PREPARACIÓN EXAMEN Orientaciones y ejercicios de autoexamen.
Ejercicios derecho a examen
20
A = 3 4 5 6 B = 6 8 -5 -7 |A| = 3(6) - 5(4) = 18 - 20 = -2 A -1 = 1 -2 6 -4 -5 3 = -3 2 5 2 - 3 2 |B| = 5(-7) - (-5)8 = -42 - (-40) = -2 B -1 = 1 -2 -7 -8 5 6 = 7 2 4 - 5 2 -3 3x 1 +4x 2 =3 5x 1 +6x 2 =7
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Transcript of Ejercicios derecho a examen
Ejercicos Unidad 3
HUMBERTO ROMERO RUBIO
November 23, 2016
1. Encuentre la inversa de
A =
[3 45 6
], B =
[6 8−5 −7
]Solución A:
|A| = 3(6)− 5(4)
= 18− 20
= −2
A−1 =1
−2
[6 −4−5 3
]=
[−3 252 − 3
2
]Solucion B:
|B| = 5(−7)− (−5)8= −42− (−40)= −2
B−1 =1
−2
[−7 −85 6
]=
[72 4− 5
2 −3
]
2. Use la inversa de la matriz A del ejercicio anterior para resolver el sistema3x1 + 4x2 = 35x1 + 6x2 = 7
.
1
Solución:
(x1
x2
)=
[−3 252 − 3
2
](37
)=
(−9+ 14152 −
212
)=
(5−3
)
Comprobación:
3x1 + 4x2 = 3
3(5) + 4(−3) = 3
15− 12 = 3
3 = 3
3. Dada la matriz A =
(3 −17 2
)hallar la matriz por la que hay que
premultiplicar A para obtener la matriz B =
(41 819 4
).
Solución: [3 −17 2
] [a bc d
]=
(41 819 4
)
3a− 1c = 41 3b− d = 8
7a+ 2c = 19 7b+ 2d = 4
3a− c = 41
(−7
3
)7a+ 2c = 19
2
−7a+7
3c = −287
37a+ 2c = 19
13
3c = −230
3
c =− 230
3133
c = −230
13
3a−(−230
13
)= 41
3a =303
13
a =30313
3
a =101
13
3b− d = 8
(−7
3
)7b+ 2d = 4
−7b+ 7
3d = −56
37b+ 2d = 4
13
3d = −44
3
d =− 44
3133
d = −44
13
3b−(−44
13
)= 8
b =20
13
3
por lo tanto la matriz a premultiplicar sera[10113
2013
− 23013 − 44
13
]por lo tanto [
3 −17 2
] [10113
2013
− 23013 − 44
13
]=
(41 819 4
)4. Dada la matriz
A =
14 0 1
40 1 014 0 1
4
hallar Andonde n ∈ N .
Solución:
A1=
14 0 1
40 1 014 0 1
4
A2=
14 0 1
40 1 014 0 1
4
14 0 1
40 1 014 0 1
4
=
216 0 2
160 1 0216 0 2
16
A3 =
216 0 2
160 1 0216 0 2
16
14 0 1
40 1 014 0 1
4
=
464 0 4
640 1 0464 0 4
64
A4=
464 0 4
640 1 0464 0 4
64
14 0 1
40 1 014 0 1
4
=
8256 0 8
2560 1 08
256 0 8256
An =
1n−1
4n 0 1n−1
4n
0 1 01n−1
4n 0 1n−1
4n
5. Calcular la inversa de las siguientes matrices por el método de Gauss-
Jordan.
