Ejercicios
4
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Resolución
Transcript of Ejercicios
7/21/2019 Ejercicios
http://slidepdf.com/reader/full/ejercicios-56da5b6d6dfba 1/4
Determinar la matriz X tal que XA=B,Sí
A=
[
1 −3 11 −38
0 1 −2 7
0
0
0
0
1
0
−2
1
] y B=
[
1 −4 −3 1
1 −5 −3 −1
−1 6 4 2
]Solución:
XA=B
XA A−1=B A
−1
XI =B A−1
X =B A−1
Hallamos A−1
A−1=[
1 −3 11 −38
0 1 −2 7
0
0
0
0
1
0
−2
1|
1 0 0 0
0 1 0 0
0
0
0
0
1
0
0
1]
A−1=[
1 −3 11 0
0 1 −2 0
0
0
0
0
1
0
0
1|
1 0 0 38
0 1 0 −7
0
0
0
0
1
0
2
1 ]
A
−1=
[1 −3 0 0
0 1 0 0
0
0
0
0
1
0
0
1|1 0 −11 16
0 1 2 −3
0
0
0
0
1
0
2
1 ] A
−1
=[1 0 0 0
0 1 0 0
0
0
0
0
1
0
0
1|
1 3 −5 7
0 1 2 −3
0
0
0
0
1
0
2
1 ]
3f 2+f
11f 3
+f 1
38f 4
+f 1
-
7/21/2019 Ejercicios
http://slidepdf.com/reader/full/ejercicios-56da5b6d6dfba 2/4
A−1=[
1 3 −5 7
0 1 2 −3
0
0
0
0
1
0
2
1 ]
X =B A−1=[
1 −4 −3 1
1 −5 −3 −1
−1 6 4 2 ] x[
1 3 −5 7
0 1 2 −3
0
0
0
0
1
0
2
1 ]
X =[ 1 −1 −16 14
1 −2 −18 15
−1 3 21 −15]
Si A=[−1 −1 −1
0 1 0
0 0 1 ] , calcular A
29
Solución:
Calculamos A2= AxA=
[−1 −1 −1
0 1 0
0 0 1 ] x
[−1 −1 −1
0 1 0
0 0 1 ]
A2=[
1 0 0
0 1 0
0 0 1]
Como vemos la matriz A2 es una matriz identidad
(I)
Entonces: A29=( A 2 )14
x A
A29=( I
2 )14
x A
A29= I x A
A29
= I x A=
[
1 0 0
0 1 0
0 0 1
] x
[
−1 −1 −1
0 1 0
0 0 1
]
7/21/2019 Ejercicios
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A29=[
−1 −1 −1
0 1 0
0 0 1 ]
Si A=
[
0 −1 0
1 1 1
0 0 −1
], calcular A
100
Solución:
Calculamos A2= A x A=[
0 −1 0
1 1 1
0 0 −1] x [
0 −1 0
1 1 1
0 0 −1]
A2=
[−1 −1 −1
1 0 0
0 0 1
]Calculamos A
3= A
2 x A=[
−1 −1 −1
1 0 0
0 0 1 ] x [
0 −1 0
1 1 1
0 0 −1]
A3=[
−1 0 0
0 −1 0
0 0 −1]
A3=−[
1 0 0
0 1 0
0 0 1]
Como vemos A3 es una matriz identidad (I)
Entonces: A100=( A 3 )33
x A
A100
=(− I 3
)33
x A
7/21/2019 Ejercicios
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A100
=− I x A
A100
=−
[
1 0 0
0 1 0
0 0 1
] x
[
0 −1 0
1 1 1
0 0 −1
] A
100
=[−1 0 0
0 −1 0
0 0 −1] x [
0 −1 0
1 1 1
0 0 −1]
A100
=[ 0 1 0
−1 −1 −1
0 0 1 ]
