Matrices

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EJERCICIOS DE MATRICES

1.-Dada las matrices A=(2¦3 (-1)¦2 ) , B=(■(0&1@4&-2)) y C= (1¦2 3¦(-1) 5¦1 ), calcula si es posible:
a) A+B
A+B =(2¦3 (-1)¦2 )+(■(0&1@4&-2))= ((2+0)¦(3+4) (-1+1)¦(2+(-2)))
b) AC=(■(0&1@4&-2))(1¦2 3¦(-1) 5¦1 )= ((2.1+(-1)2)¦(3.1+2.2) ( 2.3+(-1)(-1))¦(3.3+2(-1)) (2.5+(-1)1)¦(3.5+2.1))=(0¦7 7¦7 9¦17)
c) (2A+B) C
2A+B= (2.2¦2.3 (2(-1))¦2.2)+(■(0&1@4&-2))=(■(4&-2@6&4))+(■(0&1@4&-2))= ((4+0)¦(6+4) ( (-2)+1)¦( 4+(-2)))=(■(4&-1@10&2))
Así, (2A+B) C =(■(4&-1@10&2))(1¦2 3¦(-1) 5¦1 )=((4.1+(-1)2)¦(10.1+2.2) ( 4.3+(-1)(-1))¦(10.3+2(-1)) (4.5+(-1)1)¦(10.5+2.1))=(2¦14 13¦28 19¦52 )

2.-Dada las matrices A = (■(2&6&3@0&9&5@-6&2&1)) y B (■(1&1&1@2&-4&2@3&5&7)) , se pide:
a) Calcular AB y BA ¿coinciden los resultados?
AB=(■(2&6&3@0&9&5@-6&2&1))(■(1&1&1@2&-4&2@3&5&7))=(■(2.1+6.2+3.3&2.1+6(-4)+3.5&2.1+6.2+3.7@0.1+9.2+5.3&0.1+9(-4)+5.5&0.1+9.2+5.7@-6.1+2.2+1.3&-6.1+2.(-4)+1.5&-6.1+2.2+1.7))=
= (■(23&-7&35@33&-11&53@1&-9&5))
AB =(■(1&1&1@2&-4&2@3&5&7))(■(2&6&3@0&9&5@-6&2&1))=(■(1.2+1.0+1(-6)&1.6+1.9+1.2&1.3+1.5+1.1@2.2+(-4)0+2(-6)&2.6+(-4)+2.2&2.3+(-4)5+2.1@3.2+5.0+7(-6)&3.6+5.9+7.2&3.3+5.5+7.1))=(■(-4&17&9@-8&-20&-12@-36&77&41))

b) A+B=(■(2&6&3@0&9&5@-6&2&1))+(■(1&1&1@2&-4&2@3&5&7))=(■(2+1&6+1&3+1@0+2&9+(-4)&5+2@-6+3&2+5&1+7))=(■(3&7&4@2&5&7@-3&7&8))
(A + 〖B)〗^2=(A + B )(A + B)=(■(3&7&4@2&5&7@-3&7&8))(■(3&7&4@2&5&7@-3&7&8))=(■(3.3+7.2+4(-3)&3.7+7.5+4.7&3.4+7.7+48@2.3+5.2+7(-3)&2.7+5.5+7.7&2.4+5.7+7.8@(-3)3+7.2+8.(-3)&(-3)7+7.5+8.7&(-3)3+7.7+8.8))=
(■(11&84&93@-5&88&99@-19&70&101))
A continuación, se calcula A^2 + 2AB + B^2
A^2=AA=(■(2&6&3@0&9&5@-6&2&1))(■(2&6&3@0&9&5@-6&2&1))=(■(2.2+6.0+3(-6)&2.6+6.9+3.2&4.3+6.5+3.1@0.2+9.0+5(-6)&0.6+9.9+5.2&0.3+9.5+.5.7@-6.2+2.0+1(-6)&-6.6+2.9.+1.2&-6.3+2.5+.1.1))
(■(-14&72&39@-30&91&50@-18&-16&-7))
La matriz AB se ha calculado en el aparato a) , así2AB=2(■(23&-7&35@33&-11&53@1&-9&5))=(■(2.23&2(-7)&2.35@2.33&2(-11)&2.53@2.1&2(-9)&2.5))=(■(46&-14&70@66&-22&106@2&-18&10))
B^2=BB=(■(1&1&1@2&-4&2@3&5&7))(■(1&1&1@2&-4&2@3&5&7))=(■(1.1+1.2+1.3&1.1+1(-4)+1.5&1.1+1.2+1.7@2.1+(-4)2+2.3&2.1+(-4)(-4)+2.5&2.1+(-4)2+2.7@3.1+5.2+7.3&3.1+5(-4)+7.5&3.1+5.2+7.7))=
(■(6&2&10@0&28&8@34&18&62))
Por lo tanto: A^2 + 2AB + B^2=(■(-14&72&39@-30&91&50@-18&-16&-7))+(■(46&-14&70@66&-22&106@2&-18&10))+(■(6&2&10@0&28&8@34&18&-7))=

