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OPERACIONES ENTRE MATRICES.
DEBER. BACHILLERATO 5TO CURSO.
Dadas las siguientes matrices:
A=[−1 3 −1
0 12
2
−2 3 3 ]
B=[−1 3 5
0 12
2
−2 3 3 ]
C=[2 66 −8 ]
D=[ 1 0 23
−3 2 −5 ]E=[−1 3 −1
0 5 2−2 3 −2 ]
F=[−1 3 −1
√3 12 √2
−2 3 3 ]
G=[−1 3 120 0 2
−2 3 3 ]
H=[1 10 −122 6 −5 ]
J=[1 03 125 4 ]
K=[152359 ]
L=[12 √3 −1
0 12
2
−2 √3 3 ] M=[−5 3 √2 ]
Cada estudiante deberá resolver las operaciones entre las matrices que se le indiquen:
ESTUDIANTE RESOLVER:
SEBASTIÁN AGUILAR A2+B
A2+BT−3 E K⋅M+B/2
JORGE ALCAZAR A2+B⋅I
( A2+B )T−F
1/2B2+BMARÍA BELÉN ANDA A2+BT
A3+B
GT−L2+ I⋅BCAROL ANDRADE A2+B2
A2+B
(H⋅J )T
FELIPE ANDRADE A−BT−2 F
−3 ( A2+B )
(C⋅H )T+H
MARÍA JOSÉ BENAVIDES ( A−B )2−B
(3 A+2B )⋅A
D⋅KDAYANA CALPOPIÑA C⋅H−J /2
(F⋅G )T+ I
G2+F
ERIKA CEVALLOS F−E+5 A
( A+B ) ET
M⋅L−2M
ISABELA CORREA A2+3B
(3 A+2B )⋅A /3
( J⋅K )T⋅CMARÍA SOL DÁVILA E+F−A2
F⋅K+2MT
( A⋅B )T+5 EVÍCTOR DÍAZ B⋅A+5 F
(G2⋅L )T
( A⋅I )⋅BDARIO FLORES B2−2E+A
C⋅H−3DL2+A /3
JONATHAN GUERRERO (3 A+B )⋅AT
(H−D )⋅EL⋅I+A
ALEX MENA G⋅FT−2 BE⋅A /3K⋅M−(B2)T
DANIELA MORALES L⋅G+A /4E2⋅FT+ IC⋅D+6H
JOSELYN NARANJO ( A+2 B )⋅ATK⋅M+3 A(L⋅J )⋅CT
ROSALIA PAZMIÑO 3C⋅H−J /24 L2+A/35 (A⋅B )T+3 E
GABRIELLA ROSSETTI (G2⋅L )T−A( A⋅I )⋅3 BT
5 (H−2D )⋅ECAMILA SAÁ 2K⋅M−3 A
2 ET⋅A /3A2−8 B
MARÍA XIMENA SANTELISES ( J⋅K )T⋅C /3A⋅B2
(G2⋅L )−A+3BWILSON TITUAÑA 10B2−2E+A /3
G2+F−2 I⋅A(F⋅K )T+2M
FILIPO TOSCANO A2+4 BT
5 (E+2 F−A2 )
(C⋅H )⋅Gt
DAVID VÁSCONEZ ET+FT−2 A2
3M⋅L−6M(2 A⋅I )⋅3B
DOMENICA VELASCO ( A⋅B )T−3G−L⋅G+A /4−E(F⋅I )2+3B
TATIANA VERA (−CT⋅I )⋅HD−5H−J T
E2⋅FT⋅ITONNY YASELGA (6 J⋅K )T⋅2C
−F⋅K /3+2MT
AT⋅I−BT−2F /3ERICK YUMICEBA 2 A2−10B
3 (M⋅L )−2M(D⋅K )T
SOLUCIONES:
SEBASTIÁN AGUILAR A2+B
A2+BT−3 E K⋅M+B/2
A=[−1 3 −1
0 12
2
−2 3 3 ]
B=[−1 3 5
0 12
2
−2 3 3 ]
E=[−1 3 −10 5 2
−2 3 −2 ]
K=[152359 ]
M=[−5 3 √2 ]
A2+B
[−1 3 −1
0 12
2
−2 3 3 ] *
[−1 3 −1
0 12
2
−2 3 3 ] +
[−1 3 5
0 12
2
−2 3 3 ]=
[ 3 −92
4
−4 254
7
−4 92
17 ] +
[−1 3 5
0 12
2
−2 3 3 ]=
A2+B =
[ 2 −32
9
−4 274
9
−6 152
20 ]A2+BT−3 E
A2+BT−3 E =
[ 3 −92
4
−4 254
7
−4 92
17 ] +
[−1 0 −2
3 12
3
5 2 3 ] - 3
[−1 3 −10 5 2
−2 3 −2 ]=
A2+BT−3 E =
[ 0 −272
5
−1 −334
4
7 −52
26 ]K⋅M+B/2
[152359 ]
¿ [−5 3 √2 ]+
12 [−1 3 5
0 12
2
−2 3 3 ]=
[−75 45 15√2115 69 23√2295 177 59√2 ]
+
[−12
32
52
0 14
1
−1 32
32
]=
K⋅M+B/2 =
[−1512
932
52+15√2
115 2774
1+23√2
294 3572
32+59√2 ]
JORGE ALCAZAR A2+B⋅I
( A2+B )T−F
1/2B2+B
A=[−1 3 −1
0 12
2
−2 3 3 ]
B=[−1 3 5
0 12
2
