Download Chapter 8

Part 3 Chapter 8
Linear Algebraic
Equations and Matrices
1
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Review of Matrix Algebra
Concepts
• Let’s take a look at a problem that can be
formulated using matrices
2
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Suppose there are three
jumpers connected by
bungee cords, so that each
cord is fully extended, but
unstretched.
After they are released,
gravity takes hold and the
jumpers will eventually
come to equilibrium
Suppose you are asked to
compute the displacement
of each of the jumpers –
that is x1, x2, x3.
3
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Free body diagrams
Assume that
each cord
behaves like a
linear spring,
and follows
Hooke’s Law
4
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Force Balance
d 2 x1
m1 2  m1 g  k2 ( x2  x1 )  k1 x1  0
dt
d 2 x2
m2 2  m2 g  k3 ( x3  x2 )  k2  x1  x2   0
dt
d 2 x3
m3 2  m3 g  k3 ( x2  x3 )  0
dt
Since the jumpers are at equilibrium the
acceleration is equal to 0
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5
Rearrange
to give
3 equations
and…..
3 unknowns
m1 g  k2 ( x2  x1 )  k1 x1  0
m2 g  k3 ( x3  x2 )  k2  x1  x2   0
m3 g  k3 ( x2  x3 )  0
6
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Which
cansystem
be solved
using matrix
Linear
of equations
algebra
 k1  k2  x1  k2 x2  m1 g
k2 x1   k2  k3  x2  k3 x3  m2 g
k3 x2  k3 x3  m3 g
7
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Matrix Algebra
Applied Force
Chapter 9 in
MATLAB for
Engineers
20 feet
θ = 600
Pivot Point
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F = 200 lbf
In this chapter we’ll learn how to…
• perform the basic operations of matrix
algebra
• solve simultaneous equations using
MATLAB matrix operations
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The difference between an array
and a matrix
• Most engineers use the two terms
interchangeably
• The only time you need to be concerned
about the difference is when you perform
matrix algebra calculations
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Arrays
• Technically an array is an orderly grouping
of information
– Arrays can contain numeric information, but
they can also contain character data,
symbolic data etc.
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Matrix
• The technical definition of a matrix is a twodimensional numeric array used in linear algebra
• Not even all numeric arrays can precisely be
called matrices - only those upon which you
intend to perform linear transformations meet
the strict definition of a matrix.
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Matrix Algebra
• used extensively in engineering
applications
• Matrix algebra is different from the array
calculations we have performed thus far
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Array Operators
• A.* B
multiplies each element in
array A times the
corresponding element in
array B
• A./B
divides each element in
array A by the corresponding
element in array B
• A.^B
raises each element in array
A to the power in the
corresponding element of
array B
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Operators used in Matrix
Mathematics
•
•
•
•
•
Transpose
Multiplication
Division
Exponentiation
Left Division
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Some Matrix Algebra functions
•
•
•
•
Dot products
Cross products
Inverse
Determinants
Covered in
Applied Numerical
Methods
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Transpose
• In mathematics texts you will often see the
transpose indicated with superscript T
– AT
• The MATLAB syntax for the transpose is
– A'
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The transpose switches the rows
and columns
1 2 3
4 5 6

A
7 8 9


10 11 12
1 4 7 10
AT  2 5 8 11
3 6 9 12
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Using the transpose with complex
numbers
When used with complex
numbers, the transpose
operator returns the complex
conjugate
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Dot Products
• The dot product is sometimes called the
scalar product
• the sum of the results when you multiply
two vectors together, element by element.
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* *
*
||
||
||
+
+
Equivalent
statements
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Example 9.1
MATLAB for Engineers
Calculating the Center of Gravity
(Center of Mass)
• Finding the center of gravity of a structure
is important in a number of engineering
applications
• The location of the center of gravity can be
calculated by dividing the system up into
small components.
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xW  x1W1  x2W2  x3W3  etc...
yW  y1W1  y2W2  y3W3  etc...
zW  z1W1  z2W2  z3W3  etc...
• In a rectangular coordinate system
– x , y , and z are the coordinates of the center of gravity
– W is the total mass of the system
– x1, x2, and x3 etc are the x coordinates of each system component
– y1, y2, and y3 etc are the y coordinates of each system component
– z1, z2, and z3 etc are the z coordinates of each system component
and
– W1, W2, and W3 etc are the masses of each system component
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In this example…
• We’ll find the center of gravity of a small
collection of the components used in a
complex space vehicle
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Vehicle Component
Locations and Mass
Item
x, meters y, meters
z meters
Mass
Bolt
0.1
2
3
3.50 gram
screw
1
1
1
1.50 gram
nut
1.5
0.2
0.5
0.79 gram
bracket
2
2
4
1.75 gram
Formulate the problem using a dot product
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Input and Output
• Input
– Location of each component in an
x-y-z coordinate system – in meters
– Mass of each component, in grams
• Output
– Location of the center of gravity
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Hand Example
Find the x coordinate of the center of gravity
Item
Bolt
screw
nut
bracket
sum
x, meters
0.1
1
1.5
2
Mass,
gram
x 3.50
x 1.50
x * m, gram
meters
= 0.35
= 1.50
x 0.79
x 1.75
7.54
= 1.1850
= 3.5
6.535
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We know that…
• The x coordinate is equal to
3
x
xm
i 1
i
mTotal
i
3