=(■(-14+46+6&72+(-14)&39+70+10@-30+66+0&91+(-14)&50+106+8@-18+2+34&-16(-14)&-7+10+62))=(■(38&60&119@36&97&164@18&-16&65))

En conclusión (A + 〖B)〗^2= A^2 + 2AB + B^2

3.-Determina la matriz de cofactores de A.

A=(■(1&3&2@-1&2&0@2&4&1))

C_(11=) |■(2&0@4&1)|= 2 - 0 =2 C_(12=)-|■(-1&0@2&1)|=1 (-1 – 0)=1C_(13=)-|■(-1&2@2&4)|= - 4 – 4 = - 8

C_(21=)=|■(3&2@4&1)| = -(3 - 8) =5 C_(22=) |■(1&2@2&1)|= 1 – 4= - 3 C_(23=) |■(1&3@2&4)|=-(4 - 6)=2

C_(31=) |■(3&2@2&0)|= 0 – 4 = - 4 C_(32=) |■(1&2@-1&0)|=- (0 + 2)= -2 C_(33=) |■(1&3@-1&2)|= 2 + 3= 5

Por lo tanto A_(c=) [■(2&1&-8@5&-3&2@-4&-2&5)]4.-Resolver el siguiente sistema
{█(x+4y-z=3@x+2y 3z= -5@2x+y=0)┤
[■(1&4&-1&M&3@1&2&3&M&-5@2&1&1&M&0) ] ■(-F_1+F_2@-2F_(2+ ) F_3 ) =[■(1&4&-1&M&3@0&-2&4&M&-8@0&-7&1&M&2) ] - 1/2 F_1 =
[■(1&4&-1&M&3@0&1&-2&M&4@0&-7&3&M&-6) ] ■(-4F_2+F_1@7F_(1+ ) F_1 )= [■(1&0&7&M&-13@0&1&-2&M&4@0&0&-11&M&22) ] - 1/11 F_3 =
[■(1&0&7&M&13@0&1&-2&M&4@0&0&1&M&-2) ] ■(-7F_3+F_1@2F_(3+ ) F_2)[■(1&0&0&M&1@0&1&0&M&0@0&0&1&M&-2) ]
Por lo tanto:
x=1 y=0 z=-2

5.-hallar la inversa de A =[■(2&3@-1&1)]
[■(2&3@-1&1)][■(a&b@c&d)]=[■(1&0@0&1)] ⟹[■(2a+3c&2b+3d@-a+c&-b+d)]=[■(1&0@0&1)]
luego
2a+3c=1
(-a+c=0)/( 5c=1)
5c =1/11
a=1/5

2b+3d=0
(-b+d=1)/( 5d=2)
5c =2/5
b=-3/5
A^(-1)+
6.-Si:...
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