−2 3 3 ]
F=[−1 3 −1
√3 12 √2
−2 3 3 ]A2+B⋅I
A2=[ 3 − 9
24
−4 254
7
−4 92
17 ]
I=[1 0 00 1 00 0 1 ]
[ 3 −92
4
−4 254
7
−4 92
17 ] +
[−1 3 5
0 12
2
−2 3 3 ]*
[1 0 00 1 00 0 1 ]
=
[ 3 − 92
4
−4 254
7
−4 92
17 ] +
[ 0 3 5
0 32
2
−2 3 4 ]=
A2+B⋅I =
[ 3 −32
9
−4 314
9
−6 152
21 ]( A2+B )T−F
[ 3 −92
4
−4 254
7
−4 92
17 ]+
[−1 3 5
0 12
2
−2 3 3 ]=
[ 2 −32
9
−4 274
9
−6 152
20 ][ 2 −4 −6
−32
274
152
9 9 20 ]-
[−1 3 −1
√3 12 √2
−2 3 3 ]= [ 3 −7 −5
−32−√3 25
4152
−√211 6 17 ]
( A2+B )T−F=[ 3 −7 −5
−32−√3 25
4152
−√211 6 17 ]
1/2B2+B
1/2
[−1 3 5
0 12
2
−2 3 3 ]*
[−1 3 5
0 12
2
−2 3 3 ]+
[−1 3 5
0 12
2
−2 3 3 ]= 1/2
[−9272
16
−4 254
7
−4 92
5 ] +
[−1 3 5
0 12
2
−2 3 3 ]=
1/2B2+B =
[−112
394
13
−2 298
112
−4 214
112
]
MARÍA BELÉN ANDA A2+BT
A3+B
GT−L2+ I⋅B
A2=[ 3 − 9
24
−4 254
7
−4 92
17 ]
BT=
[−1 0 −2
3 12
3
5 2 3 ]
A2+BT=[ 2 −9
22
−1 274
10
1 132
20 ]A3+B
A3=[ 3 − 9
24
−4 254
7
−4 92
17 ] [−1 3 −1
0 12
2
−2 3 3 ]=
[−11754
0
−10 978
752
−30 1654
64 ]A3+B =
[−12874
5
−10 1018
792
−32 1774
67 ]
GT−L2+ I⋅B
G=[−1 3 120 0 2
−2 3 3 ]
GT=[−1 0 −23 0 312 2 3 ]
L=[12 √3 −1
0 12
2
−2 √3 3 ]L2=[12 √3 −1
0 12
2
−2 √3 3 ]¿ [12 √3 −1
0 12
2
−2 √3 3 ]= [146232 √3 −15+2√3
−4 14+2√3 7
−30 32 √3 11+2√3 ]
I⋅B=[1 0 00 1 00 0 1 ]¿ [−1 3 5
0 12
2
−2 3 3 ]= [−1 3 5
0 12
2
−2 3 3 ]
GT−L2+ I⋅B=[−1 0 −23 0 312 2 3 ]− [146
232 √3 −15+2√3
−4 14+2√3 7
−30 32 √3 11+2√3 ]+ [−1 3 5
0 12
2
−2 3 3 ]=
GT−L2+ I⋅B=[−148 3−23
2 √3 18−2√3
7 12−2√3 −2
40 5−32 √3 −5−2√3 ]
CAROL ANDRADE A2+B2
A2+B
(H⋅J )T
A2=[ 3 − 9
24
−4 254
7
−4 92
17 ]
B2=
[−9272
16
−4 254
7
−4 92
5 ]
A2+B2
=
[ 3 −92
4
−4 254
7
−4 92
17 ]+
[−9272
16
−4 254
7
−4 92
5 ]=
A2+B2=[−6 9 20
−8 252
14
−8 9 22 ]
A2+B =
[ 3 −92
4
−4 254
7
−4 92
17 ]+ [−1 3 5
0 12
2
−2 3 3 ]= [ 2 −32
9
−4 274
9
−6 152
20 ]A2+B=
[ 2 −32
9
−4 274
9
−6 152
20 ](H⋅J )T=
[1 10 −122 6 −5 ]∗¿ ¿[1 0
3 125 4 ]= [−29 72
−5 52 ](H⋅J )T= [−29 −5
72 52 ]FELIPE ANDRADE A−BT−2 F
−3 ( A2+B )
(C⋅H )T+H
A−BT−2 F
[−1 3 −1
0 12
2
−2 3 3 ]− [−1 0 −2
3 12
3
5 2 3 ]−2[−1 3 −1
√3 12 √2
−2 3 3 ]= [ 2 −3 3−3−2√3 −1 −1−2√2
−3 −5 −6 ]A−BT−2 F
=
[ 2 −3 3−3−2√3 −1 −1−2√2
−3 −5 −6 ]−3 ( A2+B )
A2+B=[ 3 −92
4
−4 254
7
−4 92
17 ]+ [−1 3 5
0 12
2
−2 3 3 ]= [ 2 −32
9
−4 274
9
−6 152
20 ]−3 ( A2+B )=
[−692
−27
12 −514
−27
18 −452
−60 ](C⋅H )T+H
C⋅H=[2 66 −8 ] [1 10 −12
2 6 −5 ]= [14 56 −54−10 12 −32 ]
(C⋅H )T= [14 −1056 12−54 −32 ]
(C⋅H )T+H= [14 −1056 12−54 −32 ]+ [1 10 −12
2 6 −5 ]=NO SE PUEDEN SUMAR: LAS MATRICES
NO TIENEN EL MISMO ORDEN.
