##Ejercicio 31
a1<-matrix(data=c(2,1,3,-9),
ncol = 2,nrow = 2,byrow = T) ;a1
## [,1] [,2]
## [1,] 2 1
## [2,] 3 -9
a1.1<-matrix(data=c(5,2), ncol = 1, byrow = T);a1.1
## [,1]
## [1,] 5
## [2,] 2
z<-solve(a = a1, b = a1.1)
z
## [,1]
## [1,] 2.2380952
## [2,] 0.5238095
#library(MASS)
#fractions(z)
#X= 47/21
#Y= 11/21
#Prueba A^-1*A = I
ai<-solve(a1)
ai%*%a1
## [,1] [,2]
## [1,] 1 -1.665335e-16
## [2,] 0 1.000000e+00
###########################
b1<-matrix(data=c(-8,-5,-2,7),
ncol = 2,nrow = 2,byrow = T) ;b1
## [,1] [,2]
## [1,] -8 -5
## [2,] -2 7
b1.1<-matrix(data=c(4,10), ncol = 1, byrow = T);b1.1
## [,1]
## [1,] 4
## [2,] 10
zb<-solve(a = b1, b = b1.1)
#fractions(zb)
#X= -13/11
#Y= 12/11
#Prueba A^-1*A = I
bi<-solve(b1)
bi%*%b1
## [,1] [,2]
## [1,] 1 0
## [2,] 0 1
#################################3
c1<-matrix(data=c(12,-5,0,-3,4,7,6,2,3),
ncol = 3,nrow = 3,byrow = T) ;c1
## [,1] [,2] [,3]
## [1,] 12 -5 0
## [2,] -3 4 7
## [3,] 6 2 3
c1.1<-matrix(data=c(11,-3,22), ncol = 1, byrow = T);c1.1
## [,1]
## [1,] 11
## [2,] -3
## [3,] 22
zc<-solve(a = c1, b = c1.1)
#fractions(zc)
#X= 3
#Y= 5
#X3=-2
#Prueba A^-1*A = I
ci<-solve(c1)
ci%*%c1
## [,1] [,2] [,3]
## [1,] 1.000000e+00 -5.551115e-17 0.000000e+00
## [2,] 0.000000e+00 1.000000e+00 -1.110223e-16
## [3,] 2.220446e-16 5.551115e-17 1.000000e+00
###############################
d1<-matrix(data=c(6,-3,4,12,5,-7,-5,2,6),
ncol = 3,nrow = 3,byrow = T) ;d1
## [,1] [,2] [,3]
## [1,] 6 -3 4
## [2,] 12 5 -7
## [3,] -5 2 6
d1.1<-matrix(data=c(41,-26, 14), ncol = 1, byrow = T);d1.1
## [,1]
## [1,] 41
## [2,] -26
## [3,] 14
zd<-solve(a = d1, b = d1.1)
#fractions(zd)
#X= 2
#Y= -3
#X3=-5
#Prueba A^-1*A = I
di<-solve(d1)
di%*%d1
## [,1] [,2] [,3]
## [1,] 1.000000e+00 -4.640385e-17 5.551115e-17
## [2,] -2.220446e-16 1.000000e+00 -1.110223e-16
## [3,] 0.000000e+00 -5.551115e-17 1.000000e+00
############
###Ejercicio 32
#A(BC+A)=B
#BC+A=B(A^-1)
#BC=B(A^-1)-A
#C=(B(A^-1)-A)(B^-1)
#C=(A^-1)-A(B^-1)
A<-matrix(data=c(3,9,-2,4),
ncol = 2,nrow = 2,byrow = T) ;A
## [,1] [,2]
## [1,] 3 9
## [2,] -2 4
Ai<-solve(A);Ai
## [,1] [,2]
## [1,] 0.13333333 -0.3
## [2,] 0.06666667 0.1
B<-matrix(data=c(2,-3,7,6),
ncol = 2,nrow = 2,byrow = T) ;B
## [,1] [,2]
## [1,] 2 -3
## [2,] 7 6
Bi<-solve(B);Bi
## [,1] [,2]
## [1,] 0.1818182 0.09090909
## [2,] -0.2121212 0.06060606
C<-(Ai)-(A%*%(Bi));C
## [,1] [,2]
## [1,] 1.496970 -1.11818182
## [2,] 1.278788 0.03939394
###Ejercicio 33
a1<-matrix(data=c(-2,1,-2,1),
ncol = 2,nrow = 2,byrow = T) ;a1
## [,1] [,2]
## [1,] -2 1
## [2,] -2 1
a1.1<-matrix(data=c(-5,3), ncol = 1, byrow = T);a1.1
## [,1]
## [1,] -5
## [2,] 3
det(a1)
## [1] 0
#solve(a1)
##Al ser su determinante cero no tiene solución
b1<-matrix(data=c(-2,1,-8,4),
ncol = 2,nrow = 2,byrow = T) ;b1
## [,1] [,2]
## [1,] -2 1
## [2,] -8 4
b1.1<-matrix(data=c(3,12), ncol = 1, byrow = T);b1.1
## [,1]
## [1,] 3
## [2,] 12
det(b1)
## [1] 0
##Al ser su determinante cero no tiene solución
c1<-matrix(data=c(-2,1,-2,1),
ncol = 2,nrow = 2,byrow = T) ;c1
## [,1] [,2]
## [1,] -2 1
## [2,] -2 1
c1.1<-matrix(data=c(-5,-5.000001), ncol = 1, byrow = T);c1.1
