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# Matrizes Vetores e Geometria Analítica

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```Elementares (opcional)
Definic¸a\u2dco 1.8. Uma matriz elementar \ud45b×\ud45b e´ uma matriz obtida da matriz identidade \ud43c\ud45b aplicando-se
uma, e somente uma, operac¸a\u2dco elementar.
Vamos denotar por \ud438\ud456\ud457 a matriz elementar obtida trocando-se a linha \ud456 com a linha \ud457 da matriz \ud43c\ud45b,
\ud438\ud456(\ud6fc) a matriz elementar obtida multiplicando-se a linha \ud456 da matriz \ud43c\ud45b pelo escalar \ud6fc \u2215= 0 e \ud438\ud456,\ud457(\ud6fc)
Matrizes Vetores e Geometria Anal\u131´tica Marc¸o 2010
1.2 Sistemas de Equac¸o\u2dces Lineares 57
a matriz elementar obtida da matriz \ud43c\ud45b, somando-se a` linha \ud457, \ud6fc vezes a linha \ud456.
\ud438\ud456,\ud457 =
\u23a1
\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3
1 0 \u22c5 \u22c5 \u22c5 \u22c5 \u22c5 \u22c5 0
0
.
.
. \u22c5
\u22c5 1 \u22c5
\u22c5 0 . . . 1 \u22c5
\u22c5
.
.
.
.
.
.
.
.
. \u22c5
\u22c5 1 . . . 0 \u22c5
\u22c5 1 \u22c5
\u22c5
.
.
. 0
0 \u22c5 \u22c5 \u22c5 \u22c5 \u22c5 \u22c5 0 1
\u23a4
\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6
\u2190 \ud456
\u2190\ud457
, \ud438\ud456(\ud6fc) =
\u23a1
\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3
1 0 \u22c5 \u22c5 \u22c5 \u22c5 0
0
.
.
. \u22c5
\u22c5 1 \u22c5
\u22c5 \ud6fc \u22c5
\u22c5 1 \u22c5
\u22c5 . . . 0
0 \u22c5 \u22c5 \u22c5 \u22c5 0 1
\u23a4
\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6
\u2190 \ud456
e \ud438\ud456,\ud457(\ud6fc) =
\u23a1
\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3
1 0 \u22c5 \u22c5 \u22c5 \u22c5 0
0
.
.
. \u22c5
\u22c5 1 \u22c5
\u22c5 ... . . . \u22c5
\u22c5 \ud6fc . . . 1 \u22c5
\u22c5 . . . 0
0 \u22c5 \u22c5 \u22c5 \u22c5 0 1
\u23a4
\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6
\u2190 \ud456
\u2190\ud457
Exemplo 1.17. As matrizes seguintes sa\u2dco as matrizes elementares 2× 2:
\ud4381,2 = \ud4382,1 =
[
0 1
1 0
]
, \ud4381(\ud6fc) =
[
\ud6fc 0
0 1
]
, \ud4382(\ud6fc) =
[
1 0
0 \ud6fc
]
, com \ud6fc \u2215= 0,
\ud4381,2(\ud6fc) =
[
1 0
\ud6fc 1
]
e \ud4382,1(\ud6fc) =
[
1 \ud6fc
0 1
]
.
Marc¸o 2010 Reginaldo J. Santos
58 Matrizes e Sistemas Lineares
Sejam \ud4381 =
\u23a1
\u23a2\u23a2\u23a2\u23a3
1
0
.
.
.
0
\u23a4
\u23a5\u23a5\u23a5\u23a6 , \ud4382 =
\u23a1
\u23a2\u23a2\u23a2\u23a3
0
1
.
.
.
0
\u23a4
\u23a5\u23a5\u23a5\u23a6,. . . , \ud438\ud45b =
\u23a1
\u23a2\u23a2\u23a2\u23a3
0
0
.
.
.
1
\u23a4
\u23a5\u23a5\u23a5\u23a6 matrizes \ud45a× 1.
As matrizes elementares podem ser escritas em termos das matrizes \ud438\ud456 como
\ud438\ud456,\ud457 =
\u23a1
\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3
\ud438\ud4611
.
.
.
\ud438\ud461\ud457
.
.
.
\ud438\ud461\ud456
.
.
.
\ud438\ud461\ud45a
\u23a4
\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6
\u2190 \ud456
\u2190\ud457
, \ud438\ud456(\ud6fc) =
\u23a1
\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3
\ud438\ud4611
.
.
.
