lecture 2 - matrix algebra
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FINITE ELEMENT METHODFINITE ELEMENT METHOD(BDA 4033)
Lecture Module 2: Matrix Algebra
Dr. Waluyo Adi SiswantoUniversiti Tun Hussein Onn Malaysia
BDA 4033 Dr. Waluyo Adi Siswanto 2
Topics to Topics to DiscussDiscuss
Matrix definition and notationMatrix definition and notation Definition
Transpose
Symmetric Matrix
Unit Matrix
Matrix OperationsMatrix Operations Addition
Multiplication
Inverse
Orthogonal MatrixOrthogonal Matrix
Solution of Simultaneous EquationsSolution of Simultaneous Equations
BDA 4033 Dr. Waluyo Adi Siswanto 3
Matrix Definition
A matrix is an m x n array of numbers arranged in m rows and n columns.
Matrix is then described as being of order m x n.
[a]=[a11 a12 ⋯ a1n
a21 a22 ⋯ a2n
⋮ ⋮ . ⋮
am1 am2 ⋯ amn]
Row m
Column n
BDA 4033 Dr. Waluyo Adi Siswanto 4
Example Matrix Definition
[M ]=[ 5 3 −3 2−6 5 4 −19]Rectangular matrix
[C ]=[5 3 −35 4 −192 4 −3 ]Square matrix
[B ]=[ 5 3 −3 ]Row matrix
[F ]=[53
−3]={53
−3}Column matrix (VectorVector)
VectorVector
2 x 4
3 x 3
1 x 3
3 x 1
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Transpose Matrix
The transpose matrix is obtained by interchanging rows and columnsinterchanging rows and columns.
[a ij ]T=[ a ji]
[C ]=[5 3 −35 4 −192 4 −3 ] [C ]
T=[
5 5 23 4 4
−3 −19 −3]Diagonalunchanged
BDA 4033 Dr. Waluyo Adi Siswanto 6
Transpose Matrixin OO Spreadsheet
Prepare data
Prepare area for result
Insert functionTRANSPOSETRANSPOSESelect the array data
OKOKTo showThe result
BDA 4033 Dr. Waluyo Adi Siswanto 7
Symmetric Matrix
If a square matrix is equal to its transposeequal to its transpose,
It is called a symmetric matrix
[ a ]=[ a ]T
[C ]=[5 3 −33 4 8
−3 8 −3] [C ]T=[
5 3 −33 4 8
−3 8 −3][C ] is a symmetric matrix
BDA 4033 Dr. Waluyo Adi Siswanto 8
Unit Matrix
The unitunit (or IdentityIdentity) matrix [I][I] is such that
The unit matrix is always a square matrix of any possible order with each element of the main diagonal is one and each element of the main diagonal is one and all other elements equal to zeroall other elements equal to zero.
[a] [I ]=[ I ][a ]=[a ]
[I ]=[1 0 00 1 00 0 1]For example, the 3 x 3 unit matrix is given by
BDA 4033 Dr. Waluyo Adi Siswanto 9
Addition of Matrices
Matrices of the same order can be added together by summing corresponding elements of the matricessumming corresponding elements of the matrices.
Subtraction is performed in similar manner
Matrices of unlike order cannot be added or subtracted.
[cij ]=[a ij][bij ]=[bij ][a ij ]
[A]=[−1 2−3 2] [B ]=[ 1 2
3 1] [C ]=[A][B]=[0 40 3]
BDA 4033 Dr. Waluyo Adi Siswanto 10
Multiplication of Matrices
The number of the columns of the first matrix must be equal to the number of rows of the second matrix.
