DEBER CONJUNTO PRIMER PARCIALSantiagoLozada10deagostode2015EJERCICIO1A)HallarlamatrizXtalque3 (X + 2A) 2 (2X + B) = 4 (B 3A + 2X) A =

2 34 1

; B= 2I ARESOLUCION3X 6A4X 4B = 4B 12A + 8XX + 6A2B 4B + 12A = 8X18A6B = 9X18A6 (2I A) = 9X18A12I + 6A = 9X13(18A12I + 6A = 9X)8A4I= 3X8

2 34 1

4

1 00 1

= 3

a bc d

16 2432 8

4 00 4

=

3a 3b3c 3d

12 2432 4

=

3a 3b3c 3d

PorpropiedadReexivaa = 4 b = 8 c =323d =43X =4 832343B)EncontrarunamatrizreducidaporlasR, queseaequivalentealamatrizAyunamatrizinvertiblePtalqueR=PARESOLUCIONA =1 2 1 1... 1 0 01 0 1 2... 0 1 00 2 1 1... 0 0 1f1 + f21 2 1 1... 1 0 00 2 0 3... 1 1 00 2 1 1... 0 0 1f2 1211 2 1 1... 1 0 00 1 032...121200 2 1 1... 0 0 1f1 2f2f2 2f31 0 1 2... 0 1 00 1 032...121200 0 1 4... 1 1 1f1 + f31 0 1 6... 1 2 10 1 032...121200 0 1 4... 1 1 1R =1 0 1 20 1 0320 0 1 4P=1 2 1121201 1 1EJERCICIO2Determine las matrices PyQde orden3x2, tales que satisfagan2P+3Q=A;-P-Q=BA =1 32 41 2B=1 01 31 12 [P Q = B] 2P+ 3Q = A (2)2P 2Q = 2B (1)(1)+(2)2P 2Q = 2B2P+ 3Q = AQ = 2B + AQ = 21 01 31 1+1 32 41 2Q =2 02 62 2+1 32 41 2Q =1 30 23 0P=-Q-BP= 1 30 23 01 01 31 1P=2 31 14 12EJERCICIO3HallarlasmatricesXyYsi:(2X + Y )T=

3Y T+ 4Y

T+ 2A ;

2XT+ 2Y

T= 2(X + B)AT=

2 31 0

; B=

1 20 3

RESOLUCION2XT+ Y T= 3Y+ Y T+ 2A

2Y TX= 2X + 2B

T2XT3Y= 2A (1)

2Y TX

T= (2X + 2B)T. 2Y+ XT= 2XT+ 2BT. 4Y 6XT= 4BT(2)(1) + (2)6XT9Y= 6A6XT+ 4Y= 4BT5Y= 6A + 4BT5Y= 6

2 13 0

+ 4

1 02 3

5Y=

12 618 0

+

4 08 12

15x

5Y=

16 624 12

Y=16565245125RESOLVIENDOEQ(1)2XT= 2

1 02 3

+ 3165652451252XT=

2 04 6

+485185725365XT=345145515185X=3455151451853EJERCICIO4DadalamatrizA=4 2 12 4 21 2 4yseBunamatriztriangularinferiortal queA=BxBThallarlatrazadeBa b c0 d e0 0 fxa 0 0b d 0c e f=4 2 12 4 21 2 4Porpropiedadreexivaf= 4 c =12e = 1 d =3bd + ce = 2 a2+ b2+ c2= 4b3 +12 = 2 a2+34 +14 = 4b =323a =3B=33231203 10 0 2Trz(B) = 23 + 2EJERCICIO5A)HallarunamatrizBtalqueBAB=0,siendoA =

1 22 4

RESOLUCIONB1(BAB= 0)B1BAB= B10IAB= 0AB= 0

1 22 4

a bc d

=

0 00 0

a + 2c b + 2d2a + 4c 2b + 4d

=

0 00 0

a+2c=0 b+2d=0a=-2c b=-2dB=

2c 2dc d

c, dRB)DadalamatrizA=

1 21 1

encuentredeserposibleunamatrizBtal que(A B)1= A1B1RESOLUCION4A B

(A B1) = A1B1

I= AB(A1B1)

XB

B= AB A1

ABA = AB

a bc d

1 21 1

=

1 21 1

a bc d

a b 2a + bc d 2c + d

=

a + 2c b + 2da + c b + d

B=

d 2cc d

c, dRC)seaAntalqueA2= ASeaB=(2A-I)c1)HallarB2enfunciondeAeI,luegohalleB1[B= 2AI] BB.B= (2AI)BB2= 2A.B IBB2= 2A.(2AI) BB2= 4A22A.I BB2= 4A22ABB2= 4A2A(2AI)B2= 2A2A + IB2= IB1= Ic2)HalleEnfunciondeAeIK=1(B= 2AI) .BB2= 2AB BB2= 2A(2AI) (2AI)B2= 4A22A2A + IB2= 4A4A + IB2= IK=2

B2= I

BB3= BB3= 2AIK=3

B3= 2AI

BB4= 2AB BB4= 2A(2AI) (2AI)5B4= 4A22A2A + IB4= 4A4A + IB4= IK=4(B4= I).BB5= IBB5= BB5= 2AIK=5K=KSiKespar=SK+1esimpar

BK= I

.BBK+1= B Por hipotesisBK+1= (2AI) Por hipotesisBn= 2AI si n es imparBn= I si n es parEJERCICIO6HALLARA BporparticionElrangodelamatrizAyBBT ATporparticionRESOLUCIONA =4 3 5 2 1 00 4 6 3 8 11 2 3 6 2 21 2 3 6 7 11 2 5 6 7 11 0 3 5 1 3B=0 3 9 11 3 6 32 4 6 81 4 6 32 4 8 50 2 3 2AB=

a11b11 + a12b21a11b12 + a12b22a21b11 + a22b21a21b12 + a22b22

a11b11=4 3 50 4 61 2 30 31 32 4=13 4116 365 21a12b21=2 1 03 8 16 2 21 42 40 2=4 1219 4610 366a11b12=4 3 50 4 61 2 3a 16 36 8=72 5360 6069 31a12b22=2 1 03 8 16 2 26 38 53 2=20 1185 4758 24a21b11=1 2 31 2 51 0 30 31 32 4=8 2112 296 15a22b21=6 7 16 7 15 1 31 42 40 2=20 5020 507 30a21b12=1 2 31 2 51 0 39 16 36 8=39 3151 4733 25a22b22=6 7 16 7 15 1 36 38 53 2=89 5589 5547 14a11b11 + a12b21=17 5335 8318 59a11b12 + a12b22=104 64144 10795 55a21b11 + a22b21=28 7032 7813 48a21b12 + a22b22=129 86141 10271 39AB=17 53 104 6435 82 145 10718 57 97 5528 71 128 5632 79 140 10213 45 74 497a2RESOLUCIONA =4 3 5 2 1 00 4 6 3 8 11 2 3 6 2 21 2 3 6 7 11 2 5 6 7 11 0 3 5 1 3f1 f31 2 3 6 2 20 4 6 3 8 14 3 5 2 1 01 2 3 6 7 11 2 5 6 7 11 0 3 5 1 3f3 4f1f4 f1f5 f1f6 f18