\ud6fc\ud438\ud461\ud456
.
.
.
\ud438\ud461\ud45a
\u23a4
\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6\u2190 \ud456 e \ud438\ud456,\ud457(\ud6fc) =
\u23a1
\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3
\ud438\ud4611
.
.
.
\ud438\ud461\ud456
.
.
.
\ud438\ud461\ud457 + \ud6fc\ud438
\ud461
\ud456
.
.
.
\ud438\ud461\ud45a
\u23a4
\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6
\u2190 \ud456
\u2190\ud457
Aplicar uma operac¸a\u2dco elementar em uma matriz, corresponde a multiplicar a matriz a` esquerda
por uma matriz elementar, como mostra o resultado a seguir.
Teorema 1.8. Sejam \ud438 uma matriz elementar \ud45a×\ud45a e \ud434 uma matriz qualquer \ud45a× \ud45b. Enta\u2dco, \ud438\ud434 e´
igual a` matriz obtida aplicando-se na matriz \ud434 a mesma operac¸a\u2dco elementar que originou \ud438.
Matrizes Vetores e Geometria Anal\u131´tica Marc¸o 2010
1.2 Sistemas de Equac¸o\u2dces Lineares 59
Demonstrac¸a\u2dco. Como a \ud456-e´sima linha de um produto de matrizes \ud435\ud434 e´ igual a \ud435\ud456\ud434, em que \ud435\ud456 e´ a
\ud456-e´sima linha da matriz \ud435 (Exerc\u131´cio 1.1.17 (b) na pa´gina 28) e \ud438\ud461\ud456\ud434 = \ud434\ud456, em que \ud434\ud456 e´ a linha \ud456 da
matriz \ud434 (Exerc\u131´cio 15 (b) na pa´gina 26), enta\u2dco:
\ud438\ud456,\ud457\ud434 =
\ud456\u2192
\ud457\u2192
\u23a1
\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3
\ud438\ud4611
.
.
.
\ud438\ud461\ud457
.
.
.
\ud438\ud461\ud456
.
.
.
\ud438\ud461\ud45a
\u23a4
\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6
\ud434 =
\u23a1
\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3
\ud438\ud4611\ud434
.
.
.
\ud438\ud461\ud457\ud434
.
.
.
\ud438\ud461\ud456\ud434
.
.
.
\ud438\ud461\ud45a\ud434
\u23a4
\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6
\u2190 \ud456
\u2190\ud457
=
\u23a1
\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3
\ud4341
.
.
.
\ud434\ud457
.
.
.
\ud434\ud456
.
.
.
\ud434\ud45a
\u23a4
\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6
\u2190 \ud456
\u2190\ud457
\ud438\ud456(\ud6fc)\ud434 = \ud456\u2192
\u23a1
\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3
\ud438\ud4611
.
.
.
\ud6fc\ud438\ud461\ud456
.
.
.
\ud438\ud461\ud45a
\u23a4
\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6 \ud434 =
\u23a1
\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3
\ud438\ud4611\ud434
.
.
.
\ud6fc\ud438\ud461\ud456\ud434
.
.
.
\ud438\ud461\ud45a\ud434
\u23a4
\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6\u2190 \ud456 =
\u23a1
\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3
\ud4341
.
.
.
\ud6fc\ud434\ud456
.
.
.
\ud434\ud45a
\u23a4
\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6\u2190 \ud456
Marc¸o 2010 Reginaldo J. Santos
60 Matrizes e Sistemas Lineares
\ud438\ud456,\ud457(\ud6fc)\ud434 =
\ud456\u2192
\ud457\u2192
\u23a1
\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3
\ud438\ud4611
.
.
.
\ud438\ud461\ud456
.
.
.
\ud438\ud461\ud457 + \ud6fc\ud438
\ud461
\ud456
.
.
.
\ud438\ud461\ud45a
\u23a4
\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6
\ud434 =
\u23a1
\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3
\ud438\ud4611\ud434
.
.
.
\ud438\ud461\ud456\ud434
.
.
.
\ud438\ud461\ud457\ud434+ \ud6fc\ud438
\ud461
\ud456\ud434
.
.
.
\ud438\ud461\ud45a\ud434
\u23a4
\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6
\u2190 \ud456
\u2190\ud457
=
\u23a1
\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a2\u23a3
\ud4341
.
.
.
\ud434\ud456
.
.
.
\ud434\ud457 + \ud6fc\ud434\ud456
.
.
.