xm
i 1
3
i
i
m
i 1
i1
• So…
• x =6.535/7.54 = 0.8667 meters
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This is a dot
product
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We could use a plot to evaluate our
results
Center of Gravity
4
z-axis
3
This plot
was created
using the
interactive
plotting
tools
Center of Gravity
2
1
0
2
2
1
y-axis
1
0
0
x-axis
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Matrix Multiplication
• Similar to a dot product
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• Matrix multiplication results in an array
where each element is a dot product.
• In general, the results are found by taking
the dot product of each row in matrix A
with each column in Matrix B
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N
Ci , j   Ai ,k Bk , j
k 1
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• Because matrix multiplication is a
series of dot products
– the number of columns in matrix A
must equal the number of rows in
matrix B
– For an mxn matrix multiplied by an
nxp matrix
These dimensions must match
mxn
nxp
The resulting matrix will have
these dimensions
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We could use matrix
multiplication to solve the
problem in Example 9.1, in a
single step
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Matrix Powers
• Raising a matrix to a power is equivalent to
multiplying it times itself the requisite number of
times
– A2 is the same as A*A
– A3 is the same as A*A*A
• Raising a matrix to a power requires it to have the
name number of rows and columns
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Matrix Inverse
• MATLAB offers two approaches
– The matrix inverse function
• inv(A)
– Raising a matrix to the -1 power
• A-1
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Equivalent
approaches to
finding the
inverse of a
matrix
A matrix times its
inverse is the
identity matrix
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Not all matrices have an inverse
• Called
– Singular
– Ill-conditioned matrices
• Attempting to take the inverse of a singular
matrix results in an error statement
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Determinants
• Related to the matrix inverse
• If the determinant is equal to 0, the matrix
does not have an inverse
• The MATLAB function to find a
determinant is
– det(A)
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We’ll talk about
determinants
more in the next
chapter of
Applied
Numerical
Methods
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Cross Products
• sometimes called vector products
• the result of a cross product is a vector
• always at right angles (normal) to the
plane defined by the two input vectors
– orthogonality
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Consider two vectors



A  Ax i  Ay j  Az k



B  Bx i  B y j  Bz k
The cross product is equal to…



A  B  ( Ay * Bz  Az * B y )i  ( Az * Bx  Ax * Bz ) j  ( Ax B y  Ay Bx )k
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i

 Ax
 Bx

j
Ay
By
k

Az 
Bz 
 j

 Ay
 By

k
Az
Bz
i 

Ax 
Bx 
k

 Az
 Bz

i
Ax
Bx
j

Ay 
By 



A  B  ( Ay * Bz  Az * B y )i  ( Az * Bx  Ax * Bz ) j  ( Ax B y  Ay Bx )k
45
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Cross Products are Widely Used
• Cross products find wide use in statics,
dynamics, fluid mechanics and electrical
engineering problems
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Solutions to Systems of Linear
Equations
3x 2 y
x
x
 z  10
3y 2 z  5
 y  z  1
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Using Matrix Nomenclature
 3 2  1
 x
A   1 3 2  X   y 
 z 
 1  1  1
10 
B   5 
 1
and
AX=B
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Symbolically
 a11 a12
A   a21 a22
 a31 a32
a13 
 x1 
 b1 
a23  X   x2  B  b2 
 b3 
 x3 
a33 
and
AX=B
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We can solve this problem using the matrix
inverse approach
This approach is easy
to understand, but its
not the most efficient
computationally
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Matrix left division
uses Gaussian
elimination, which
is much more
efficient, and less
prone to round-off
error
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Example 8.2
Applied Numerical Methods
Bungee Jumpers
53
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Linear System of 3 Equations
 k1  k2  x1
 k2 x2
 m1 g
 k2 x1   k2  k3  x2  k3 x3  m2 g
 k3 x2  k3 x3  m3 g
Jumper
Mass (kg)
Spring Constant Unstretched Cord
(N/m)
Length (m)
Top (1)
60
50
20
Middle (2)
70
100
20
Bottom (3)
80
50
20
54
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 150 100 0   x1  588.6 
 100 150 50  x   686.7 


 2 
 0
50 50   x3  784.8
55
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56
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Solution using the matrix
inverse strategy
Solution using left division
57
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Linear Algebraic Equations find
wide use in Engineering
58
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