Matrizes  Vetores e Geometria Analítica
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Matrizes Vetores e Geometria Analítica


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\ud434(\ud435 + \ud436) = \ud434\ud435 + \ud434\ud436 e (\ud435 + \ud436)\ud434 = \ud435\ud434+ \ud436\ud434;
(k) \ud6fc(\ud434\ud435) = (\ud6fc\ud434)\ud435 = \ud434(\ud6fc\ud435);
(l) (\ud434\ud461)\ud461 = \ud434;
(m) (\ud434+ \ud435)\ud461 = \ud434\ud461 + \ud435\ud461;
(n) (\ud6fc\ud434)\ud461 = \ud6fc\ud434\ud461;
(o) (\ud434\ud435)\ud461 = \ud435\ud461\ud434\ud461;
Demonstrac¸a\u2dco. Para provar as igualdades acima, devemos mostrar que os elementos da matriz do
lado esquerdo sa\u2dco iguais aos elementos correspondentes da matriz do lado direito. Sera\u2dco usadas
va´rias propriedades dos nu´meros sem cita´-las explicitamente.
Marc¸o 2010 Reginaldo J. Santos
12 Matrizes e Sistemas Lineares
(a) [\ud434+ \ud435]\ud456\ud457 = \ud44e\ud456\ud457 + \ud44f\ud456\ud457 = \ud44f\ud456\ud457 + \ud44e\ud456\ud457 = [\ud435 + \ud434]\ud456\ud457 ;
(b) [\ud434+ (\ud435 + \ud436)]\ud456\ud457 = \ud44e\ud456\ud457 + [\ud435 + \ud436]\ud456\ud457 = \ud44e\ud456\ud457 + (\ud44f\ud456\ud457 + \ud450\ud456\ud457) = (\ud44e\ud456\ud457 + \ud44f\ud456\ud457) + \ud450\ud456\ud457 = [\ud434+\ud435]\ud456\ud457 + \ud450\ud456\ud457 =
[(\ud434+ \ud435) + \ud436]\ud456\ud457 ;
(c) Seja \ud44b uma matriz \ud45a× \ud45b tal que
\ud434+\ud44b = \ud434 (1.2)
para qualquer matriz A, \ud45a× \ud45b. Comparando os elementos correspondentes, temos que
\ud44e\ud456\ud457 + \ud465\ud456\ud457 = \ud44e\ud456\ud457 ,
ou seja, \ud465\ud456\ud457 = 0, para \ud456 = 1 . . . ,\ud45a e \ud457 = 1 . . . , \ud45b. Portanto, a u´nica matriz que satisfaz (1.2) e´
a matriz em que todos os seus elementos sa\u2dco iguais a zero. Denotamos a matriz \ud44b por 0¯.
(d) Dada uma matriz \ud434, \ud45a× \ud45b, seja \ud44b uma matriz \ud45a× \ud45b, tal que
\ud434+\ud44b = 0¯ . (1.3)
Comparando os elementos correspondentes, temos que
\ud44e\ud456\ud457 + \ud465\ud456\ud457 = 0 ,
ou seja, \ud465\ud456\ud457 = \u2212\ud44e\ud456\ud457 , para \ud456 = 1 . . . ,\ud45a e \ud457 = 1 . . . , \ud45b. Portanto, a u´nica matriz que satisfaz
(1.3) e´ a matriz em que todos os seus elementos sa\u2dco iguais aos sime´tricos dos elementos de
\ud434. Denotamos a matriz \ud44b por \u2212\ud434.
(e) [\ud6fc(\ud6fd\ud434)]\ud456\ud457 = \ud6fc[\ud6fd\ud434]\ud456\ud457 = \ud6fc(\ud6fd\ud44e\ud456\ud457) = (\ud6fc\ud6fd)\ud44e\ud456\ud457 = [(\ud6fc\ud6fd)\ud434]\ud456\ud457 .
Matrizes Vetores e Geometria Anal\u131´tica Marc¸o 2010
1.1 Matrizes 13
(f) [(\ud6fc + \ud6fd)\ud434]\ud456\ud457 = (\ud6fc + \ud6fd)\ud44e\ud456\ud457 = (\ud6fc\ud44e\ud456\ud457) + (\ud6fd\ud44e\ud456\ud457) = [\ud6fc\ud434]\ud456\ud457 + [\ud6fd\ud434]\ud456\ud457 = [\ud6fc\ud434+ \ud6fd\ud434]\ud456\ud457 .
(g) [\ud6fc(\ud434+ \ud435)]\ud456\ud457 = \ud6fc[\ud434+\ud435]\ud456\ud457 = \ud6fc(\ud44e\ud456\ud457 + \ud44f\ud456\ud457) = \ud6fc\ud44e\ud456\ud457 + \ud6fc\ud44f\ud456\ud457 = [\ud6fc\ud434]\ud456\ud457 + [\ud6fc\ud435]\ud456\ud457
= [\ud6fc\ud434+ \ud6fc\ud435]\ud456\ud457 .