\ud434\ud45a
\u23a4
\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a5\u23a6
\u2190 \ud456
\u2190\ud457
\u25a0
Assim, aplicar uma sequ¨e\u2c6ncia de operac¸o\u2dces elementares em uma matriz, corresponde a multipli-
car a matriz a` esquerda por um produto de matrizes elementares.
Exemplo 1.18. Quando usamos o me´todo de Gauss-Jordan para resolver o sistema do Exemplo 1.11
na pa´gina 39, aplicamos uma sequ¨e\u2c6ncia de operac¸o\u2dces elementares na matriz aumentada do sistema.
Isto corresponde a multiplicar a matriz aumentada
[\ud434 \u2223\ud435 ] =
\u23a1
\u23a3 1 1 1 10002 1 4 2000
2 3 5 2500
\u23a4
\u23a6
a` esquerda pelas matrizes elementares
\ud4381,2(\u22122) =
\u23a1
\u23a3 1 0 0\u22122 1 0
0 0 1
\u23a4
\u23a6 , \ud4381,3(\u22122) =
\u23a1
\u23a3 1 0 00 1 0
\u22122 0 1
\u23a4
\u23a6 ,
Matrizes Vetores e Geometria Anal\u131´tica Marc¸o 2010
1.2 Sistemas de Equac¸o\u2dces Lineares 61
\ud4382(\u22121) =
\u23a1
\u23a3 1 0 00 \u22121 0
0 0 1
\u23a4
\u23a6 , \ud4382,1(\u22121) =
\u23a1
\u23a3 1 \u22121 00 1 0
0 0 1
\u23a4
\u23a6 , \ud4382,3(\u22121) =
\u23a1
\u23a3 1 0 00 1 0
0 \u22121 1
\u23a4
\u23a6
\ud4383(
1
5
) =
\u23a1
\u23a3 1 0 00 1 0
0 0 1
5
\u23a4
\u23a6 , \ud4383,1(\u22123) =
\u23a1
\u23a3 1 0 \u221230 1 0
0 0 1
\u23a4
\u23a6 , \ud4383,2(2) =
\u23a1
\u23a3 1 0 00 1 2
0 0 1
\u23a4
\u23a6 ,
ou seja,
\ud4383,2(2)\ud4383,1(\u22123)\ud4383(15)\ud4382,3(\u22121)\ud4382,1(\u22121)\ud4382(\u22121)\ud4381,3(\u22122)\ud4381,2(\u22122) [\ud434 \u2223\ud435 ]=
\u23a1
\u23a3 1 0 0 7000 1 0 200
0 0 1 100
\u23a4
\u23a6 .
Marc¸o 2010 Reginaldo J. Santos
62 Matrizes e Sistemas Lineares
Exerc\u131´cios Nume´ricos (respostas na pa´gina 549)
1.2.1. Quais das seguintes matrizes esta\u2dco na forma escalonada reduzida:
\ud434 =
\u23a1
\u23a3 1 0 0 0 30 0 1 0 \u22124
0 0 0 1 2
\u23a4
\u23a6,
\ud436 =
\u23a1
\u23a2\u23a2\u23a3
1 0 0 0 3
0 0 1 0 0
0 0 0 1 2
0 0 0 0 0
\u23a4
\u23a5\u23a5\u23a6,
\ud435 =
\u23a1
\u23a3 0 1 0 0 \u221240 0 1 0 5
0 0 0 \u22121 2
\u23a4
\u23a6,
\ud437 =
\u23a1
\u23a2\u23a2\u23a3
0 0 0 0 0
0 0 1 2 \u22124
0 0 0 1 0
0 0 0 0 0
\u23a4
\u23a5\u23a5\u23a6.
dente.
(a)
\u23a1
\u23a3 1 0 0 \u22127 80 1 0 3 2
0 0 1 1 \u22125
\u23a4
\u23a6;
(b)
\u23a1
\u23a2\u23a2\u23a3
1 \u22126 0 0 3 \u22122
0 0 1 0 4 7
0 0 0 1 5 8
0 0 0 0 0 0
\u23a4
\u23a5\u23a5\u23a6;
(c)
\u23a1
\u23a3 1 0 0 0 60 1 0 0 3
0 0 1 1 2
\u23a4
\u23a6;
(d)
\u23a1
\u23a2\u23a2\u23a3
1 7 0 0 \u22128 \u22123
0 0 1 0 6 5
0 0 0 1 3 9
0 0 0 0 0 0
\u23a4
\u23a5\u23a5\u23a6.
