6 12 

cA    9 0 21 
 27 3 15 
 Multiplication by a
scalar
Matrix operations
Special case

If A is a matrix and c =-1 is a scalar, then the product

(-1)A =-A is a matrix whose entries are obtained by
multiplying each of the entries of A by -1

 1 2 4

A    3 0 7  c  1
 9 1 5
 1  2  4

cA -A   3 0  7
  9 1  5
Matrix operations Subtraction

If A and B are two square matrices of the same

size, then A-B is defined as the sum A+(-1)B

 1 2 4   1 3 10 
  
A    3 0 7 B    3 1 0  
 9 1 5  1 0 6 
 2  1  6

C A  B  0  1 7  
 8 1  1
Note that A - A 0 and 0 - A -A
 Transpose
Special
operations
If A is a mxn matrix, then the transpose of A is
the nxm matrix whose first column is the first
row of A, whose second column is the second
column of A and so on.

 1 2 4 1  3 9

A   3 0 7  T 
 A  2 0 1 
 9 1 5   4 7 5 
 Transpose
Special
operations

If A is a square matrix (mxm), it is called

symmetric if
T
A A
Matrix operations Scalar (dot) product of
two vectors
If a and b are two vectors of the same size
 a1   b1 
  
a   a2 ; b   b2  
 a 3   b 3 
The scalar (dot) product of a and b is a scalar
obtained by adding the products of
corresponding entries of the two vectors
a b  a 1 b 1  a 2 b 2  a 3 b 3 
T
Matrix operations Matrix multiplication

For a product to be defined, the number of columns

of A must be equal to the number of rows of B.

A B = AB
mxr rxn mxn
inside

outside
 Matrix operations Matrix multiplication

If A is a mxr matrix and B is a rxn matrix, then the

product C=AB is a mxn matrix whose entries are
obtained as follows. The entry corresponding to row ‘i’
and column ‘j’ of C is the dot product of the vectors
formed by the row ‘i’ of A and column ‘j’ of B
 1 2 4   1 3
A 3x3   3 0 7  B3x2   3 1 
 9 1 5   1 0 
T
3 5  1   1
C 3x2 AB  10  9  notice  2    3   3
 
  7 28   4   1 
 Multiplication of
Matrix operations matrices
Properties

Properties of matrix multiplication:

1. Matrix multiplication is noncommutative
(order of addition does matter)
A B  B A in g e n e ra l
 It may be that the product AB exists but BA
does not (e.g. in the previous example
C=AB is a 3x2 matrix, but BA does not
exist)
 Even if the product exists, the products AB
and BA are not generally the same
 Multiplication of
Matrix operations matrices
Properties

2. Matrix multiplication is associative

A B C   A B C
3. Distributive law
A B  C   AB  AC
B  C A  BA  CA
4. Multiplication by identity matrix
A I  A ; IA  A
5. Multiplication by zero matrix A 0  0 ; 0 A  0
6.  
AB B A
T T T
 Miscellaneous
Matrix operations properties

1. If A , B and C are square matrices of the

same size, and A  0 then A B  A C
does not necessarily mean that B  C
2. A B  0 does not necessarily imply that
either A or B is zero
 Definition
Inverse of a
matrix

If A is any square matrix and B is another

square matrix satisfying the conditions

AB BA I
Then
(a)The matrix A is called invertible, and
(b) the matrix B is the inverse of A and is
denoted as A-1.

