Dinámica De Sistemas Modelaje Y Simulación

Tarea: Dinámica de sistemas modelaje y simulación
Tarea #2

Fecha: 11 de abril de 2011

2.1.1) Using the component forms of vectors a, b, and c, show that Eq. 2.1.5 holds.
(a+ b)+ c= a+ (b+ c ) = a+ b+ c
a=i+j
b= -i-2j
c=i
a+ b+ c=i+j+ -i-2j+ i=i-j
2.1.2) Show that the component form of Eq. 2.1.15 is
θa, b= sgn-aybx+ axbyArcos(axbx+ aybyax2+ay2bx2+by2)
θa, b=sgna⫠bArcosa .bax2+ay2bx2+by2
θa, b=sgn-aybx+ axbyArcosaxbx+ aybyax2+ay2bx2+by2

Section 2.2
2.2.1) Use the definitions of equality and addition of matrices of equal dimension to show that Eq. 2.2.5 is valid.
Dos matrices son iguales si tienen el mismo tamaño y los mismos valores. Al sumar dos matrices del mismo tamaño se obtiene una tercera matriz de las mismas dimensiones.
Aij+ Bij= Cij
Por lo tanto:A+B+C=A+(B+C)
2.2.2) Show that Eq. 2.2.6 is valid.
Aij+ Bij= Bij+ Aij
Cij= Cij

2.2.3) Calculate the sum C=A+B, difference D=A-B, and product E=AB of the 3 x 3 matrices
C= A + B
C =110-10101-1+121130210=23103122-1
D= A – B
D=110-10101-1-121130210=0-1-1-2-31-20-1
E= AB
E=110-10101-1121130210=2501-1-106-1
2.2.4) Carry out the operations on both sides of Eqs. 2.2.12 and 2.2.13 with thematrices
A=1001, B=1001, C=1111
a) (A + B)C= AC + BC
(1001+1221)1111=10011111+12211111
22221111=1+01+00+10+1+1+21+22+12+1
1111=1+01+00+10+1+1+21+22+12+1
2+22+22+22+2=1111+3333
4444=4444
b) (AB)C= A(BC)
(10011221)1111=1001(12211111)
1+01+20+20+11111=10011+21+22+12+1
1+21+22+12+1=3+03+00+30+3
3333=3333
2.2.5) Show that the expansion of A∝ as a linear combination of columns of A inEq. 2.2.17 is valid. Similarly verify that the expansion in Eq. 2.2.18 is valid.
Considerando una matriz A=[a1,a2…,an]. Si una combinación lineal de las columnas ai de A es Zero.
A∝ = j=0nαjaj=0
Para α=[α1,α2,…,αn]T ≠0 entonces las columnas de A son linealmente dependientes de lo contrario son linealmente independientes.
Las filas di de A son linealmente dependientes si
BTA=i=0nBidi=0
ParaB=[B1,B2,…Bn]≠0 de otra forma es linealmente independiente.
2.2.6) Show that the Rank of matrix B of Example 2.2.5 is 2 by finding a 2x2 submatrix with nonzero determinant and showing that |B| = 0.
b1=110, b2=101, b3=12-1
B=[b1,b2,b3]
B=11110201-1
Submatriz 1110 det=1*0 - 1*1= -1
B11=(-1)2021-1= 0 - 2= -2
B12=(-1)3120-1= 1 - 0= 1
B13=(-1)41001= 1 - 0= 1
lBl= 1*(-2) + 1(1) + 1(1)= 02.2.7) Use Eqs. 2.2.16 and 2.2.19 to show that Eq. 2.2.20 is valid
AA-1= A-1A= I
Y ATA= AAT= I
A-1=AT
(A-1)T= (AT)-1
2.2.8) Show that ABB-1A-1= I, to verify that Eq. 2.2.21 is valid.
(AB)(B-1A-1)= A(BB-1)A-1= AIA-1= AA-1= I
(AB)-1= B-1A-1
2.2.9) Show that the following 3x3 matrix is orthogonal for any value of φ:
A=1000CosƟ-SenƟ0SenƟCosƟ
ATA=1000CosƟ-SenƟ0SenƟCosƟ1000CosƟSenƟ0-SenƟCosƟATA=10000+Cos2Ɵ+Sen2Ɵ0+CosƟSenƟ-CosƟSenƟ00+CosƟSenƟ-CosƟSenƟ0+Sen2Ɵ+Cos2Ɵ=100010001= I
2.2.10) Show that if nxn matrices A and B are orthogonal then A⫠B is orthogonal.
Section 2.3
2.3.1 Show that c of Eq. 2.3.3 is algebraic representation of c= a+b.
a=axay
b=bxby
c=[ax, ay]T + [bx, by]T
2.3.2 Show that aTa⫠= aTRa = 0:
aT= [ax, ay]
aT= axay
aTa˔= [ax, ay] axay0
aTRa=[ax, ay] axay=02.3.3 Show that RRa=R2a, RRRa=R3a˔ and RRRRa = R4a
RRa=R2a=-a
a=axay
Ra=-ayax
R=0-110
0-110-ayax=0-1100-110axay
0-110-ayax=-a= -axay
RRRa= R3a=-a˔
0-1100-110-ayax=0-1100-1100-110axay
0-1100-110-ayax=-a˔=-[ax, ay]
RRRRa=R4a=a
0-1100-1100-110-ayax=0-1100-1100-1100-110axay
0-1100-1100-110-ayax=a=axay
2.3.4 Show that (a+b)⫠=a⫠+ b⫠
(a+b)˔=a˔+b˔
(axay+bxby)˔=-ayax+-bybx-axay+-bxby=-ayax+-bybx
Section 2.4
2.4.1. Show that R=Aπ2 and, more generally, Rn=Anπ2, n=2,3,4.
R=0-110 Aφ=cosφ-sinφsinφcosφ
R=Aπ2
R=cosπ2-sinπ2sinπ2cosπ2
R≡ 0-110
Entonces queda de la siguiente manera:
R=Aπ2

Rn=Anπ2
Para n=2,3,4; queda:
R2=R∙R=0-110∙0-110= -100-1
R3=R2∙R=-100-1∙0-110= 01-10
R4=R3∙R=01-10∙0-110= 1001

Para n=2,3,4; para Anπ2 queda:
Aπ= cosπ-sinπsinπcosπ= -100-1
A3π2=...