A =
[−6 49 15
], B =
1 −4 02 3 −12 −3 1
, C =
0 −1 −21 −0 −34 −3 8
,D =
3240 −1669 −299 951328 −648 −116 37−217 106 19 −635 −17 −3 1
Solución A:
4
A =
[−6 4 1 09 15 0 1
]F1 → F1
(−1
6
)=
[1 − 2
3 − 16 0
9 15 0 1
]F2 → F2 − 9(F1)
=
[1 − 2
316 0
0 21 32 1
]F2 → F2
(1
21
)=
[1 − 2
3 − 16 0
0 1 114
121
]F1 → F1 +
2
3(F2)
=
[1 0 − 5
42263
0 1 114
121
]Por lo tanto la matriz inversa es:
A−1 =
[− 5
42263
114
121
]Solución B:
5
B =
1 −4 0 1 0 02 3 −1 0 1 02 −3 1 0 0 1
F2 → F2 − 2(F1)
=
1 −4 0 1 0 00 11 −1 −2 1 02 −3 1 0 0 1
F3 → F3 − 2(F1)
=
1 −4 0 1 0 00 11 −1 −2 1 00 5 1 −2 0 1
F2 → F2
(1
11
)
=
1 −4 0 1 0 00 1 − 1
11 − 211 1 0
0 5 1 −2 0 1
F1 → F1 + 4(F2)
=
1 0 − 411
311
411 0
0 1 − 111 − 2
11111 0
0 5 1 −2 0 1
F3 → F3 − 5(F2)
=
1 0 − 411
311
411 0
0 1 − 111 − 2
11111 0
0 0 1611 − 12
11 − 511 1
F3 → F3
(11
16
)
=
1 0 − 411
311
411 0
0 1 − 111 − 2
11111 0
0 0 1 − 34 − 5
161116
F1 → F1 +4
11(F3)
=
1 0 0 0 14
45
0 1 − 111 − 2
11111 0
0 0 1 − 34 − 5
161116
F2 → F2 +1
11(F3)
=
1 0 0 0 14
14
0 1 0 − 14
116
116
0 0 1 − 34 − 5
161116
Por lo tanto la matriz inversa es:
B−1 =
0 14
14
− 14
116
116
− 34 − 5
161116
Solución C:
6
C =
0 −1 −2 1 0 01 0 −3 0 1 04 −3 8 0 0 1
=
1 0 −3 0 1 00 −1 −2 1 0 04 −3 8 0 0 1
F3 → F3 − 4(F1)
=
1 0 −3 0 1 00 −1 −2 1 0 00 −3 20 0 −4 1
F2 → F2(−1)
=
1 0 −3 0 1 00 1 2 −1 0 00 −3 20 0 −4 1
F3 → F3 + 3(F2)
=
1 0 −3 0 1 00 1 2 −1 0 00 0 26 −3 −4 1
F3 → F3
(1
26
)
=
1 0 −3 0 1 00 1 2 −1 0 00 0 1 − 3
26 − 213
126
F1 → F1 + 3 (F3)
=
1 0 0 − 926
713
326
0 1 2 −1 0 00 0 1 − 3
26 − 213
126
F2 → F2 − 2 (F3)
=
1 0 0 − 926
713
326
0 1 0 − 1013
413 − 1
130 0 1 − 3
26 − 213
126
Por lo tanto la matriz inversa es:
C−1 =
− 926
713
326
− 1013
413 − 1
13− 3
26 − 213
126
Solución D:
D =
3240 −1669 −299 951328 −648 −116 37−217 106 19 −635 −17 −3 1
7
3240 −1669 −299 95 1 0 0 01328 −648 −116 37 0 1 0 0−217 106 19 −6 0 0 1 035 −17 −3 1 0 0 0 1
=
35 −17 −3 1 0 0 0 1−217 106 19 −6 0 0 1 01328 −648 −116 37 0 1 0 03240 −1669 −299 95 1 0 0 0
F1 → F1
(1
35
)
=
1 −17
35−335
135 0 0 0 1
35−217 106 19 −6 0 0 1 01328 −648 −116 37 0 1 0 03240 −1669 −299 95 1 0 0 0
F2 → F2 + 217 (F1)
=
1 −17
35−335
135 0 0 0 1
350 3
525
15 0 0 1 31
51328 −648 −116 37 0 1 0 03240 −1669 −299 95 1 0 0 0
F3 → F3 − 1328 (F1)
=
1 −17
35−335