MARÍA JOSÉ BENAVIDES ( A−B )2−B
(3 A+2B )⋅A
D⋅K
( A−B )2−B
( A−B )=[−1 3 −1
0 12
2
−2 3 3 ]− [−1 3 5
0 12
2
−2 3 3 ]= [0 0 −60 0 00 0 0 ]
( A−B )2= [0 0 −60 0 00 0 0 ]¿ [0 0 −6
0 0 00 0 0 ]= [0 0 0
0 0 00 0 0 ]
( A−B )2−B= [1 −3 −5
0 −12
−2
2 −3 −3 ]
(3 A+2B )⋅A
(3 A+2B )=
3[−1 3 −1
0 12
2
−2 3 3 ]+ 2[−1 3 5
0 12
2
−2 3 3 ]= [ −5 15 7
0 52
10
−10 15 15 ]
[ −5 15 7
0 52
10
−10 15 15 ]∗¿ ¿[−1 3 −1
0 12
2
−2 3 3 ]= [ −9 272
56
−20 1254
35
−20 452
85 ](3 A+2B )⋅A=
[ −9 272
56
−20 1254
35
−20 452
85 ]D⋅K
D⋅K=[ 1 0 2
3−3 2 −5 ]∗¿ ¿[152359 ]= [ 1632−294]
DAYANA CALPOPIÑA C⋅H−J /2
(F⋅G )T+ I
G2+FC⋅H−J /2
C⋅H=[2 66 −8 ] [1 10 −12
2 6 −5 ]= [14 56 −54−10 12 −32 ]
EL ORDEN DE LA MARTIZ J ES 3X2
NO TIENE SOLUCIÓN PUES LAS MATRICES NO TIENEN EL MISMO ORDEN
(F⋅G )T+ I
F=[−1 3 −1
√3 12 √2
−2 3 3 ]
G=[−1 3 120 0 2
−2 3 3 ]
(F⋅G )=[−1 3 −1
√3 12 √2
−2 3 3 ]∗¿ ¿[−1 3 120 0 2
−2 3 3 ]= [ 3 −6 −9−√3−2√2 3√3+3√2 1+12√3+3√2
−4 3 −9 ](F⋅G )T+ I= [ 3 −6 −9
−√3−2√2 3√3+3√2 1+12√3+3√2−4 3 −9 ]+ [1 0 0
0 1 00 0 1 ]=
(F⋅G )T+ I= [ 4 −6 −9−√3−2√2 1+3 √3+3√2 1+12√3+3√2
−4 3 −8 ]G2+F
G2=[−1 3 120 0 2
−2 3 3 ]¿ [−1 3 120 0 2
−2 3 3 ]= [−25 33 30−4 6 6−4 3 −9 ]
G2+F=[−25 33 30−4 6 6−4 3 −9 ]
+
[−1 3 −1
√3 12 √2
−2 3 3 ]= [ −26 36 29
−4+√3 132
6+√2−6 6 −6 ]
G2+F=[ −26 36 29
−4+√3 132
6+√2−6 6 −6 ]
ERIKA CEVALLOS F−E+5 A
( A+B ) ET
M⋅L−2MF−E+5 A
F−E+5 A=[−1 3 −1
√3 12 √2
−2 3 3 ]− [−1 3 −1
0 12
2
−2 3 3 ]+5[−1 3 −10 5 2
−2 3 −2 ]= [ −5 15 −5√3 +25 8+√2
−10 15 −10 ]F−E+5 A=
[ −5 15 −5√3 +25 8+√2
−10 15 −10 ]( A+B ) ET
ET=[−1 0 −2
3 12
3
−1 2 3 ]
( A+B )= [−1 3 −10 5 2
−2 3 −2 ]+ [−1 3 5
0 12
2
−2 3 3 ]= [−2 6 4
0 112
4
−4 6 1 ]( A+B ) ET=
[−2 6 4
0 112
4
−4 6 1 ]∗¿ ¿[−1 0 −2
3 12
3
−1 2 3 ]( A+B ) ET=
[16 11 34252
434
572
21 6 29 ]M⋅L−2M
M=[−5 3 √2 ]
L=[12 √3 −1
0 12
2
−2 √3 3 ]M⋅L= [−5 3 √2 ]∗¿ ¿
[12 √3 −1
0 12
2
−2 √3 3 ]= [−60−2√2 32−5√3+√6 11+3√2 ]
M⋅L−2M= [−60−2√2 32−5√3+√6 11+3√2 ]−2 [−5 3 √2 ]=
M⋅L−2M= [−50−2√2 −92−5√3+√6 11+√2 ]
ISABELA CORREA A2+3B
(3 A+2B )⋅A /3
( J⋅K )T⋅CA2+3B
A2=[ 3 −92
4
−4 254
7
−4 92
17 ]
A2+3B=[ 3 −9
24
−4 254
7
−4 92
17 ]+ 3[−1 3 5
0 12
2
−2 3 3 ]= [ 092
19
−4 314
13
−10 272
26 ]A2+3B=
[ 092
19
−4 314
13
−10 272
26 ](3 A+2B )⋅A /3
(3 A+2B )=3[−1 3 −10 5 2
−2 3 −2 ]+ 2[−1 3 5
0 12
2
−2 3 3 ]= [ −5 15 70 16 10
−10 15 0 ]
(3 A+2B )⋅A /3= [ −5 15 70 16 10
−10 15 0 ]∗1/3[−1 3 −10 5 2
−2 3 −2 ]= [−3 27 7
−203
1103
4
103
15 403
](3 A+2B )⋅A /3=
[−3 27 7
−203
1103
4
103
15 403
]( J⋅K )T⋅C
J=[1 03 125 4 ]
K=[152359 ]
C=[2 66 −8 ]
( J⋅K )=
[1 03 125 4 ]∗¿ ¿
[152359 ]=
NO SE PUEDE RESOLVER: EL NÚMERO DE COLUMNAS DE LA PRIMERA MATRIZ NO ES
IGUAL NÚMERO DE FILAS DE LA SEGUNDA.MARÍA SOL DÁVILA E+F−A2
F⋅K+2MT
( A⋅B )T+5 EE+F−A2
A2=[ 3 −92
4
−4 254
7
−4 92
17 ]E+F−A2=
[−1 3 −1
0 12
2
−2 3 3 ]+ [−1 3 −1
√3 12 √2
−2 3 3 ]− [ 3 −92
4
−4 254
7
−4 92
17 ]=E+F−A2=
[ −5 252