(h) A demonstrac¸a\u2dco deste item e´ a mais trabalhosa. Sejam \ud434, \ud435 e \ud436 matrizes \ud45a× \ud45d, \ud45d× \ud45e e \ud45e×\ud45b
respectivamente. A notac¸a\u2dco de somato´rio aqui pode ser muito u´til, pelo fato de ser compacta.
[\ud434(\ud435\ud436)]\ud456\ud457 =
\ud45d\u2211
\ud458=1
\ud44e\ud456\ud458[\ud435\ud436]\ud458\ud457 =
\ud45d\u2211
\ud458=1
\ud44e\ud456\ud458(
\ud45e\u2211
\ud459=1
\ud44f\ud458\ud459\ud450\ud459\ud457) =
\ud45d\u2211
\ud458=1
\ud45e\u2211
\ud459=1
\ud44e\ud456\ud458(\ud44f\ud458\ud459\ud450\ud459\ud457) =
=
\ud45d\u2211
\ud458=1
\ud45e\u2211
\ud459=1
(\ud44e\ud456\ud458\ud44f\ud458\ud459)\ud450\ud459\ud457 =
\ud45e\u2211
\ud459=1
\ud45d\u2211
\ud458=1
(\ud44e\ud456\ud458\ud44f\ud458\ud459)\ud450\ud459\ud457 =
\ud45e\u2211
\ud459=1
(
\ud45d\u2211
\ud458=1
\ud44e\ud456\ud458\ud44f\ud458\ud459)\ud450\ud459\ud457 =
=
\ud45e\u2211
\ud459=1
[\ud434\ud435]\ud456\ud459\ud450\ud459\ud457 = [(\ud434\ud435)\ud436]\ud456\ud457 .
(i) Podemos escrever a matriz identidade em termos do delta de Kronecker que e´ definido por
\ud6ff\ud456\ud457 =
{
1, se \ud456 = \ud457
0, se \ud456 \u2215= \ud457
como [\ud43c\ud45b]\ud456\ud457 = \ud6ff\ud456\ud457 . Assim,
[\ud434\ud43c\ud45b]\ud456\ud457 =
\ud45b\u2211
\ud458=1
\ud44e\ud456\ud458[\ud43c\ud45b]\ud458\ud457 =
\ud45b\u2211
\ud458=1
\ud44e\ud456\ud458\ud6ff\ud458\ud457 = \ud44e\ud456\ud457.
A outra igualdade e´ ana´loga.
Marc¸o 2010 Reginaldo J. Santos
14 Matrizes e Sistemas Lineares
(j) [\ud434(\ud435 + \ud436)]\ud456\ud457 =
\ud45d\u2211
\ud458=1
\ud44e\ud456\ud458[\ud435 + \ud436]\ud458\ud457 =
\ud45d\u2211
\ud458=1
\ud44e\ud456\ud458(\ud44f\ud458\ud457 + \ud450\ud458\ud457) =
\ud45d\u2211
\ud458=1
(\ud44e\ud456\ud458\ud44f\ud458\ud457 + \ud44e\ud456\ud458\ud450\ud458\ud457) =
=
\ud45d\u2211
\ud458=1
\ud44e\ud456\ud458\ud44f\ud458\ud457 +
\ud45d\u2211
\ud458=1
\ud44e\ud456\ud458\ud450\ud458\ud457 = [\ud434\ud435]\ud456\ud457 + [\ud434\ud436]\ud456\ud457 = [\ud434\ud435 + \ud434\ud436]\ud456\ud457 .
A outra igualdade e´ inteiramente ana´loga a anterior e deixamos como exerc\u131´cio.
(k) [\ud6fc(\ud434\ud435)]\ud456\ud457 = \ud6fc
\ud45d\u2211
\ud458=1
\ud44e\ud456\ud458\ud44f\ud458\ud457 =
\ud45d\u2211
\ud458=1
(\ud6fc\ud44e\ud456\ud458)\ud44f\ud458\ud457 = [(\ud6fc\ud434)\ud435]\ud456\ud457 e
[\ud6fc(\ud434\ud435)]\ud456\ud457 = \ud6fc
\ud45d\u2211
\ud458=1
\ud44e\ud456\ud458\ud44f\ud458\ud457 =
\ud45d\u2211
\ud458=1
\ud44e\ud456\ud458(\ud6fc\ud44f\ud458\ud457) = [\ud434(\ud6fc\ud435)]\ud456\ud457 .
(l) [(\ud434\ud461)\ud461]\ud456\ud457 = [\ud434\ud461]\ud457\ud456 = \ud44e\ud456\ud457 .
(m) [(\ud434+ \ud435)\ud461]\ud456\ud457 = [\ud434+ \ud435]\ud457\ud456 = \ud44e\ud457\ud456 + \ud44f\ud457\ud456 = [\ud434\ud461]\ud456\ud457 + [\ud435\ud461]\ud456\ud457 .
