lista6(sistemas)

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Lista de Exerc´ıcios 6
Jorge C. Lucero
21 de Abril de 2009
1. Resolva os seguintes sistemas lineares usando fatorac¸a˜o LU:
(a)
⎡
⎣
1 0 0
2 1 0
−1 0 −1
⎤
⎦
⎡
⎣
2 3 −1
0 −2 1
0 0 3
⎤
⎦
⎡
⎣
x1
x2
x3
⎤
⎦ =
⎡
⎣
2
−1
1
⎤
⎦
(b)
⎡
⎣
2 0 0
−1 1 0
3 2 −1
⎤
⎦
⎡
⎣
1 1 1
0 1 2
0 0 1
⎤
⎦
⎡
⎣
x1
x2
x3
⎤
⎦ =
⎡
⎣
−1
3
0
⎤
⎦
2. Aplique o algoritmo de fatorac¸a˜o LU nas seguintes matrizes.
(a) A =
⎡
⎣
2 −1 1
3 3 9
3 3 5
⎤
⎦
(b) A =
⎡
⎣
1,012 −2,132 3,104
−2,132 4,906 −7,013
3,104 −7,013 0,014
⎤
⎦
(c) A =
⎡
⎢⎢⎣
2 0 0 0
1 1,5 0 0
0 −3 0.5 0
2 −2 1 1
⎤
⎥⎥⎦
(d) A =
⎡
⎢⎢⎣
2,1756 4,0231 −2,1732 5,1967
−4,0231 6,0000 0 1,1973
−1,0000 −5,2107 1,1111 0
6,0235 7,0000 1 −4,1561
⎤
⎥⎥⎦
3. Resolva os seguintes sistemas lineares usando Fatorac¸a˜o LU
(a)
⎧⎨
⎩
2x1 − x2 + x3 = −1,
3x1 + 3x2 + 9x3 = 0,
3x1 + 3x2 + 5x3 = 4.
(b)
⎧⎨
⎩
1,012x1 − 2,132x2 + 3,104x3 = 1,984,
−2,132x1 + 4,096x2 − 7,013x3 = −5,049,
3,104x1 − 7,013x2 + 0,014x3 = −3,895.
(c)
⎧⎪⎪⎨
⎪⎪⎩
2x1 = 3,
x1 + 1,5x2 = 4,5,
− 3x2 + 0,5x3 = −6,6,
2x1 − 2x2 + x3 + x4 = 0,8,
(d)
⎧⎪⎪⎨
⎪⎪⎩
2,1756x1 + 4,0231x2 − 2,1732x3 + 5,1967x4 = 17,102,
−4,0231x1 + 6,0000x2 + 1,1973x4 = −6,1593,
−1,0000x1 − 5,2107x2 + 1,1111x3 = 3,0004,
6,0235x1 + 7,0000x2 − 4,1561x4 = 0,0000
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4. Obtenha fatorac¸o˜es da forma PA = LU para as seguintes matrizes.