−6
√3+4 −214
−5+√2
0 32
−11 ]F⋅K+2MT
F⋅K=[−1 3 −1
√3 12 √2
−2 3 3 ]∗¿ ¿[152359 ]= [ −5232
+15√3+59√2216 ]
M=[−5 3 √2 ]
MT=[−53√2 ]
F⋅K+2MT=[ −5232
+15√3+59√2216 ]+2[−53√2 ]=
F⋅K+2MT=[ −15352
+15√3+59√2
216+2√2 ]
( A⋅B )T+5 E
( A⋅B )=[−1 3 −1
0 12
2
−2 3 3 ]∗¿ ¿[−1 3 5
0 12
2
−2 3 3 ]= [ 3 −92
−2
−4 254
7
−4 92
5 ]( A⋅B )T+5 E= [ 3 −4 −4
−92
254
92
−2 7 5 ]+5
[−1 3 −1
0 12
2
−2 3 3 ]= [ −2 11 −9
−92
354
292
−12 22 20 ]( A⋅B )T+5 E= [ −2 11 −9
−92
354
292
−12 22 20 ]VÍCTOR DÍAZ B⋅A+5 F
(G2⋅L )T
( A⋅I )⋅BB⋅A+5 F
B⋅A=[−1 3 5
0 12
2
−2 3 3 ]∗¿ ¿[−1 3 −1
0 12
2
−2 3 3 ]= [−9272
22
−4 254
7
−4 92
17 ]
B⋅A+5 F=[−9
272
22
−4 254
7
−4 92
17 ]+5
[−1 3 −1
√3 12 √2
−2 3 3 ]= [ −14 572
17
−4+5√3 354
7+5√2
−14 392
32 ]B⋅A+5 F=
[ −14 572
17
−4+5√3 354
7+5√2
−14 392
32 ](G2⋅L )T
G2=[−1 3 120 0 2
−2 3 3 ]¿ [−1 3 120 0 2
−2 3 3 ]= [−25 33 30−4 6 6−4 3 −9 ]
(G2⋅L )= [−25 33 30−4 6 6−4 3 −9 ]
*
[12 √3 −1
0 12
2
−2 √3 3 ]= [−360332
+5 √3 181
−60 3+2√3 34
−30 32−13√3 −17 ]
(G2⋅L )T= [ −360 −60 −30332
+5√3 3+2√3 32−13√3
181 34 −17 ]( A⋅I )⋅B
( A⋅I )= [−1 3 −1
√3 12 √2
−2 3 3 ]∗¿ ¿[1 0 00 1 00 0 1 ]= [−1 3 −1
√3 12 √2
−2 3 3 ]
( A⋅I )⋅B= ( A⋅B )=[−1 3 −1
0 12
2
−2 3 3 ]∗¿ ¿[−1 3 5
0 12
2
−2 3 3 ]= [ 3 − 92
−2
−4 254
7
−4 92
5 ]( A⋅I )⋅B=
[ 3 −92
−2
−4 254
7
−4 92
5 ]DARIO FLORES B2−2E+A
C⋅H−3DL2+A /3
B2−2E+A
B2=[−9272
16
−4 254
7
−4 92
5 ]−2[−1 3 −1
0 12
2
−2 3 3 ]+ [−1 3 −10 5 2
−2 3 −2 ]= [−8212
17
−4 414
5
−2 32
−3 ]B2−2E+A
=[−8212
17
−4 414
5
−2 32
−3 ]C⋅H−3D
C⋅H=[2 66 −8 ] [1 10 −12
2 6 −5 ]= [14 56 −54−10 12 −32 ]
C⋅H−3D =[14 56 −54−10 12 −32 ]−3[ 1 0 2
3−3 2 −5 ]= [11 56 −56
−1 6 −17 ]C⋅H−3D
=[11 56 −56−1 6 −17 ]
L2+A /3
L2=[146232 √3 −15+2√3
−4 14+2√3 7
−30 32 √3 11+2√3 ]
L2+A /3[146
232 √3 −15+2√3
−4 14+2√3 7
−30 32 √3 11+2√3 ]+ 13 [−1 3 −1
0 5 2−2 3 −2 ]=
[4353
232 √3−13 −46
3+2√3
−4 2312
+2√3 233
−923
32 √3+1 31
3+2√3 ]
L2+A /3=[4353
232 √3−13 −46
3+2√3
−4 2312
+2√3 233
−923
32 √3+1 31
3+2√3 ]
JONATHAN GUERRERO (3 A+B )⋅AT
(H−D )⋅EL⋅I+A
(3 A+B )⋅AT
3[−1 3 −10 5 2
−2 3 −2 ]+ [−9272
22
−4 254
7
−4 92
17 ]= [−12452
19
−4 854
13
−10 272
11 ]
AT=[−1 0 −23 5 3
−1 2 −2 ]
(3 A+B )⋅AT=[−12
452
19
−4 854
13
−10 272
11 ] [−1 0 −23 5 3
−1 2 −2 ]= [−452
3012
1212
2194
5294
1834
792
1792
772
](3 A+B )⋅AT=
[−452
3012
1212
2194
5294
1834
792
1792
772
](H−D )⋅E
(H−D )= [1 10 −122 6 −5 ]− [ 1 0 2
3−3 2 −5 ]= [0 10 −38
35 4 0 ]=
(H−D )⋅E= [0 10 −383
5 4 0 ] [−1 3 −1
0 12
2
−2 3 3 ]= [762 −33 −18
−5 17 3 ](H−D )⋅E= [762 −33 −18
−5 17 3 ]L⋅I+AL⋅I=I⋅L=L
L⋅I+A =
[12 √3 −1
0 12
2
−2 √3 3 ]+ [−1 3 −10 5 2
−2 3 −2 ]= [11 √3+3 −2
0 112
4
−4 √3+3 1 ]L⋅I+A =
[11 √3+3 −2
0 112
4
−4 √3+3 1 ]ALEX MENA G⋅FT−2 B
E⋅A /3K⋅M−(B2)T
G⋅FT−2 B
F=[−1 3 −1
√3 12 √2
−2 3 3 ]
FT=[−1 √3 −2
3 12
3
−1 √2 3 ]G⋅FT=
[−1 3 120 0 2
−2 3 3 ][−1 √3 −2
3 12
3
−1 √2 3 ]= [−2 12√2−√3+ 32 47
−2 2√2 6
8 3√2−2√3+ 32 22 ]G⋅FT−2 B =