The inverse of a matrix is unique

 Uniqueness
Inverse of a
matrix

The inverse of a matrix is unique

Assume that B and C both are inverses of A

AB BA I
AC CA I
(BA)C IC C
B(AC) BI B
 B C
Hence a matrix cannot have two or more
inverses.
 Some properties
Inverse of a
matrix

Property 1: If A is any invertible square

matrix the inverse of its inverse is the matrix A
itself
A -1  1
A
Property 2: If A is any invertible square
matrix and k is any scalar then

1 -1
k A 1
 A
k
 Properties
Inverse of a
matrix
Property 3: If A and B are invertible square
matrices then
1
A B  1 -1
B A
(AB) AB  I
1

Premultiplying both sides by A-1

A (AB) AB   A  1
-1 1

A ABAB 
-1 1
A 1
BAB   A  1
1

Premultiplying both sides by B-1

AB  1 B  1A  1
 What is a determinant?
The determinant of a square matrix is a number
obtained in a specific manner from the matrix.

For a 1x1 matrix:

A  a 1 1  ; d e t ( A )  a 1 1
For a 2x2 matrix:

 a 11 a 1 2 
A   ; det( A )  a 11a 2 2  a 1 2 a 2 1
 a 21 a 22 

Product along red arrow minus product along blue arrow

Example 1
 1 3
Consider the matrix A  
 5 7 

Notice (1) A matrix is an array of numbers

(2) A matrix is enclosed by square brackets

1 3
det( A )   1  7  3 5   8
5 7

Notice (1) The determinant of a matrix is a number

(2) The symbol for the determinant of a matrix is
a pair of parallel lines
Computation of larger matrices is more difficult
Duplicate column method for 3x3 matrix
For ONLY a 3x3 matrix write down the first two
columns after the third column

 a 11 a 12 a 13   a11 a12 a13  a11 a12

A   a 21 a 22 a 23  a 
 21 a 22 a 23  a 21 a 22
 a 31 a 32 a 33   a 31 a 32 a 33  a 31 a 32
Sum of products along red arrow
minus sum of products along blue arrow

det( A )  a 11a 2 2 a 3 3  a 1 2 a 2 3a 3 1  a 1 3a 2 1a 3 2
 a 1 3a 2 2 a 3 1  a 11a 2 3a 3 2  a 1 2 a 2 1a 3 3
This technique works only for 3x3 matrices
Example

2 4 -3  2 4  3 2 4
A   1 0 4 

1 0 
4 1 0
 2 - 1 2  
 2  1 2  2 1

0 -8 8 0 32 3

Sum of red terms = 0 + 32 + 3 = 35

Sum of blue terms = 0 – 8 + 8 = 0
Determinant of matrix A= det(A) = 35 – 0 = 35
 Finding determinant using inspection
Special case. If two rows or two columns are
proportional (i.e. multiples of each other), then the
determinant of the matrix is zero

2 7 8
3 2 4 0
 2  7  8

because rows 1 and 3 are proportional to each other

If the determinant of a matrix is zero, it is called a

singular matrix
 What is a cofactor?
Cofactor method
If A is a square matrix
 a 11 a 12 a 13 
A   a 21 a 22 a 23 
 a 31 a 32 a 33 
The minor, Mij, of entry aij is the determinant of the submatrix
that remains after the ith row and jth column are deleted from A.
The cofactor of entry aij is Cij=(-1)(i+j) Mij

a 21 a 23 a 21 a 23
M 12   a 2 1a 3 3  a 2 3a 3 1 C 12   M 12  
a 31 a 33 a 31 a 33
What is a cofactor?

Sign of cofactor  - 
-  - 
 
  - 
Find the minor and cofactor of a33
2 4 -3 
A   1 0 4 
 Minor 2 4
 2 - 1 2  M 33   2 0  4 1   4
1 0

Cofactor C  (  1 ) ( 3  3 ) M
33 33  M 33   4
 Cofactor method of obtaining the
determinant of a matrix

The determinant of a n x n matrix A can be computed by

multiplying ALL the entries in ANY row (or column) by
their cofactors and adding the resulting products. That is,
for each 1  i  n and 1  j  n