(a) A =
⎡
⎣
0 2 3
1 1 −1
0 −1 1
⎤
⎦
(b) A =
⎡
⎣
1 2 −1
1 2 3
2 −1 4
⎤
⎦
(c) A =
⎡
⎢⎢⎣
1 −2 3 0
3 −6 9 3
2 1 4 1
1 −2 2 −2
⎤
⎥⎥⎦
(d) A =
⎡
⎢⎢⎣
1 −2 3 0
1 −2 3 1
1 −2 2 −2
2 1 3 −1
⎤
⎥⎥⎦
5. Calcule a inversa das matrizes A no exerc´ıcio 2, utilizando os fatores L e U ja´ obtidos.
6. Calcule o determinante das matrizes A no exerc´ıcio 4.
Respostas
1. (a) x1 = −3, x2 = 3, x3 = 1
(b) x1 = 1/2, x2 = −9/2, x3 = 7/2
2. (a) L =
⎡
⎣
1 0 0
1,5 1 0
1,5 1 1
⎤
⎦ U =
⎡
⎣
2 −1 1
0 4,5 7,5
0 0 −4
⎤
⎦
(b) L =
⎡
⎣
1 0 0
−2,106719 1 0
3,067193 1,197756 1
⎤
⎦ U =
⎡
⎣
1,012 −2,132 3,104
0 −0,3955257 −0,4737443
0 0 −8,939141
⎤
⎦
(c) L =
⎡
⎢⎢⎣
1 0 0 0
0,5 1 0 0
0 −2 1 0
1 −1,33333 2 1
⎤
⎥⎥⎦ U =
⎡
⎢⎢⎣
2 0 0 0
0 1,5 0 0
0 0 0,5 0
0 0 0 1
⎤
⎥⎥⎦
(d) L =
⎡
⎢⎢⎣
1 0 0 0
−1,849190 1 0 0
−0,4596433 −0,2501219 1 0
2,768661 −0,3079435 −5,352283 1
⎤
⎥⎥⎦ U =
⎡
⎢⎢⎣
2,175600 4,023099 −2,173199 5,196
0 13,43947 −4,0186600 10,80
0 0 −0,8929510 5,09
0 0 0 12,03
3. (a) x1 = 1, x2 = 2, x3 = −1.
(b) x1 = 1, x2 = 1, x3 = 1.
(c) x1 = 1,5, x2 = 2, x3 = −1,199998, x4 = 3.
(d) x1 = 2,939851, x2 = 0,07067770, x3 = 5,677735, x4 = 4,379812.
2
4. (a) P =
⎡
⎣
0 1 0
1 0 0
0 0 1
⎤
⎦ L =
⎡
⎣
1 0 0
0 1 0
0 −1/2 0
⎤
⎦ U =
⎡
⎣
1 1 −1
0 2 3
0 0 5/2
⎤
⎦
(b) P =
⎡
⎣
1 0 0
0 0 1
0 1 0
⎤
⎦ L =
⎡
⎣
1 0 0
2 1 0
1 0 1
⎤
⎦ U =
⎡
⎣
1 2 −1
0 −5 6
0 0 4
⎤
⎦
(c) P =
⎡
⎢⎢⎣
1 0 0 0
0 0 1 0
0 0 0 1
0 1 0 0
⎤
⎥⎥⎦ L =
⎡
⎢⎢⎣
1 0 0 0
2 1 0 0
1 0 1 0
3 0 0 1
⎤
⎥⎥⎦ U =
⎡
⎢⎢⎣
1 −2 3 0
0 5 −2 1
0 0 −1 −2
0 0 0 3
⎤
⎥⎥⎦
(d) P =
⎡
⎢⎢⎣
1 0 0 0
0 0 0 1
0 0 1 0
0 1 0 0
⎤
⎥⎥⎦ L =
⎡
⎢⎢⎣
1 0 0 0
2 1 0 0
1 0 1 0
1 0 0 1
⎤
⎥⎥⎦ U =
⎡
⎢⎢⎣
1 −2 3 0
0 5 −3 −1
0 0 −1 −2
0 0 0 1
⎤
⎥⎥⎦
5. (a) A−1 =
⎡
⎣
0,3333 −0,2222 0,3333
−0,3333 −0,1944 0,4167
0 0,2500 −0,2500
⎤
⎦
(b) A−1 =
⎡
⎣
11,6531 5,1579 0,0656
5,1579 2,2827 −0,1138
0,0656 −0,1138 −0,0995
⎤
⎦
(c) A−1 =
⎡
⎢⎢⎣
0,5000 0 0 0
−0,3333 0,6667 0 0
−2,0000 4,0000 2,0000 0
0,3333 −2,6667 −2,0000 1,0000
⎤
⎥⎥⎦
(d) A−1 =
⎡
⎢⎢⎣
0,0934 0,0842 0,0994 0,0925
0,0202 0,0885 −0,0061 0,0508
0,1789 0,3393 0,9606 0,3215
0,2125 0,1086 0,3649 0,0564
⎤
⎥⎥⎦
6. (a) 5; (b) 20; (c) -15; (d) 5.
Fonte: Richard L. Burden e J. Douglas Faires. Ana´lise Nume´rica. Cengage Learning, Sa˜o Paulo, 2008.
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