135 0 0 0 1
350 3
525
15 0 0 1 31
50 − 104
35 − 7635 − 33
35 0 1 0 − 132835
3240 −1669 −299 95 1 0 0 0
F4 → F4 − 3240 (F1)
=
1 −17
35−335
135 0 0 0 1
350 3
525
15 0 0 1 31
50 − 104
35 − 7635 − 33
35 0 1 0 − 132835
0 − 6677 − 149
7177 1 0 0 − 648
7
F2 → F2
(5
3
)
=
1 −17
35−335
135 0 0 0 1
350 1 2
313 0 0 5
3313
0 − 10435 − 76
35 − 3335 0 1 0 − 1328
350 − 667
7 − 1497
177 1 0 0 − 648
7
F1 → F1 +17
35(F2)
=
1 0 5
21421 0 0 17
2110621
0 1 23
13 0 0 5
3313
0 − 10435 − 76
35 − 3335 0 1 0 − 1328
350 − 667
7 − 1497
177 1 0 0 − 648
7
F3 → F3 +104
35(F2)
8
=
1 0 5
21421 0 0 17
2110621
0 1 23
13 0 0 5
3313
0 0 − 421 − 1
21 0 1 10421 − 152
210 − 667
7 − 1497
177 1 0 0 − 648
7
F4 → F4 +667
7(F2)
=
1 0 5
21421 0 0 17
2110621
0 1 23
13 0 0 5
3313
0 0 − 421 − 1
21 0 1 10421 − 152
210 0 887
2171821 1 0 3335
211873321
F3 → F3
(−21
4
)
=
1 0 5
21421 0 0 17
2110621
0 1 23
13 0 0 5
3313
0 0 1 − 14 0 − 21
4 −26 380 0 887
2171821 1 0 3335
211873321
F1 → F1 −4
21(F3)
=
1 0 0 1
4 0 54 7 −4
0 1 23
13 0 0 5
3313
0 0 1 − 14 0 − 21
4 −26 380 0 887
2171821 1 0 3335
211873321
F2 → F2 −2
3(F3)
=
1 0 0 1
4 0 54 7 −4
0 1 0 12 0 7
2 19 −150 0 1 − 1
4 0 − 214 −26 38
0 0 88721
71821 1 0 3335
211873321
F4 → F4 −887
21(F3)
=
1 0 0 1
4 0 54 7 −4
0 1 0 12 0 7
2 19 −150 0 1 − 1
4 0 − 214 −26 38
0 0 0 1794 1 887
179 1257 −713
F4 → F4
(4
179
)
=
1 0 0 1
4 0 54 7 −4
0 1 0 12 0 7
2 19 −150 0 1 − 1
4 0 − 214 −26 38
0 0 0 1 4179
887179
5028179 − 2852
179
F1 → F1 −1
4(F4)
=
1 0 0 0 − 1
1792
179 − 4179 − 3
1790 1 0 1
2 0 72 19 −15
0 0 1 − 14 0 − 21
4 −26 380 0 0 1 4
179887179
5028179 − 2852
179
F2 → F2 −1
2(F4)
9
=
1 0 0 0 − 1
1792
179 − 4179 − 3
1790 1 0 0 − 2
179183179
887179 − 1259
1790 0 1 − 1
4 0 − 214 −26 38
0 0 0 1 4179
887179
5028179 − 2852
179
F3 → F3 +1
4(F4)
=
1 0 0 0 − 1
1792
179 − 4179 − 3
1790 1 0 0 − 2
179183179
887179 − 1259
1790 0 1 0 1
179 − 178179 − 3397
1796089179
0 0 0 1 4179
887179
5028179 − 2852
179
Asi que su matriz inversa sera
A−1 =
− 1
1792
179 − 4179 − 3
179− 2
179183179
887179 − 1259
1791
179 − 178179 − 3397
1796089179
4179
887179
5028179 − 2852
179
6. Dada las Matrices
A =
11 −7 −6−5 3 −1919 −8 17
,B =
3240 −1669 −299 951328 −648 −116 37−217 106 19 −635 −17 −3 1
, hallar su inversa por medio de su ad-
junta.