Cofactor expansion along the jth column

d e t ( A )  a 1 jC 1 j  a 2 jC 2j    a n jC n j

Cofactor expansion along the ith row

d e t ( A )  a i 1C i 1  a i 2 C i 2    a i n C i n
 Example: evaluate det(A) for:
1 0 2 -3
A= 3 4 0 1
det(A) = a11C11 +a12C12 + a13C13 +a14C14
-1 5 2 -2
0 1 1 3
4 0 1 3 0 1 3 4 1
det(A)=(1) 5 2 -2 - (0) -1 2 -2 +2 -1 5 -2
1 1 3 0 1 3 0 1 3
3 4 0
- (-3) -1 5 2 = (1)(35)-0+(2)(62)-(-3)(13)=198
0 1 1
 Example : evaluate
1 5 -3
det(A)= 2
1 0
2
By a cofactor along the third column
3 -1
det(A)=a13C13 +a23C23+a33C33

1 0 1 5 1 5
det(A)= -3* (-1)4 +2*(-1)5 +2*(-1)6
3 -1 3 -1 1 0

= det(A)= -3(-1-0)+2(-1)5(-1-15)+2(0-5)=25
Quadratic form

The scalar T
U d k d d  v ector
k  square m atrix
Is known as a quadratic form

If U>0: Matrix k is known as positive definite

If U≥0: Matrix k is known as positive semidefinite
Quadratic form

 d1   k 11 k 12 
Let d   k   Symmetric
d 2   k 21 k 22  matrix

Then
 k11 k12   d 1 
U  d k d d 1 d 2 
T
  
 k12 k 22  d 2 
 k11 d 1  k12 d 2 
d 1 d 2  
k12 d 1  k 22 d 2 
 d 1 ( k11 d 1  k12 d 2 )  d 2 ( k12 d 1  k 22 d 2 )
2 2
 k11 d 1  2k12 d 1 d 2  k 22 d 2
Differentiation of quadratic form

Differentiate U wrt d1
U
 2 k 11 d 1  2 k 1 2 d 2
d 1
Differentiate U wrt d2
U
 2 k12 d 1  2 k 22 d 2
d 2
Differentiation of quadratic form

Hence

 U 
U  d 1   k11 k12   d 1 
  2   
 d  U   k12 k 22  d 2 
 d 2 
2 k d
Outline

• Role of FEM simulation in Engineering

Design
• Course Philosophy
 Role of simulation in design:
Boeing 777

Source: Boeing Web site (http://www.boeing.com/companyoffices/gallery/images/commercial/).

Another success ..in failure:
Airbus A380

http://www.airbus.com/en/aircraftfamilies/a380/
Drag Force Analysis
of Aircraft

• Question
What is the drag force distribution on the aircraft?
• Solve
– Navier-Stokes Partial Differential Equations.
• Recent Developments
– Multigrid Methods for Unstructured Grids
 San Francisco Oakland Bay Bridge

Before the 1989 Loma Prieta earthquake

San Francisco Oakland Bay Bridge

After the earthquake

San Francisco Oakland Bay Bridge

A finite element model to analyze the

bridge under seismic loads
Courtesy: ADINA R&D
Crush Analysis of
Ford Windstar

• Question
– What is the load-deformation relation?
• Solve
– Partial Differential Equations of Continuum Mechanics
• Recent Developments
– Meshless Methods, Iterative methods, Automatic Error Control
Engine Thermal
Analysis

Picture from
http://www.adina.com

• Question
– What is the temperature distribution in the engine block?
• Solve
– Poisson Partial Differential Equation.
• Recent Developments
– Fast Integral Equation Solvers, Monte-Carlo Methods
Electromagnetic
Analysis of Packages

Thanks to
Coventor
http://www.cov
entor.com

• Solve
– Maxwell’s Partial Differential Equations
• Recent Developments
– Fast Solvers for Integral Formulations
Micromachine Device
Performance Analysis

From www.memscap.com

• Equations
– Elastomechanics, Electrostatics, Stokes Flow.
• Recent Developments
– Fast Integral Equation Solvers, Matrix-Implicit Multi-level Newton
Methods for coupled domain problems.
 Radiation Therapy of
Lung Cancer

http://www.simulia.com/academics/research_lung.html
Virtual Surgery
 General scenario..
Engineering design

Physical Problem

Question regarding the problem

...how large are the deformations?
...how much is the heat transfer?