[−2 12√2−√3+ 32 47
−2 2√2 6
8 3√2−2√3+ 32 22 ]−2[−9272
22
−4 254
7
−4 92
17 ]= [16 12√2−√3−512 3
6 2√2−252 −8
16 3√2−2√3−152 −12 ]G⋅FT−2 B =
[16 12√2−√3−512 3
6 2√2−252 −8
16 3√2−2√3−152 −12 ]E⋅A /3
E=[−1 3 −1
0 12
2
−2 3 3 ]
A=[−1 3 −10 5 2
−2 3 −2 ]
A/3=[−13
1 −13
0 53
23
−23
1 −23
]
E⋅A /3=[−1 3 −1
0 12
2
−2 3 3 ][−13
1 −13
0 53
23
−23
1 −23
]= [6 3 3
−43
176
−1
−43
6 23
]E⋅A /3=
[6 3 3
−43
176
−1
−43
6 23
]
K⋅M−(B2)T
K⋅M=[152359 ] [−5 3 √2 ]= [ −75 45 15√2
−115 69 23√2−295 177 59√2 ]
(B2 )T= [−9 −4 −4272
254
92
16 7 5 ]K⋅M−(B2)T= [ −75 45 15√2
−115 69 23√2−295 177 59√2 ]− [−9 −4 −4
272
254
92
16 7 5 ]= [ −66 49 15√2+4−2572
2514
23√2−92−279 170 59√2−5 ]
K⋅M−(B2)T= [ −66 49 15√2+4−2572
2514
23√2−92−279 170 59√2−5 ]
DANIELA MORALES L⋅G+A /4E2⋅FT+ IC⋅D+6H
L⋅G+A /4
L⋅G=[12 √3 −1
0 12
2
−2 √3 3 ] [−1 3 120 0 2
−2 3 3 ]= [−10 33 141+2√3−4 6 7−4 3 −15+2√3 ]
L⋅G+A /4=[−10 33 141+2√3−4 6 7−4 3 −15+2√3 ]+ 14 [−1 3 −1
0 5 2−2 3 −2 ]= [−
414
1354
5634
+2√3
−4 294
152
−92
154
−312
+2√3 ]L⋅G+A /4=
[−414
1354
5634
+2√3
−4 294
152
−92
154
−312
+2√3 ]E2⋅FT+ I
E2=[−1 3 −10 5 2
−2 3 −2 ][−1 3 −10 5 2
−2 3 −2 ]= [ 3 9 9−4 31 66 3 12 ]
E2⋅FT=[ 3 9 9−4 31 66 3 12 ]∗¿ ¿[−1 √3 −2
3 12
3
−1 √2 3 ]= [1592+3√3+9√2 48
61 312
−4 √3+6√2 89
−9 32+6 √3+12√2 33 ]
E2⋅FT+ I=[15
92+3√3+9√2 48
61 312
−4 √3+6√2 89
−9 32+6 √3+12√2 33 ]
+
[1 0 00 1 00 0 1 ]= [16
92+3√3+9√2 48
61 332
−4√3+6√2 89
−9 32+6√3+12√2 34 ]
E2⋅FT+ I =
[1692+3√3+9√2 48
61 332
−4√3+6√2 89
−9 32+6√3+12√2 34 ]
C⋅D+6H
C⋅D= [2 66 −8 ] [ 1 0 2
3−3 2 −5 ]= [−16 12 −86
330 −16 44 ]
C⋅D+6H=[−16 12 −86
330 −16 44 ]+ 6 [1 10 −12
2 6 −5 ]= [−10 72 −3023
42 20 14 ]C⋅D+6H=
[−10 72 −3023
42 20 14 ]JOSELYN NARANJO ( A+2 B )⋅AT
K⋅M+3 A(L⋅J )⋅CT
( A+2 B )⋅AT
A+2 B=[−1 3 −10 5 2
−2 3 −2 ]+2 [−1 3 5
0 12
2
−2 3 3 ]= [−3 9 90 6 6
−6 9 4 ]AT=[−1 0 −2
3 5 3−1 2 −2 ]
( A+2 B )⋅AT= [−3 9 90 6 6
−6 9 4 ] [−1 0 −23 5 3
−1 2 −2 ]= [21 63 1512 42 629 53 31 ]
( A+2 B )⋅AT= [21 63 1512 42 629 53 31 ]
K⋅M+3 A
K⋅M=[152359 ] [−5 3 √2 ]= [ −75 45 15√2
−115 69 23√2−295 177 59√2 ]
K⋅M+3 A =
[ −75 45 15√2−115 69 23√2−295 177 59√2 ]+3 [−1 3 −1
0 5 2−2 3 −2 ]= [ −78 54 15√2−3
−115 84 23√2+6−301 186 59√2−6 ]
K⋅M+3 A=[ −78 54 15√2−3−115 84 23√2+6−301 186 59√2−6 ]
(L⋅J )⋅CT
L=[12 √3 −1
0 12
2
−2 √3 3 ]
J=[1 03 125 4 ]
(L⋅J )= [12 √3 −1
0 12
2
−2 √3 3 ] [1 03 125 4 ]= [ 7+3√3 12√3−4
232
14
13+3√3 12√3+12 ]C=[2 6
6 −8 ]
CT=[2 66 −8 ]
(L⋅J )⋅CT=[ 7+3√3 12√3−4232
14
13+3√3 12√3+12 ][2 66 −8 ]= [−20+78√3 74−78√3
97 −4398+78√3 18−78√3 ]
(L⋅J )⋅CT=[−20+78√3 74−78√397 −4398+78√3 18−78√3 ]
ROSALIA PAZMIÑO 3C⋅H−J /24 L2+A/3
5 (A⋅B )T+3 E3C⋅H−J /2
C=[2 66 −8 ]
3C=[ 6 1818 −24 ]
3C⋅H=[2 66 −8 ][1 10 −12
2 6 −5 ]= [14 56 −54−10 12 −32 ]
3C⋅H−J /2= [14 56 −54−10 12 −32 ]−12 [1 0
3 125 4 ]
NO SE PUEDE RESOLVER PORQUE LAS MATRICES NO SON DEL MISMO ORDEN.