Solución A:
AT =
11 −5 19−7 3 −8−6 −19 17
(AT)∗
=
(3 −8−19 17
)−(−7 −8−6 −17
) (−7 3−6 −19
)−(−5 19−19 17
) (11 19−6 17
)−(11 −5−7 −19
)(−5 193 −8
)−(11 19−7 −8
) (11 −5−7 3
)
=
−101 167 151−276 301 239−17 −45 −2
4 = (561 + 2527− 240)− (−342 + 595 + 1672)
= 2848− 1925
= 923
10
A−1 =1
|A|(AT)∗
=1
923
−101 167 151−276 301 239−17 −45 −2
A−1 =
− 101923
167923
151923
− 276923
301923
239923
− 17923 − 45
923 − 2923
Solución B:
3240 ∗
−648 −116 37106 19 −6−17 −3 1
− (−1669) ∗
1328 −116 37−217 19 −635 −3 1
+ (−299)
∗
1328 −648 37−217 106 −635 −17 1
− 95 ∗
1328 −648 −116−217 106 1935 −17 −3
= −179
11
a1,1 =
−648 −116 37106 19 −6−17 −3 1
= 1
a1,2 =
1328 −116 37−217 19 −635 −3 1
= −2
a1,3 =
1328 −648 37−217 106 −635 −17 1
= −1
a1,4 =
1328 −648 −116−217 106 1935 −17 −3
= 4
a2,1 =
−648 −299 95106 19 −6−17 −3 1
= 2
a2,2 =
3240 −299 95−217 19 −635 −3 1
= −183
a2,3 =
3240 −1669 95−217 106 −635 −17 1
= −718
a2,4 =
3240 −1669 −299−217 106 1935 −17 −3
= −887
a3,1 =
−1669 −299 95−648 116 37−17 −3 1
= 4
a3,2 =
3240 −299 951328 −116 3735 −3 1
= −887
a3,3 =
3240 −1669 951328 −648 3735 −17 1
= 3397
a3,4 =
3240 −1669 −2991328 −648 −11635 −17 −3
= 5028
a4,1 =
−1669 −299 95−648 −116 37106 19 −6
= −3
a4,2 =
3240 −299 951328 −116 37−217 19 −6
= 1259
a4,3 =
3240 −1669 951328 −648 37−217 106 −6
= 6089
a4,4 =
3240 −648 −1161328 106 19−217 −17 −3
= 2852
A∗ =
1 2 −1 −4−2 −183 718 −8874 −887 3397 −50283 1259 −6089 2852
(A∗)T
=
1 −2 4 32 −183 −887 1259−1 718 3397 −6089−4 −887 −5028 2852
A−1 =1
Det(A∗)
T
=1
−179
1 −2 4 32 −183 −887 1259−1 718 3397 −6089−4 −887 −5028 2852
A−1 =
− 1
1792
179 − 4179 − 3
179− 2
179183179
887179 − 1259
1791
179 − 178179 − 3397
1796089179
4179
887179
5028179 − 2852
179
12
7) Dada la matriz
A =
3240 −1669 −299 951328 −648 −116 37−217 106 19 −635 −17 −3 1
hallar su determinante por medio de
expansion de cofactores
A =
3240 −1669 −299 951328 −648 −116 37−217 106 19 −635 −17 −3 1
4 = 3240 ∗
−648 −116 37106 19 −6−17 −3 1
− (−1669) ∗
1328 −116 37−217 19 −635 −3 1
+
(−299)*
1328 −648 37−217 106 −635 −17 1
− 95 ∗
1328 −648 −116−217 106 1935 −17 −3
= −179
4 = (3240)(1)− (−1669)(−2) + (−299)(−1)− (95)(4)
4 = −179
8) Hallar el determinante de la matriz
A =
0 −1 −21 0 34 −3 8
expandido por cofactores con respecto a la segunda columna. Halle tambien
los cofactores A31, A32, A33
