Algebra Lineal
1.1
2008/09/18 09:02:16.837 GMT-5
2008/09/18 10:29:04.600 GMT-5
Daniel Felipe
Gonzalez Obando
danyfel80@gmail.com
Daniel Felipe
Gonzalez Obando
danyfel80@gmail.com
Algebra
Algebra Lineal
Solucionario de Algebra
Solucionario de ejercicios de algebra lineal
Ejercicios 4.2
n-vectores.
1. Determine u+v,u-v,2u,3u-2v si
a)
u
=
(
1
,
2
,
-
3
)
,
v
=
(
0
,
1
,
-
2
)
b)
u
=
(
4
,
-
2
,
1
,
3
)
,
v
=
(
-
1
,
2
,
5
,
-
4
)
Soluci'on
a)
u = vector([1,2,-3])
v = vector([0,1,-2])
redsage]
blueu+v
1
,
3
,
-
5
redsage]
blueu-v
1
,
1
,
-
1
redsage]
blue2*u
2
,
4
,
-
6
redsage]
blue3*u-2*v
3
,
4
,
-
5
b)
redsage]
blueu = vector([4,-2,1,3])
redsage]
bluev = vector([-1,2,5,-4])
redsage]
blueu+v
3
,
0
,
6
,
-
1
redsage]
blueu-v
5
,
-
4
,
-
4
,
7
redsage]
blue2*u
8
,
-
4
,
2
,
6
redsage]
blue3*u-2*v
14
,
-
10
,
-
7
,
17
3. Sean
u
=
(
1
,
-
2
,
3
)
,
v
=
(
-
3
,
-
1
,
3
)
,
w
=
(
a
,
-
1
,
b
)
,
x
=
(
3
,
c
,
2
)
determine a, b y c de modo que
a)
w
=
1
2
u
b)
w
+
v
=
u
c)
w
-
u
=
v
Solucion.
Sabiendo las propiedades de la suma, resta y multiplicacion de matrices sabemo
que para
a)
w
=
1
2
u
(
a
,
-
1
,
b
)
=
1
2
(
1
,
-
2
,
3
)
(
a
,
-
1
,
b
)
=
(
1
2
(
1
)
,
1
2
(
-
2
)
,
1
2
(
3
)
)
(
a
,
-
1
,
b
)
=
(
1
2
,
-
1
,
3
2
)
por lo tanto, podemos decir que
a
=
1
2
y
b
=
3
2
b)
w
+
v
=
u
(
a
,
-
1
,
b
)
+
(
-
3
,
-
1
,
3
)
=
(
1
,
-
2
,
3
)
(
a
+
(
-
3
)
,
-
1
+
(
-
1
)
,
b
+
3
)
=
(
1
,
-
2
,
3
)
eso es
a
-
3
=
1
b
+
3
=
3
por lo tanto
a
=
4
b
=
0
c)
w
+
x
=
v
(
a
,
-
1
,
b
)
+
(
3
,
c
,
2
)
=
(
-
3
,
-
1
,
3
)
(
a
+
3
,
-
1
+
c
,
b
+
2
)
=
(
-
3
,
-
1
,
3
)
eso es
a
+
3
=
-
3
c
-
1
=
-
1
b
+
2
=
3
por lo tanto
a
=
-
6
c
=
0
b
=
1
5. Sean
u
=
(
4
,
5
,
-
2
,
3
)
,
v
=
(
3
,
-
2
,
-
4
,
1
)
,
x
=
(
-
3
,
2
-
5
,
3
)
,
c
=
2
y
d
=
3
Verifique las propiedades (a) a (h) del teorema 4.2
Solucion
a)
u
+
v
=
v
+
u
redsage]
blueu=vector([4,5,-2,3])
redsage]
bluev=vector([3,-2,-4,1])
redsage]
blueu+v
7
,
3
,
-
6
,
4
redsage]
bluev+u
7
,
3
,
-
6
,
4
redsage]
blueu+v==v+u
1
b)
u
+
(
v
+
x
)
=
(
u
+
v
)
+
x
redsage]
blueu = vector([4,5,-2,3])
redsage]
bluev = vector([3,-2,-4,1])
redsage]
bluex = vector([-3,2,-5,3])
redsage]
blueu+(v+x)
4
,
5
,
-
11
,
7
redsage]
blue(u+v)+x
4
,
5
,
-
11
,
7
c)
Existe
un
vector
0
en
R
n
tal
que
u
+
0
=
0
+
u
=
u
para
toda
u
en
R
n
.
redsage]
blueu = vector([4,5,-2,3])
redsage]
bluea=0
redsage]
blueu+0
4
,
5
,
-
2
,
3
redsage]
blue0+u
4
,
5
,
-
2
,
3
redsage]
blueu
4
,
5
,
-
2
,
3
redsage]
blueu+0==0+u
1
d)
para
cada
vector
u
en
R
n
,
existe
un
vector
-
u
en
R
n
tal
que
u
+
(
-
u
)
=
0
.
redsage]
blueu = vector([4,5,-2,3])
redsage]
blueu
4
,
5
,
-
2
,
3
redsage]
blue-u
-
4
,
-
5
,
2
,
-
3
redsage]
blueu+(-u)
0
,
0
,
0
,
0
redsage]
blueu+(-u)==0
1
e)
c
(
u
+
v
)
=
c
u
+
v
u
redsage]
blueu = vector([4,5,-2,3])
red
bluev = vector([3,-2,-4,1])
red
bluec=2
red
bluec*(u+v)
14
,
6
,
-
12
,
8
red
bluec*u+c*v
14
,
6
,
-
12
,
8
red
bluec*(u+v)==c*u+c*v
1
f)
(
c
+
d
)
u
=
cu
+
du
red
blueu = vector([4,5,-2,3])
red
bluec=2
red
blued=3
red
blue(c+d)*u
20
,
25
,
-
10
,
15
red
bluec*u+d*u
20
,
25
,
-
10
,
15
red
blue(c+d)*u==c*u+d*u
1
g)
c
(
du
)
=
(
cd
)
u
red
blueu = vector([4,5,-2,3])
red
bluec=2
red
blued=3
red
bluec*(d*u)
24
,
30
,
-
12
,
18
red
blue(c*d)*u
24
,
30
,
-
12
,
18
red
bluec*(d*u)==(c*d)*u
1
h)
1
u
=
u
red
blueu = vector([4,5,-2,3])
red
blue1*u
4
,
5
,
-
2
,
3
red
blue1*u==u
1