Embed Size (px)
DESCRIPTION
calculo de MATRIZ
Citation preview
[−2 −1 0 23 1 −2 −2
−4 −1 2 33 1 −1 −2]
1 0 0 00 1 0 00 0 1 00 0 0 1
|L1 .(−12 )
¿¿
L1((−2 ).(−12 )=1 )
((−1) .(−12 )=1 /2)((0) .(−12 )=0)(2.(−12 )=−1)
((1 ).(−12 )=−1/2)((0 ) .(−12 )=0)((0 ) .(−12 )=0)((0 ) .(−12 )=0)
[ 1 1/2 0 −13 1 −2 −2
−4 −1 2 33 1 −1 −2]
−1/2 0 0 00 1 0 00 0 1 00 0 0 1
|L2=L2+L1.(−3)L3=L3+L1.(4)L4=L4+ L1.(−3)
L2 ( (3 )+(1.(−3) )=0)
((1)+( 12 .(−3 ))=−1 /2)((−2) +(0. (−3) )=−2)
( (−2 )+((−1) . (−3 ))=1)
((O )+(−12 .(−3))=3 /2)( (1) +(0. (−3 ))=1)((0) +(0. (−3 ))=0)((0) +(0. (−3 ))=0)
L3 ( (−4 )+(1.(4) )=0 )
( (−1 )+( 12 .(4))=1)((2)+(0. (4) )=2)
((3)+ ((−1 ) . (4) )=−1)
((O )+(−12 .(4 ))=−2)((0)+(0. (4) )=0)((1)+(0. (4) )=1)((0)+(0. (4) )=0)
L4 ( (3 )+(1.(−3) )=0)
((1) +( 12 .(−3))=−1 /2)( (−1 )+(0. (−3 ))=−1 )
( (−2)+((−1). (−3 ))=1)
((O )+(−12 .(−3))=3/2)( (0 )+(0. (−3 ))=0 )( (0 )+(0. (−3 ))=0 )( (1 )+(0. (−3 ))=1)
[1 1 /2 0 −10 −1 /2 −2 10 1 2 −10 −1 /2 −1 1
]−1/2 0 0 03 /2 1 0 0−2 0 1 03 /2 0 0 1
|L2=L2.−2¿
L2 ((0 ) . (−2)=0)((−1/2) . (−2)=1)
((−2) . (−2)=4 )(1 . (−2)=−2 )
( (3 /2 ). (−2 )=−3)((1) . (−2)=−2 )( (0 ). (−2 )=0)( (0 ). (−2 )=0)
[1 1 /2 0 −10 1 4 −20 1 2 −10 −1 /2 −1 1
]−1/2 0 0 0−3 −2 0 0−2 0 1 03 /2 0 0 1
| L1=L1+L2.(−12
)
¿ L3=L3+L2.(−1)
L4=L4+L2.( 12)
L1 ((1)+(0.(−1/2))=1)( (1 /2 )+(1.(−1/2) )=0)((0) +(4. (−1 /2 ))=−2)
((−1)+( (−2 ) . (−1 /2 ))=0 )
((−1/2)+ (−3.(−1/2))=1 )( (0 )+(−2. (−1/2) )=1 )( (0 )+(0. (−1 /2) )=0)( (0 )+(0. (−1 /2) )=0)
L3 ((0)+ (0.(−1))=0 )( (1)+(1.(−1))=0 )
( (2) +(4. (−1 ))=−2)((−1) +((−2) . (−1))=1 )
((−2)+ (−3.(¿1) )=1 )( (0 )+(−2. (−1 ))=2)((1)+(0. (−1))=1 )( (0)+(0. (−1))=0 )
L4 ( (0) +(0.(1/2))=0 )( (−1 /2 )+(1.(1/2))=0 )
( (−1 )+(4. (1 /2 ))=1)( (1 )+((−2) . (1/2) )=0)
((3 /2)+(−3.(1 /2) )=1/2)((0)+ (−2. (1/2))=−1)
((0) +(0. (1/2))=0 )( (1) +(0. (1/2))=1 )
[1 0 −2 00 1 4 −20 0 −2 10 0 1 0
] 1 1 0 0−3 1 0 01 2 1 00 −1 0 1
|¿
L3 ((0) . (−1/2)=0 )((0) . (−1/2)=0 )
( (−2 ) . (−1/2)=1)(1. (−1/2)=−1 /2 )
((1 ) . (−1 /2 )=−1/2)( (2) . (−1/2)=−1 )