1.2.3. Resolva, usando o me´todo de Gauss-Jordan, os seguintes sistemas:
(a)
\u23a7\u23a8
\u23a9
\ud4651 + \ud4652 + 2\ud4653 = 8
\u2212\ud4651 \u2212 2\ud4652 + 3\ud4653 = 1
3\ud4651 \u2212 7\ud4652 + 4\ud4653 = 10
;
Matrizes Vetores e Geometria Anal\u131´tica Marc¸o 2010
1.2 Sistemas de Equac¸o\u2dces Lineares 63
(b)
\u23a7\u23a8
\u23a9
2\ud4651 + 2\ud4652 + 2\ud4653 = 0
\u22122\ud4651 + 5\ud4652 + 2\ud4653 = 1
8\ud4651 + \ud4652 + 4\ud4653 = \u22121
;
(c)
\u23a7\u23a8
\u23a9
\u2212 2\ud4652 + 3\ud4653 = 1
3\ud4651 + 6\ud4652 \u2212 3\ud4653 = \u22122
6\ud4651 + 6\ud4652 + 3\ud4653 = 5
.
1.2.4. Os sistemas lineares seguintes possuem a mesma matriz \ud434. Resolva-os usando o me´todo de
Gauss-Jordan. Observe que os dois sistemas podem ser resolvidos ao mesmo tempo escalo-
nando a matriz aumentada [\ud434 \u2223\ud4351 \u2223\ud4352 ].
(a)
\u23a7\u23a8
\u23a9
\ud4651 \u2212 2\ud4652 + \ud4653 = 1
2\ud4651 \u2212 5\ud4652 + \ud4653 = \u22122
3\ud4651 \u2212 7\ud4652 + 2\ud4653 = \u22121
; (b)
\u23a7\u23a8
\u23a9
\ud4651 \u2212 2\ud4652 + \ud4653 = 2
2\ud4651 \u2212 5\ud4652 + \ud4653 = \u22121
3\ud4651 \u2212 7\ud4652 + 2\ud4653 = 2
.
1.2.5. Seja \ud434 =
\u23a1
\u23a3 1 0 51 1 1
0 1 \u22124
\u23a4
\u23a6
.
(a) Encontre a soluc¸a\u2dco geral do sistema (\ud434+ 4\ud43c3)\ud44b = 0¯;
(b) Encontre a soluc¸a\u2dco geral do sistema (\ud434\u2212 2\ud43c3)\ud44b = 0¯.
1.2.6. Para cada sistema linear dado, encontre todos os valores de \ud44e para os quais o sistema na\u2dco tem
soluc¸a\u2dco, tem soluc¸a\u2dco u´nica e tem infinitas soluc¸o\u2dces:
(a)
\u23a7\u23a8
\u23a9
\ud465 + 2\ud466 \u2212 3\ud467 = 4
3\ud465 \u2212 \ud466 + 5\ud467 = 2
4\ud465 + \ud466 + (\ud44e2 \u2212 14)\ud467 = \ud44e+ 2
;
Marc¸o 2010 Reginaldo J. Santos
64 Matrizes e Sistemas Lineares
(b)
\u23a7\u23a8
\u23a9
\ud465 + \ud466 + \ud467 = 2
2\ud465 + 3\ud466 + 2\ud467 = 5
2\ud465 + 3\ud466 + (\ud44e2 \u2212 1)\ud467 = \ud44e+ 1
.
1.2.7. Uma indu´stria produz tre\u2c6s produtos, X, Y e Z, utilizando dois tipos de insumo, A e B. Para a
manufatura de cada kg de X sa\u2dco utilizados 2 gramas do insumo A e 1 grama do insumo B; para
cada kg de Y, 1 grama de insumo A e 3 gramas de insumo B e, para cada kg de Z, 3 gramas de A
e 5 gramas de B. O prec¸o de venda do kg de cada um dos produtos X, Y e Z e´ R\$ 3,00, R\$ 2,00
e R\$ 4,00, respectivamente. Com a venda de toda a produc¸a\u2dco de X, Y e Z manufaturada com
1,9 kg de A e 2,4 kg de B, essa indu´stria arrecadou R\$ 2900,00. Determine quantos kg de cada
um dos produtos X, Y e Z foram vendidos. (Sugesta\u2dco: veja o Exemplo 1.11 na pa´gina 39.)
1.2.8. Determine os coeficientes \ud44e, \ud44f, \ud450 e \ud451 da func¸a\u2dco```