13
4 = −(−1)(
1 34 8
)+ 0
(0 −24 8
)− (−3)
(0 −21 3
)4 = 1(8− 12) + 0 + 3(0− (−2))4 = −4 + 6
4 = 2
C31 = |A31|
C31 = 4
(−1 −20 3
)C31 = 4(−3)C31 = −12
C32 = |A32|
C32 = −(−3)(
0 −21 3
)C32 = 3(2)
C32 = 6
C33 = |A33|
C33 = 8
(0 −11 0
)C33 = 8(1)
C33 = 8
9) Halle los valores de x.y.w y z para que la operacion con matrices resulteverdadera
A =
(w xy x
)=
(−7 56 −2
)+ 3
(y wz x
),escriba los resultados en el
orden x, y, w, z.(w xy x
)=
(−7 56 −2
)+ 3
(y wz x
)=
(−7 56 −2
)+
(3y 3w3z 3x
)=
(−7 + 3y 5 + 3w6 + 3z −2 + 3x
)Entonces
14
3y − 7 = w
3z + 6 = y
3w + 5 = x
3x− 2 = x
3x− 2 = x
3x− x = 2
2x = 2
x = 1
Por lo tanto
3w + 5 = x
3w = 1− 5
3w = −4
w = −4
3
3y − 7 = w
3y = −4
3+ 7
3y =17
3
y =17
9
3z + 6 = y
3z =17
9− 6
3z = −37
9
z = −37
27
Comprobamos los valores
15
(w xy x
)=
(−7 + 3y 5 + 3w6 + 3z −2 + 3x
)(− 4
3 1179 1
)=
(−7 + 3( 179 ) 5 + 3(− 4
3. )6 + 3(− 37
27 ) −2 + 3(1)
)(− 4
3 1179 1
)=
(−7 + 17
3 5− 46− 37
9 −2 + 3
)(− 4
3 1179 1
)=
(− 4
3 1179 1
)Asi que los valores que satisfacen a la incognitas para que la matriz sea
verdadera son
x = 1
y =17
9
w = −4
3
z = −37
27
10) ¾Cual es la matriz cuya adjunta es la matriz A =
10 −4 200 3 20 0 5
? 10 0 0−4 3 020 2 5
R S TU V WX Y Z
=
1 0 00 1 00 0 1
(10R+ 0U + 0X) (10S + 0V + 0Y ) (10T + 0W + 0Z)
(−4R+ 3U + 0X) (−4S + 3V + 0Y ) (−4T + 3W + 0Z)(20R+ 2U + 5X) (20S + 2V + 5Y ) (20T + 2W + 5Z)
=
1 0 00 1 00 0 1
De lo cual obtendremos 9 ecuaciones a solucionar para obtener los valores
de de la matriz
10R+ 0U + 0X = 1
−4R+ 3U + 0X = 0
20R+ 2U + 5X = 0
4 =
10 0 0−4 3 020 2 5
= 150 4R =
1 0 00 3 00 2 5
= 15
4U =
10 1 0−4 0 020 0 5
= 20 4X =
10 0 1−4 3 020 2 0
= −68
16
R = 15150 = 1
10 U = 20150 = 2
15 X = −68150 = −34
75
10S + 0V + 0Y = 0
−4S + 3V + 0Y = 1
20S + 2V + 5Y = 0
4 =
10 0 0−4 3 020 2 5
= 150 4S =
0 0 01 3 00 2 5
= 0
4V =
10 0 0−4 1 020 0 5
= 50 4Y =
10 0 0−4 3 120 2 0
= −20
S = 0150 = 0 V = 50
150 = 13 Y = −20
150 = −215
10T + 0W + 0Z = 0
−4T + 3W + 0Z = 0
20T + 2W + 5Z = 1
4 =
10 0 0−4 3 020 2 5
= 150 4T =
0 0 00 3 01 2 5
= 0
4W =
10 0 0−4 0 020 1 5
= 0 4Z =
10 0 0−4 3 020 2 1
= 30
T = 0150 = 0 W = 0
150 = 0 Z = 30150 = 1
5
Por lo consiguiente la matriz B =
110 0 0215
13 0
− 3475 − 2
1515
cuya adjunta es A = 10 −4 200 3 20 0 5
11) Calcule el determinante de la siguiente matriz de numros complejos.