## [,1]
## [1,] -5.000000
## [2,] -5.000001
det(c1)
## [1] 0
##Al ser su determinante cero no tiene solución
####Ejercicio 34
conversionpm<-function(pies=0,metros=0){
m<-paste(round(pies*0.3048,digits = 3),"metros")
p<-paste(round(metros/0.3048,digits = 3),"pies")
if(pies == 0) return(p)
else(return(m))
}
conversionpm(pies=24)
## [1] "7.315 metros"
conversionpm(metros=65)
## [1] "213.255 pies"
conversionln<-function(libras=0,newtons=0){
n<-paste(round(libras*4.4482216,digits = 3),"newtons")
l<-paste(round(newtons/4.4482216,digits = 3),"libras")
if(libras == 0) return(l)
else(return(n))
}
conversionln(newtons = 18)
## [1] "4.047 libras"
conversionln(libras= 5)
## [1] "22.241 newtons"
conversionks<-function(kg=0,slugs=0){
n<-paste(round(kg*0.0685218,digits = 3),"slugs")
l<-paste(round(slugs/0.0685218,digits = 3),"kg")
if(kg == 0) return(l)
else(return(n))
}
conversionks(kg=15)
## [1] "1.028 slugs"
conversionks(slugs = 6)
## [1] "87.563 kg"
########## Ejercicio 36
temp<-function(f=0,C=0){
ce<-paste(round((5/9)*(f-32),digits = 3),"°C")
fa<-paste(round(((1.8)*C)+32,digits = 3),"°F")
if(f == 0) return(fa)
else(return(ce))
}
temp(C = 150)
## [1] "302 °F"
##Ejercicio 37
t<-function(h,v0,g=9.81){
tiempo1<-(-v0+sqrt(v0^2-(4*(-0.5*g*-h))))/-g
tiempo2<-(-v0-sqrt(v0^2-(4*(-0.5*g*-h))))/-g
return(list(tiempo1,tiempo2))
}
t(h=100, v0=50)
## [[1]]
## [1] 2.732434
##
## [[2]]
## [1] 7.461246
##El valor mas bajo es el tiempo necesario para alcanzar la altura
#Mientras asciende y el otro mientras desciende
##Ejercicio 38
x4<-function(w,y){
m<-matrix(nrow = 4,ncol = 4)
for(i in 1:4){
m[i,1]=w[i]^3
m[i,2]=w[i]^2
m[i,3]=w[i]^1
m[i,4]=w[i]^0
}
r<-matrix(data=y,nrow = 4,ncol = 1);r
solve(a = m,b = r)
}
##Sea el vector w, el vector de las x
##Sea el vector y, el vecttor de las y
x4(w=c(-2,0,2,4),y=c(-20,4,68,508))
## [,1]
## [1,] 7
## [2,] 5
## [3,] -6
## [4,] 4
##Ejercicio 39
plot(function(x)(10*exp(1))^-(2*x),xlim = c(0,2))
### Ejercicio 40
plot(function(x)((20*(x^2))-(200*x)+3),xlim = c(-20,20), ylim=c(-500,1000))
##Utilizando la primera derivada e igulanado a cero, encontramos que
##El minimo es cuando x= 5
points(x=5,y=-497)
##Ejercicio 41
x<-6
z=(x<10);z
## [1] TRUE
z=(x==10);z
## [1] FALSE
z=(x>=4);z
## [1] TRUE
##Ejercicio 42
x<-c(3,0,0,2,6,8)
y<-c(-5,-2,0,3,4,10)
p<-which(x>y)
x[p]
## [1] 3 0 6
###Ejercicio 43
price<-c(19,18,22,21,25,19,17,21,27,29)
p<-which(price>20);p
## [1] 3 4 5 8 9 10
10*length(p)
## [1] 60
##60 días
## Ejercicio 44
price_A<-c(19,18,22,21,25,19,17,21,27,29)
price_B<-c(22,17,20,19,24,28,16,25,28,27)
p<-which(price_A>price_B);p
## [1] 2 3 4 5 7 10
10*length(p)
## [1] 60
## 60 días
##Ejercicio 45
price_A<-c(19,18,22,21,25,19,17,21,27,29)
price_B<-c(22,17,20,19,24,28,16,25,28,27)
price_C<-c(17,13,22,23,19,17,20,21,24,28)
##a
p1<-which(price_A>price_B & price_A>price_C);p1
## [1] 2 5 10
price_A[p]
## [1] 18 22 21 25 17 29
10*length(p)
## [1] 60
## 30 días
p<-which(price_A>price_B | price_A>price_C);p
## [1] 1 2 3 4 5 6 7 9 10
10*length(p)
## [1] 90
##| operador o
p<-which(price_A>price_B | price_A>price_C);p
## [1] 1 2 3 4 5 6 7 9 10
10*length(p)
## [1] 90
#pos 1,3,4,6,7,10