4 L2+A/3
L2=[146232 √3 −15+2√3
−4 14+2√3 7
−30 32 √3 11+2√3 ]
4 L2+A/3 =
4 [146232 √3 −15+2√3
−4 14+2√3 7
−30 32 √3 11+2√3 ]+ 13 [−1 3 −1
0 5 2−2 3 −2 ]=
[17513
1+26√3 −1813
+8√3
−16 83+8√3 86
3
−3623
1+6√3 1303
+8√3 ]4 L2+A/3=
[17513
1+26√3 −1813
+8√3
−16 83+8√3 86
3
−3623
1+6√3 1303
+8√3 ]5 (A⋅B )T+3 E
( A⋅B )=[−1 3 −1
0 12
2
−2 3 3 ]∗¿ ¿[−1 3 5
0 12
2
−2 3 3 ]= [ 3 −92
−2
−4 254
7
−4 92
5 ]( A⋅B )T= [ 3 −4 −4
−92
254
92
−2 7 5 ]5 (A⋅B )T+3 E= [ 3 −4 −4
−92
254
92
−2 7 5 ]+3[−1 3 −10 5 2
−2 3 −2 ]=
[ 0 5 −7
−92
854
212
−8 16 −1 ]5 (A⋅B )T+3 E= [ 0 5 −7
−92
854
212
−8 16 −1 ]GABRIELLA ROSSETTI (G2⋅L )T−A
( A⋅I )⋅3 BT
5 (H−2D )⋅E(G2⋅L )T−A
G2=[−1 3 120 0 2
−2 3 3 ]¿ [−1 3 120 0 2
−2 3 3 ]= [−25 33 30−4 6 6−4 3 −9 ]
(G2⋅L )= [−25 33 30−4 6 6−4 3 −9 ] [
12 √3 −1
0 12
2
−2 √3 3 ]=[−360332
+5 √3 181
−60 3+2√3 34
−30 32−13√3 −17 ]
(G2⋅L )T==[ −360 −60 −30332
+5√3 3+2√3 32−13√3
181 34 −17 ](G2⋅L )T−A=
=[ −360 −60 −30332
+5√3 3+2√3 32−13√3
181 34 −17 ]−[−1 3 −1
0 12
2
−2 3 3 ]==[ −359 −63 −29332
+5√3 52+2√3 −1
2−13√3
183 31 −20 ]
(G2⋅L )T−A=
=[ −359 −63 −29332
+5√3 52+2√3 −1
2−13√3
183 31 −20 ]( A⋅I )⋅3 BT
( A⋅I )=A
3BT=3 [−1 0 −2
3 12
3
5 2 3 ]= [−3 0 −6
9 32
9
15 6 9 ]
( A⋅I )⋅3 BT=[−1 3 −1
0 12
2
−2 3 3 ][−3 0 −6
9 32
9
15 6 9 ]= [15 −32
24
692
514
452
78 452
66 ]( A⋅I )⋅3 BT=
[15 −32
24
692
514
452
78 452
66 ]5 (H−2D )⋅E
(H−2D )= [1 10 −122 6 −5 ]−2 [ 1 0 2
3−3 2 −5 ]= [−1 10 −40
38 2 5 ]
5 (H−2D )= [−5 50 −2003
40 10 25 ]5 (H−2D )⋅E= [−5 50 −200
340 10 25 ][−1 3 −1
0 5 2−2 3 −2 ]
=
[ 4153 35 7153
−90 245 −70 ]5 (H−2D )⋅E= [ 4153 35 715
3−90 245 −70 ]
CAMILA SAÁ 2K⋅M−3 A2 ET⋅A /3A2−8 B
2K⋅M−3 A
K⋅M=[152359 ] [−5 3 √2 ]= [ −75 45 15√2
−115 69 23√2−295 177 59√2 ]
2K⋅M−3 A =
2[ −75 45 15√2−115 69 23√2−295 177 59√2 ]−3 [−1 3 −1
0 12
2
−2 3 3 ]= [−147 81 30√2+3−230 273
246√2−6
−584 345 118√2−9 ]2K⋅M−3 A=
[−147 81 30√2+3−230 273
246√2−6
−584 345 118√2−9 ]2 ET⋅A /3
E=[−1 3 −10 5 2
−2 3 −2 ]
ET=[−1 0 −23 5 3
−1 2 −2 ]
2 ET=[−2 0 −46 10 6
−2 4 −4 ]
A /3=13 [−1 3 −10 5 2
−2 3 −2 ]= [−13
1 −13
0 53
23
−23
1 −23
]2 ET⋅A /3=
[−2 0 −46 10 6
−2 4 −4 ][−13
1 −13
0 53
23
−23
1 −23
]= [103
−6 103
−6 863
23
103
23
6 ]2 ET⋅A /3=
[103
−6 103
−6 863
23
103
23
6 ]A2−8 B
A2=[ 3 −92
4
−4 254
7
−4 92
17 ]A2−8 B=
[ 3 −92
4
−4 254
7
−4 92
17 ]−8 [−1 3 5
0 12
2
−2 3 3 ]= [11 −572
−36
−4 94
−9
12 −392
−7 ]
A2−8 B=[ 3 −9
24
−4 254
7
−4 92
17 ]MARÍA XIMENA SANTELISES ( J⋅K )T⋅C /3
A⋅B2
(G2⋅L )−A+3B( J⋅K )T⋅C /3( J⋅K )T⋅C
J=[1 03 125 4 ]
K=[152359 ]
C=[2 66 −8 ]
( J⋅K )=
[1 03 125 4 ]∗¿ ¿
[152359 ]=
NO SE PUEDE RESOLVER: EL NÚMERO DE COLUMNAS DE LA PRIMERA MATRIZ NO ES
IGUAL NÚMERO DE FILAS DE LA SEGUNDA.