Matrix multiplication is not cumulativeis not cumulative [A][B]≠[B][A]
[cij ]=∑e=1
n
a iebej
[A]=[2 13 2] [B ]=[ 1 −1
2 0 ] [C ]=[A] [B]=[ 4 −27 −3]
BDA 4033 Dr. Waluyo Adi Siswanto 11
Multiplication of Matricesin OO Calc Spreadsheet
Prepare data[A] and [B]
Prepare area for result Insert function MMULTMMULTSelect the array data [A] and [B]
OKOKTo showThe result
BDA 4033 Dr. Waluyo Adi Siswanto 12
Inverse of a Matrix
The inverse of a matrix is a matrix such that
[A]−1
[A]=[A] [A]−1
=[I ]
[A]−1
=[C ]
T
∣A∣=adj [ A]
∣A∣
[C ]
∣A∣
: Cofactor of matrix
: Determinant of matrix
[A]
[A]
BDA 4033 Dr. Waluyo Adi Siswanto 13
Cofactor Matrix
[A]=[−1 3 22 −4 20 4 1]
[C ]=[∣−4 2
4 1∣ −∣2 20 1∣ ∣2 −4
0 4 ∣−∣3 −2
4 1 ∣ ∣−1 −20 1 ∣ −∣−1 3
0 4∣∣ 3 −2
−4 2 ∣ −∣−1 −22 2 ∣ ∣−1 3
2 −4∣]=[−12 −22 8−11 −1 4−2 −2 −2]
BDA 4033 Dr. Waluyo Adi Siswanto 14
Adjoint and Determinant Matrix
Continuing the previous result
adj [A]=[C ]T=[−12 −11 −2
−2 −1 −28 4 −2 ]
det [ A]=∣A∣=−11∣−4 24 1∣3−1∣2 2
0 1∣−21∣2 −40 4 ∣=−10
BDA 4033 Dr. Waluyo Adi Siswanto 15
Determinant of a Matrixin OO Calc Spreadsheet
Prepare data [A] and place for result
Insert function MDETERM MDETERM Select the array data [A]
OKOKTo showThe result
BDA 4033 Dr. Waluyo Adi Siswanto 16
Inverse of a Matrixin OO Calc Spreadsheet
Prepare data [A]
Insert function MINVERSE MINVERSE Select the array data [A]
OKOKTo showThe result
Prepare area for result
BDA 4033 Dr. Waluyo Adi Siswanto 17
Orthogonal Matrix
A matrix [A] is an orthogonal matrix if
[A]−1=[ A]T
[A]=[ cos 60deg sin 60deg−sin 60deg cos 60deg]=[ 0.5 0.87
−0.87 0.5 ]
[A]T=[ 0.5 −0.87
0.87 0.5 ] [A]−1
=[ 0.5 −0.870.87 0.5 ]
Matrix [A] is an orthogonal matrix
BDA 4033 Dr. Waluyo Adi Siswanto 18
Solution Methodsof Simultaneous Linear Equations
Representation of linear equations in Matrix Matrix Inversion Method Classical Methods
Cramer's Rule Method Gaussian Elimination Method Gauss-Siedel Iteration Method
BDA 4033 Dr. Waluyo Adi Siswanto 19
Linear Equations in Matrix
a11 x1a12 x2⋯a1n xn=c1
a21 x1a22 x2⋯a2n xn=c2
⋮an1 x1an2 x2⋯ann xn=cn
The equation can be written in Matrix
[a11 a12 ⋯ a1n
a21 a22 ⋯ a2n
⋮ ⋮ . ⋮an1 an2 ⋯ ann
]{x1
x2
⋮xn
}={c1
c2
⋮cn
}[K ]{x }={F }or {x }=[K ]
−1{F }then
BDA 4033 Dr. Waluyo Adi Siswanto 20
Example Problem 2.1
a. Write the simultaneous equation in Matrixb. Find the unknown variables x1, x2 and x3 from the following equations
−x13 x2−2 x3=22 x1−4 x22 x3=1
4 x2 x3=3
[−1 3 −22 −4 20 4 1 ]{
x1
x2
x3}={
213}
[K ]{x }={F }or {x }=[K ]−1 {F }then
BDA 4033 Dr. Waluyo Adi Siswanto 21
Solving Simultaneous Equationin OO Calc Spreadsheet
{x1
x2
x3}={
4.11.1
−1.4}The result:
BDA 4033 Dr. Waluyo Adi Siswanto 22
Cramer's Rule Method
[−1 3 −22 −4 20 4 1 ]{
x1
x2
x3}={
213}
x1=
∣2 3 −21 −4 23 4 1 ∣
∣−1 3 22 −4 20 4 1∣
=4.1
x2=
∣−1 2 −22 1 20 3 1 ∣
∣−1 3 22 −4 20 4 1∣
=1.1 x3=
∣−1 3 22 −4 10 4 3∣
∣−1 3 22 −4 20 4 1∣
=1.4
BDA 4033 Dr. Waluyo Adi Siswanto 23