((1 ) . (−1 /2 )=−1/2)((0 ) . (−1/2)=0)
[1 0 −2 00 1 4 −20 0 1 −1/20 0 1 0
] 1 1 0 0−3 −2 0 0
−1 /2 −1 −1/2 00 −1 0 1
| L1=L1+L3.(2)L2=L2+L3.(−4)
¿ L4=L4+ L3.(−1)
L1 ( (1 )+(0.(2) )=1 )((0)+(0 .(2 ))=0 )
( (−2 )+(1 . (2) )=0)((0)+((−1 /2) . (2 ))=−1)
( (1) +(−1 /2 .(2) )=0 )( (1)+ (−1 . (2) )=−1)
( (0 )+(−1/2 . (2 ))=−1 )( (0 )+(0. (2 ))=0 )
L2 ( (0 )+(0.(−4 ))=0 )( (1 )+(0.(−4 ))=1 )( (4 )+(1. (−4) )=0)
((−2)+ ((−1/2) . (−4 ))=0)
( (−3 )+(−1/2.(−4) )=−1)( (−2 )+(−1. (−4 ))=2)( (0 )+(−1/2. (−4 ))=2)
( (0 )+(0. (−4) )=0)
L4 ((0) +(0.(−1) )=0 )((0) +(0.(−1) )=0 )( (1)+ (1. (−1))=0 )
( (0 )+((−1 /2) . (−1 ))=1/2)
( (0 )+(−1/2.(−1 ))=1/2)( (−1)+(−1. (−1) )=0 )
( (0 )+(−1/2. (−1 ))=1/2)((1)+(0. (−1))=1 )
[1 0 0 −10 1 0 00 0 1 −1/20 0 0 1/2 ] 0 −1 −1 0
−1 2 2 0−1/2 −1 −1/2 01/2 0 1/2 1
|L4=L4.2
L4 ( (0 ). (2)=0 )( (0 ). (2)=0 )( (0 ). (2)=0 )(1 /2. (2)=1)
((1/2) . (2)=1)( (0 ). (2 )=0 )
((1/2) . (2)=1)( (1 ). (2 )=2 )
[1 0 0 −10 1 0 00 0 1 −1/20 0 0 1
] 0 −1 −1 0−1 2 −2 0
−1/2 −1 −1/2 01 0 1 2
| L1=L1+L4.(1)¿ L3=L3+L4.( 12)
¿
L1 ( (1 )+(0.(1))=1 )( (0 )+(0.(1))=0 )((0) +(0. (1))=0 )
((−1)+( (1) . (1) )=0)
( (0 )+(1.(1) )=1)( (−1 )+(0. (1) )=−1)((−1)+ (1. (1 ))=0 )( (0 )+(2. (1) )=2)
L3 ((0)+ (0.(1 /2 ))=0 )((0)+ (0.(1 /2 ))=0 )((1)+(0. (1 /2 ))=1)
((−1/2) +((1 ) . (1 /2 ))=0)
( (−1 /2 )+(1.(1 /2) )=0)( (−1 )+(0. (1 /2) )=−1)( (−1 /2 )+(1. (1/2) )=0)
( (0 )+(2. (1 /2 ))=1)
[1 0 0 00 1 0 00 0 1 00 0 0 1
] 1 −1 0 2−1 2 −2 00 −1 0 11 0 1 2
|
[−2 −1 0 23 1 −2 −2
−4 −1 2 33 1 −1 −2] X [ 1 −1 0 2
−1 2 −2 00 −1 0 11 0 1 2
]=1 0 0 00 1 0 00 0 1 00 0 0 1
|
PROVAR a22
[−2 −1 0 23 1 −2 −2
−4 −1 2 33 1 −1 −2] X [ 1 −1 0 2
−1 2 −2 00 −1 0 11 0 1 2
] = (-3+2+2+0)= 1
PROVAR a34
[−2 −1 0 23 1 −2 −2
−4 −1 2 33 1 −1 −2] X [ 1 −1 0 2
−1 2 −2 00 −1 0 11 0 1 2
] = (-8+0+2+6)= 0
PROVAR a41
[−2 −1 0 23 1 −2 −2
−4 −1 2 33 1 −1 −2] X [ 1 −1 0 2
−1 2 −2 00 −1 0 11 0 1 2
] = (3-1+0-2)= 0
PROVAR a13
[−2 −1 0 23 1 −2 −2
−4 −1 2 33 1 −1 −2] X [ 1 −1 0 2
−1 2 −2 00 −1 0 11 0 1 2
] – (0+2+0+2)= 0