A⋅B2
A⋅B2=[−1 3 −1
0 12
2
−2 3 3 ][−9 27
216
−4 254
7
−4 92
5 ]= [ 134
0
−10 438
272
−6 214
4 ]A⋅B2=
[ 134
0
−10 438
272
−6 214
4 ](G2⋅L )−A+3B
G2=[−1 3 120 0 2
−2 3 3 ]¿ [−1 3 120 0 2
−2 3 3 ]= [−25 33 30−4 6 6−4 3 −9 ]
(G2⋅L )= [−25 33 30−4 6 6−4 3 −9 ] [
12 √3 −1
0 12
2
−2 √3 3 ]=[−360332
+5 √3 181
−60 3+2√3 34
−30 32−13√3 −17 ]
(G2⋅L )−A+3B=
=[−360 332
+5 √3 181
−60 3+2√3 34
−30 32−13√3 −17 ]−[−1 3 −1
0 12
2
−2 3 3 ]+3 [−1 3 5
0 12
2
−2 3 3 ]=
(G2⋅L )−A+3B=[362 45
2+5√3 197
−60 4+2√3 38
−34 152
−13√3 −11 ]WILSON TITUAÑA 10B2−2E+A /3
G2+F−2 I⋅A(F⋅K )T+2M
10B2−2E+A /3
10B2=10[−9
272
16
−4 254
7
−4 92
5 ]= [−90 135 160−40 50 70−40 45 50 ]
10B2−2E+A /3=[−90 135 160−40 50 70−40 45 50 ]−2[−1 3 −1
0 5 2−2 3 −2 ]+[−
13
1 −13
0 53
23
−23
1 −23
]=
10B2−2E+A /3=[−2653
130 4853
−40 1253
2003
−1103
40 1603
]G2+F−2 I⋅A
G2=[−1 3 120 0 2
−2 3 3 ]¿ [−1 3 120 0 2
−2 3 3 ]= [−25 33 30−4 6 6−4 3 −9 ]
G2+F−2 I⋅A=G2+F−2 A=[−25 33 30−4 6 6−4 3 −9 ]+[−1 3 −1
√3 12 √2
−2 3 3 ]− [−2 6 −20 1 4
−4 6 6 ]=G2+F−2 I⋅A=
[ −24 30 31
−4+√3 92
2+√2−2 0 −12 ]
(F⋅K )T+2M
F⋅K=[−1 3 −1
√3 12 √2
−2 3 3 ]∗¿ ¿[152359 ]= [ −5232
+15√3+59√2216 ]
(F⋅K )T= [−5 232
+15√3+59√2 216 ](F⋅K )T+2M=
[−5 23
2+15√3+59√2 216 ]+ 2 [−5 3 √2 ]=
(F⋅K )T+2M= [−10 352
+15√3+59√2 216 +2√2]FILIPO TOSCANO A2+4 BT
5 (E+2 F−A2 )(C⋅H )⋅Gt
A2+4 BT
A2=[ 3 −92
4
−4 254
7
−4 92
17 ]
BT=[−1 0 −2
3 12
3
5 2 3 ]
A2+4 BT=[ 3 −9
24
−4 254
7
−4 92
17 ]+ 4 [−1 0 −2
3 12
3
5 2 3 ]= [−1 −92
−4
8 334
19
16 252
29 ]A2+4 BT=
[−1 −92
−4
8 334
19
16 252
29 ]5 (E+2 F−A2 )
(E+2 F−A2 )= [−1 3 −10 5 2
−2 3 −2 ]+2 [−1 3 −1
√3 12 √2
−2 3 3 ]− [ 3 −92
4
−4 254
7
−4 92
17 ]=(E+2 F−A2 )=
[ −6 272
−7
4+2√3 − 14
−5+2√2
−2 92
−13 ]5 (E+2 F−A2 )=
[ −30 1352
−35
20+10√3 −54
−25+10√2
−10 452
−65 ](C⋅H )⋅Gt
C⋅H=[2 66 −8 ] [1 10 −12
2 6 −5 ]= [14 56 −54−10 12 −32 ]
GT=[−1 0 −23 0 312 2 3 ]
(C⋅H )⋅Gt= [14 56 −54−10 12 −32 ][−1 0 −2
3 0 312 2 3 ]= [−494 108 −22
−338 −36 −40 ](C⋅H )⋅Gt= [−494 108 −22
−338 −36 −40 ]
DAVID VÁSCONEZ ET+FT−2 A2
3M⋅L−6M(2 A⋅I )⋅3B
ET+FT−2 A2
ET+FT−2 A2=[−1 0 −23 5 3
−1 2 −2 ]+ [−1 √3 −2
3 12
3
−1 √2 3 ]−2 [3 −9
24
−4 254
7
−4 92
17 ]=ET+FT−2 A2=
[−8 9+√3 −1214 −7 −86 −7+√2 −33 ]
3M⋅L−6M
3M=3 [−5 3 √2 ]= [−15 9 3√2 ]=
3M⋅L= [−15 9 3√2 ] [12 √3 −1
0 12
2
−2 √3 3 ]= [−180−6 √2 92−15√3 33+9√2 ]
6M= [−30 18 6 √2 ]
3M⋅L−6M= [−180−6 √2 92−15√3 33+9√2 ]−[−30 18 6√2 ]=
3M⋅L−6M= [−150−6 √2 −252
−15√3 33+3√2 ](2 A⋅I )⋅3B
(2 A⋅I )=2A= [−2 6 −20 1 4
−4 6 6 ]3B=
3[−1 3 5
0 12
2
−2 3 3 ]= [−3 9 15
0 32
6
−6 9 9 ](2 A⋅I )⋅3B= [−2 6 −2
0 1 4−4 6 6 ][−3 9 15
0 32
6
−6 9 9 ]= [18 −45 −12
−24 752
42
−24 9 30 ](2 A⋅I )⋅3B= [18 −45 −12
−24 752
42
−24 9 30 ]DOMENICA VELASCO ( A⋅B )T−3G
−L⋅G+A /4−E(F⋅I )2+3B
( A⋅B )T−3G
( A⋅B )=[−1 3 −1
0 12
2
−2 3 3 ]∗¿ ¿[−1 3 5
0 12
2
−2 3 3 ]= [ 3 −92
−2
−4 254
7
−4 92
5 ]( A⋅B )T= [ 3 −4 −4
−92
254
92
−2 7 5 ]3G=3 [−1 3 12
0 0 2−2 3 3 ]= [−3 9 36
0 0 6−6 9 9 ]
( A⋅B )T−3G= [ 3 −4 −4