Gauss Elimination
Gauss Elimination is based on triangularisation of the coefficient matrix and evaluation of the unknowns by back-substitution starting from the last equation
The form can be summarised in general form by
a ij=aij−akjaikakk
k=1,2,⋯, n−1i=k1,⋯ , nj=k ,⋯, n1
x i=1a ii ai , n1− ∑
r=i1
n
a ir xr
BDA 4033 Dr. Waluyo Adi Siswanto 24
Example Problem 2.2
2 x12 x21 x3=92 x11 x2=4
1 x11 x21 x3=6
a. Find the unknown variables x1, x2 and x3 from the following equations, using Gauss Elimination Methodb. Verify the result in Speadsheet
n=3k=1,2i=2,3j=1,2,3,4
Since the order of matrix is n=3, the index for solving this problem
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The running index will be (for n=3)k = 1
i = 2j = 1 to 4
Calculate aij
i.e. a21
a22
a23
a24
i = 3
j = 1 to 4Calculate a
ij
i.e. a31
a32
a33
a34
k = 2i = 3
j = 2 to 4Calculate a
ij
i.e. a32
a33
a34
k=1,2,⋯, n−1i=k1,⋯ , nj=k ,⋯, n1
BDA 4033 Dr. Waluyo Adi Siswanto 26
The running index will be (for n=3)k = 1
i = 2j = 1 to 4
Calculate aij
i.e. a21
a22
a23
a24
a ij=aij−akjaikakk
[a]=[2 2 12 1 01 1 1]
a21=a21−a11
a21
a11
=2−222=0
a22=a22−a12
a21
a11
=1−222=−1
a23=a23−a13
a21
a11
=0−122=−1
a24=a24−a14
a21
a11
=4−9 22=−5
BDA 4033 Dr. Waluyo Adi Siswanto 27
i = 3
j = 1 to 4Calculate a
ij
i.e. a31
a32
a33
a34
a ij=aij−akjaikakk
[a]=[2 2 12 1 01 1 1]
a31=a31−a11
a31
a11
=1−212=0
a32=a32−a12
a31
a11
=1−212=0
a33=a33−a13
a31
a11
=1−112=0.5
a34=a34−a14
a31
a11
=6−912=1.5
BDA 4033 Dr. Waluyo Adi Siswanto 28
k = 2i = 3
j = 2 to 4Calculate a
ij
i.e. a32
a33
a34
a ij=aij−akjaikakk
[a]=[2 2 12 1 01 1 1]
a32=a32−a22
a32
a22
=0−−10
−1=0
a33=a33−a23
a32
a22
=0.5−−10
−1=0.5
a34=a34−a24
a32
a22
=1.5−−50
−1=1.5
BDA 4033 Dr. Waluyo Adi Siswanto 29
x i=1a ii ai , n1− ∑
r=i1
n
a ir xr
x3=1a33
a34−0 =1
0.51.5=3
x2=1a22
a24−a23 x3=1
−1[−5−−13]=2
x1=1a11
a14−a12 x2−a13 x3 =12[9−2 2−13]=1
BDA 4033 Dr. Waluyo Adi Siswanto 30
Verification using inverse matrix in OO Spreadsheet
BDA 4033 Dr. Waluyo Adi Siswanto 31
Gauss Siedel Iteration Method
Gauss-Siedel method is based on iterative approach
x1=1a11
c1−a12 x2−a13 x3−⋯−a1n xn
x2=1a22
c2−a21 x1−a23 x3−⋯−a2n xn
⋮
xn=1ann
cn−an1 x1−an2 x2−⋯−an , n−1 xn−1
A good initial guess x i=cia ii
BDA 4033 Dr. Waluyo Adi Siswanto 32
Example Problem 2.3
4x1− x2=2−x14x2−x3=5−x24x3−x4=6−x32x4=−2
a. Find the unknown variables x1, x2 and x3 from the following equations, using Gauss-Siedel Iteration approachb. Verify the result in Speadsheet
BDA 4033 Dr. Waluyo Adi Siswanto 33
A good initial guess x i=cia ii
[a]=[4 −1 0 0
−1 4 −1 00 −1 4 −10 0 −1 2
] {c }={256
−2} x1=0.5 x2=1 x3=1 x4=−1
x1=142 x2=0.75
x2=145x1x3=1.68
x3=146x2x4=1.672
x4=12−2 x3=−0.16
First trial Second trial
x1=1421.68=0.992
x2=1450.9921.672=1.899
x3=1461.899±0.16=1.944
x4=12−21.944=−0.028
BDA 4033 Dr. Waluyo Adi Siswanto 34
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