A =
5− 7i 3− 4i 2i12 + 5i 1 4− 3i9 + 9i 13− 78i 7 + 80i
det
5− 7i 3− 4i 2i12 + 5i 1 4− 3i9 + 9i 13− 78i 7 + 80i5− 7i 3− 4i 2i12 + 5i 1 4− 3i
= ((5− 7i)(1)(7+80i)+ (12+5i)(13−
17
78i)(2i) + (9 + 9i)(3− 4i)(4− 3i))− ((12 + 5i)(3− 4i)(7 + 80i) + (5− 7i)(13−78i)(4− 3i) + (9 + 9i)(1)(2i))
= (595+351i)+(1092i+1742)+(225−225i)−(3032+4249i) + (−3367− 481i) + (18i− 18)
= (2562 + 1218i)− (−353 + 3786i)= 2915− 2568i
12) Basandose en el concepto de Matriz Elemental. halleA =
1 0 00 1 0−4 0 1
I =
1 0 00 1 0−4 0 1
F3 → 14 (F3 + F1(4))
I =
1 0 00 1 00 0 1
13)¾Por que matriz hay que premultiplicar A =
(3 −54 −7
)para que resulte
la matriz B =
(8 63 7
)?
(3 −54 −7
)(a bc d
)=
(8 63 7
)
3a− 5c = 8 3b− 5d = 64a− 7c = 3 4b− 7d = 7
3a− 5c = 8(− 4
3
)3b− 5d = 6
(− 4
3
)4a− 7c = 3 4b− 7d = 7−4a+ 20
3 c = − 323 −4a+ 20
3 d = −84a− 7c = 3 4b− 7d = 7− 1
3c = −233 − 1
3d = −1c =
− 233
− 13
d = −1− 1
3
c = 23 d = 3
3a− 5c = 8 3b− 5d = 63a− 5(23) = 8 3b− 5(3) = 63a− 115 = 8 3b = 6 + 153a = 123 b = 7a = 41
18
(3 −54 −7
)(41 723 3
)=
(8 63 7
)
14) Encuntre una factorizacion LU paraA =
3 −7 −2−3 5 16 −4 0
, b =
−752
U =
3 −7 −2−3 5 16 −4 0
F2 → F2 − (−1)(F1)
U =
3 −7 −20 −2 −16 −4 0
F3 → F3 − 2(F1)
U =
3 −7 −20 −2 −10 10 4
F3 → F3 − (−5)(F2)
U =
3 −7 −20 −2 −10 0 −1
L =
1 0 0−1 1 02 −5 1
Por lo consiguiente A = LXU 3 −7 −2−3 5 16 −4 0
=
1 0 0−1 1 02 −5 1
X
3 −7 −20 −2 −10 0 −1
por lo tantox = −7−x+ y = 52x− 5y + z = 2
x = −7y = 5− 7 = −2z = 2 + 14− 10 = 6
3x− 7y − 2z = −7−2y − z = −2−z = 6
z = −6y = 2+6
2 = 4x = −7+28−12
3 = 3
Obteniendo asi los valores de x,y,zx = 3
19
y = 4z = −6
20