−92
254
92
−2 7 5 ]− [−3 9 360 0 6
−6 9 9 ]( A⋅B )T−3G= [ 6 −13 −40
−92
254
−32
4 −2 −4 ]−L⋅G+A /4−E
L⋅G=[12 √3 −1
0 12
2
−2 √3 3 ] [−1 3 120 0 2
−2 3 3 ]= [−10 33 141+2√3−4 6 7−4 3 −15+2√3 ]
A4
=[−14
34
−14
0 18
12
−12
34
34
]−L⋅G+A /4−E=
[10 −33 −141−2√34 −6 −74 −3 15−2√3 ]+ [−
14
34
−14
0 18
12
− 12
34
34
]− [−1 3 −10 5 2
−2 3 −2 ]=
−L⋅G+A /4−E=[434
−1414
−5614
−2√3
4 −878
−172
112
−214
714
−2√3 ](F⋅I )2+3B
F=[−1 3 −1
√3 12 √2
−2 3 3 ]
F2=[−1 3 −1
√3 12 √2
−2 3 3 ]¿ [−1 3 −1
√3 12 √2
−2 3 3 ]= [ 3+3√3 −92
−2+3√2
−√32
−2√2 3√3+3√2+ 14 −√3+ 72 √2
−4+3√3 92
11+3√2 ]3B=
3[−1 3 5
0 12
2
−2 3 3 ]= [−3 9 15
0 32
6
−6 9 9 ]
(F⋅I )2+3B=[ 3+3√3 −9
2−2+3√2
−√32
−2√2 3√3+3√2+ 14 −√3+ 72 √2
−4+3√3 92
11+3√2 ]+ [−3 9 15
0 32
6
−6 9 9 ]=
(F⋅I )2+3B=[ 3√3 9
213+3√2
−√32
−2√2 3√3+3√2+ 74 6−√3+ 72 √2
−10+3 √3 272
20+3√2 ]TATIANA VERA (−CT⋅I )⋅H
D−5H−J T
E2⋅FT⋅I
(−CT⋅I )⋅H
C=[2 66 −8 ]
CT=[2 66 −8 ]
(−CT⋅I )=−C=
[−2 −6−6 8 ]
(−CT⋅I )⋅H= [−2 −6−6 8 ]∗¿ ¿[1 10 −12
2 6 −5 ]=
(−CT⋅I )⋅H= [−14 −56 5410 −12 102 ]
D−5H−J T
5H=5 [1 10 −122 6 −5 ]= [ 5 50 −60
10 30 −25 ]J=[1 0
3 125 4 ]
JT= [1 3 50 12 4 ]
D=[ 1 0 23
−3 2 −5 ]D−5H−J T=
[ 1 0 23
−3 2 −5 ]− [ 5 50 −6010 30 −25 ]− [1 3 5
0 12 4 ]=
D−5H−J T=[ −5 −53 167
3−13 −40 16 ]
E2⋅FT⋅I
E2=[−1 3 −10 5 2
−2 3 −2 ][−1 3 −10 5 2
−2 3 −2 ]= [ 3 9 9−4 31 66 3 12 ]
FT=[−1 √3 −2
3 12
3
−1 √2 3 ]E2⋅FT⋅I=E2⋅FT=
[ 3 9 9−4 31 66 3 12 ]∗¿ ¿[−1 √3 −2
3 12
3
−1 √2 3 ]=
E2⋅FT⋅I=[15
92+3√3+9√2 48
61 312
−4 √3+6√2 89
−9 32+6 √3+12√2 33 ]
TONNY YASELGA (6 J⋅K )T⋅2C−F⋅K /3+2MT
AT⋅I−BT−2F /3
(6 J⋅K )T⋅2C
NO SE PUEDE RESOLVER: EL NÚMERO DE COLUMNAS DE LA PRIMERA MATRIZ NO ES
IGUAL NÚMERO DE FILAS DE LA SEGUNDA.
−F⋅K /3+2MT
K3
=13 [152359 ]= [
5233593
]F=[−1 3 −1
√3 12 √2
−2 3 3 ]
−F=[ 1 −3 1
−√3 −12
−√22 −3 −3 ]
−F⋅K /3=[ 1 −3 1
−√3 −12
−√22 −3 −3 ]∗¿ ¿[
5233593
]=
−F⋅K /3+2MT=[
53
−236
−5√3−593 √2−72
]AT⋅I−BT−2F /3AT⋅I=AT
A=[−1 3 −1
0 12
2
−2 3 3 ]
AT=[−1 0 −2
3 12
3
−1 2 3 ]
B=[−1 3 5
0 12
2
−2 3 3 ]
BT=[−1 0 −2
3 12
3
5 2 3 ]
F=[−1 3 −1
√3 12 √2
−2 3 3 ]
23F=[ −
23
2 −23
23 √3 1
323 √2
−43
2 2 ]AT⋅I−BT−2F /3=
[−1 0 −2
3 12
3
−1 2 3 ]−[−1 0 −2
3 12
3
5 2 3 ]−[ −23
2 −23
23 √3 1
323 √2
−43
2 2 ]=
AT⋅I−BT−2F /3=[23
2 23
−23 √3 −1
3−23 √2
−143
−2 −2 ]ERICK YUMICEBA 2 A2−10B
3 (M⋅L )−2M(D⋅K )T
2 A2−10B
A2=[ 3 −92
4
−4 254
7
−4 92
17 ]
2 A2=2[ 3 −92
4
−4 254
7
−4 92
17 ]= [ 6 −9 8
−8 252
14
−8 9 34 ]B=[−1 3 5
0 12
2
−2 3 3 ]
10B=[−10 30 500 5 20
−20 30 30 ]
2 A2−10B=[ 6 −9 8
−8 252
14
−8 9 34 ]−[−10 30 500 5 20
−20 30 30 ]=2 A2−10B=
[16 −39 −42
−8 152
−6
12 −21 4 ]3 (M⋅L )−2M
3M=3 [−5 3 √2 ]= [−15 9 3√2 ]=
3M⋅L= [−15 9 3√2 ] [12 √3 −1
0 12
2
−2 √3 3 ]= [−180−6 √2 92−15√3 33+9√2 ]
2M=2 [−5 3 √2 ]= [−10 6 2√2 ]
3 (M⋅L )−2M= [−180−6 √2 92−15√3 33+9√2 ]−[−10 6 2√2 ]=
3 (M⋅L )−2M= [−170−6 √2 −32−15√3 33+7√2 ]
(D⋅K )T
D=[ 1 0 23
−3 2 −5 ]
K=[152359 ](D⋅K )= [ 1 0 2
3−3 2 −5 ]∗¿ ¿[152359 ]= [ 1633−294]
(D⋅K )T=[1633 −294 ]