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 ]