Mathematical model Assumptions regarding

Geometry
Governed by differential Kinematics
equations Material law
Loading
Boundary conditions
Etc.
 Example: A bracket
Engineering design Physical problem

Questions:
1. What is the bending moment at section AA?
2. What is the deflection at the pin?
Finite Element Procedures, K J Bathe
 Example: A bracket
Engineering design Mathematical model 1:
beam

Moment at section AA M W L
 27,500 N cm
1 W (L  rN )3 W (L  rN )
Deflection at load  at load W  
3 EI 5
AG
6
How reliable is this model? 0.053 cm

How effective is this model?

 Example: A bracket
Engineering design Mathematical model 2:
plane stress

Difficult to solve by hand!

 ..General scenario..
Engineering design

Physical Problem

Mathematical model
Governed by differential
equations

Numerical model
e.g., finite element
model
 ..General scenario..
Engineering design Finite element analysis

PREPROCESSING
1. Create a geometric model
2. Develop the finite element model

Solid model Finite element model

 ..General scenario..
Engineering design Finite element analysis
FEM analysis scheme

Step 1: Divide the problem domain into non

overlapping regions (“elements”) connected to
each other through special points (“nodes”)
Element
Node

Finite element model

 ..General scenario..
Engineering design Finite element analysis

FEM analysis scheme

Step 2: Describe the behavior of each element

Step 3: Describe the behavior of the entire body by

putting together the behavior of each of the
elements (this is a process known as “assembly”)
 ..General scenario..
Engineering design Finite element analysis

POSTPROCESSING

Compute moment at section AA

 ..General scenario..
Engineering design Finite element analysis

Preprocessing
Step 1

Step 2
Analysis
Step 3

Postprocessing
 Example: A bracket
Engineering design Mathematical model 2:
plane stress
FEM solution to mathematical model 2 (plane stress)
Moment at section AA M  27 ,500 N c m
Deflection at load  at load W  0 . 064 c m

Conclusion: With respect to the questions we posed, the

beam model is reliable if the required bending moment is to be
predicted within 1% and the deflection is to be predicted within
20%. The beam model is also highly effective since it can be
solved easily (by hand).
What if we asked: what is the maximum stress in the bracket?
would the beam model be of any use?
 Example: A bracket
Engineering design Summary

1. The selection of the mathematical model

depends on the response to be
predicted.
2. The most effective mathematical model
is the one that delivers the answers to
the questions in reliable manner with
least effort.
3. The numerical solution is only as
accurate as the mathematical model.
 Example:
...GeneralAscenario
bracket
Modeling a physical
problem
Physical Problem Change
physical
problem

Mathematical Improve
Model mathematical
model

Numerical model

No!
Does answer
Refine analysis
make sense?

YES! Design improvements

Structural optimization
Happy 
 Verification
Example:and
A bracket
validation
Modeling a physical
problem

Physical Problem
Validation
Mathematical
Model
Verification

Numerical model
 Critical assessment of the FEM
Reliability:
For a well-posed mathematical problem the numerical
technique should always, for a reasonable discretization,
give a reasonable solution which must converge to the
accurate solution as the discretization is refined.
e.g., use of reduced integration in FEM results in an
unreliable analysis procedure.

Robustness:
The performance of the numerical method should not be
unduly sensitive to the material data, the boundary
conditions, and the loading conditions used.
e.g., displacement based formulation for incompressible
problems in elasticity

Efficiency:
Gaussian Elimination
 Naïve Gaussian Elimination
A method to solve simultaneous linear
equations of the form [A][X]=[C]

Two steps
1. Forward Elimination
2. Back Substitution
 Forward Elimination
The goal of forward elimination is to
transform the coefficient matrix into an
upper triangular matrix
 25 5 1  x1   106.8 
 64 8 1  x   177.2 
   2  
 144 12 1  x3   279.2

 25 5 1   x1   106.8 
 0  4.8  1.56  x    96.21
   2  
 0 0 0.7   x3   0.735 
 Forward Elimination
A set of n equations and n unknowns
a11 x1  a12 x2  a13 x3  ...  a1n xn b1
a21 x1  a22 x2  a23 x3  ...  a2 n xn b2
. .
. .
. .
an1 x1  an 2 x2  an 3 x3  ...  ann xn bn

(n-1) steps of forward

elimination
 Forward Elimination
Step 1
For Equation 2, divide Equation 1a11by and
multiply bya21 .

 a21 
 a  ( a11 x1  a12 x2  a13 x3  ...  a1n xn b1 )
 11 
a21 a21 a21
a21 x1  a12 x2  ...  a1n xn  b1
a11 a11 a11
 Forward Elimination
Subtract the result from Equation 2.
a21 x1  a22 x2  a23 x3  ...  a2 n xn b2
a21 a21 a21
− a21 x1  a a12 x2  ...  a a1n xn  a b1
11 11 11
_________________________________________________
 a21   a21  a21
 a22  a12  x2  ...   a2 n  a1n  xn b2  b1
 a11   a11  a11

' ' '

o a x  ...  a x b
22 2 2n n 2
r
 Forward Elimination
Repeat this procedure for the remaining
equations to reduce the set of equations
as a11 x1  a12 x2  a13 x3  ...  a1n xn b1
' '
a22 x2  a23 x3  ...  a2' n xn b2'
' '
a32 x2  a33 x3  ...  a3' n xn b3'
. . .
. . .
. . .

an' 2 x2  an' 3 x3  ...  ann

'
xn bn'

End of Step 1
 Forward Elimination
Step 2
Repeat the same procedure for the 3rd
term of Equation 3.
a11 x1  a12 x2  a13 x3  ...  a1n xn b1
' '
a22 x2  a23 x3  ...  a2' n xn b2'
"
a33 x3  ...  a3" n xn b3"
. .
. .
. .

an" 3 x3  ...  ann

"
xn bn"
End of Step 2
 Forward Elimination
At the end of (n-1) Forward Elimination steps, the
system of equations will look like
a11 x1  a12 x 2  a13 x3  ...  a1n x n b1
' '
a22 x2  a23 x3  ...  a2' n xn b2'
" " "
a x  ...  a x b
33 3 3n n 3
. .
. .
. .

n  1 n  1 
ann xn bn

End of Step (n-1)

 Matrix Form at End of Forward
Elimination
 a11 a12 a13  a1n   x1   b1 
 0 a '
a ' '
 a 2 n x2     b2' 
 22 23    
" " "
 0 0 a 33  a3n   x3   b3 
    
           
 0 0 0 0 ann   xn   bn 
(n  1 )
    (n-1 )
 Back Substitution
Solve each equation starting from the last
equation
 25 5 1   x1   106.8 
 0  4.8  1.56  x    96.21
   2  
 0 0 0.7   x3   0.735 

Example of a system of 3
equations
Back Substitution Starting Eqns
a11 x1  a12 x 2  a13 x3  ...  a1n x n b1
' '
a22 x2  a23 x3  ...  a2' n xn b2'
"
a33 x3  ...  an" xn b3"
. .
. .
. .

n  1 n  1 
ann xn bn
 Back Substitution
Start with the last equation because it has only one unknown
( n  1)
b n
xn  ( n  1)
a nn
Solve the linear system by Gauss elimination method.
Example 1.
x+y+z=5
2x + 3y + 5z = 8
4x + 5z = 2
Example 2
Consider the system of equations