(n) [(\ud6fc\ud434)\ud461]\ud456\ud457 = [\ud6fc\ud434]\ud457\ud456 = \ud6fc\ud44e\ud457\ud456 = \ud6fc[\ud434\ud461]\ud456\ud457 = [\ud6fc\ud434\ud461]\ud456\ud457 .
(o) [(\ud434\ud435)\ud461]\ud456\ud457 = [\ud434\ud435]\ud457\ud456 =
\ud45d\u2211
\ud458=1
\ud44e\ud457\ud458\ud44f\ud458\ud456 =
\ud45d\u2211
\ud458=1
[\ud434\ud461]\ud458\ud457[\ud435
\ud461]\ud456\ud458 =
\ud45d\u2211
\ud458=1
[\ud435\ud461]\ud456\ud458[\ud434
\ud461]\ud458\ud457 = [\ud435
\ud461\ud434\ud461]\ud456\ud457 .
\u25a0
A diferenc¸a entre duas matrizes de mesmo tamanho \ud434 e \ud435 e´ definida por
\ud434\u2212 \ud435 = \ud434+ (\u2212\ud435),
Matrizes Vetores e Geometria Anal\u131´tica Marc¸o 2010
1.1 Matrizes 15
ou seja, e´ a soma da matriz \ud434 com a sime´trica da matriz \ud435.
Sejam \ud434 uma matriz \ud45b×\ud45b e \ud45d um inteiro positivo. Definimos a pote\u2c6ncia \ud45d de \ud434, por \ud434\ud45d = \ud434 . . . \ud434\ufe38 \ufe37\ufe37 \ufe38
\ud45d vezes
.
E para \ud45d = 0, definimos \ud4340 = \ud43c\ud45b.
Exemplo 1.8. Vamos verificar se para matrizes \ud434 e \ud435, quadradas, vale a igualdade
(\ud434+ \ud435)(\ud434\u2212\ud435) = \ud4342 \u2212\ud4352. (1.4)
Usando a propriedade (i) do teorema anterior obtemos
(\ud434+ \ud435)(\ud434\u2212\ud435) = (\ud434+ \ud435)\ud434+ (\ud434+ \ud435)(\u2212\ud435)
= \ud434\ud434+ \ud435\ud434\u2212 \ud434\ud435 \u2212\ud435\ud435 = \ud4342 + \ud435\ud434\u2212 \ud434\ud435 \u2212 \ud4352
Assim, (\ud434 + \ud435)(\ud434 \u2212 \ud435) = \ud4342 \u2212 \ud4352 se, e somente se, \ud435\ud434 \u2212 \ud434\ud435 = 0, ou seja, se, e somente se,
\ud434\ud435 = \ud435\ud434. Como o produto de matrizes na\u2dco e´ comutativo, a conclusa\u2dco e´ que a igualdade (1.4), na\u2dco
vale para matrizes em geral. Como contra-exemplo basta tomarmos duas matrizes que na\u2dco comutem
entre si. Sejam
\ud434 =
[
0 0
1 1
]
e \ud435 =
[
1 0
1 0
]
.
Para estas matrizes
\ud434+\ud435 =
[
1 0
2 1
]
, \ud434\u2212\ud435 =
[ \u22121 0
0 1
]
, \ud4342 = \ud434 =
[
0 0
1 1
]
, \ud4352 = \ud435 =
[
1 0
1 0
]
.
Assim,
(\ud434+ \ud435)(\ud434\u2212\ud435) =
[ \u22121 0
\u22122 1
]
\u2215=
[ \u22121 0
0 1
]
= \ud4342 \u2212 \ud4352.
Marc¸o 2010 Reginaldo J. Santos
16 Matrizes e Sistemas Lineares
1.1.3 Aplicac¸a\u2dco: Cadeias de Markov
Vamos supor que uma populac¸a\u2dco e´ dividida em tre\u2c6s estados (por exemplo: ricos, classe me´dia e
pobres) e que em cada unidade de tempo a probabilidade de mudanc¸a de um estado para outro seja
constante no tempo, so´ dependa dos estados. Este processo e´ chamado cadeia de Markov.
Seja \ud461\ud456\ud457 a probabilidade de mudanc¸a do estado \ud457 para o estado \ud456 em uma unidade de tempo
(gerac¸a\u2dco). Tome cuidado com a ordem dos \u131´ndices. A matriz
\ud447 =
1\u20dd 2\u20dd 3\u20dd\u23a1
\u23a3 \ud46111 \ud46112 \ud46113\ud46121 \ud46122 \ud46123
\ud46131 \ud46132 \ud46133
\u23a4
\u23a6 1\u20dd2\u20dd
3\u20dd
e´ chamada matriz de transic¸a\u2dco. A distribuic¸a\u2dco da populac¸a\u2dco inicial entre os tre\u2c6s estados pode ser
descrita pela seguinte matriz:
\ud4430 =
\u23a1
\u23a3 \ud45d1\ud45d2
\ud45d3
\u23a4
\u23a6 esta´ no estado 1esta´ no estado 2
esta´ no estado 3
A matriz \ud4430 caracteriza a distribuic¸a\u2dco inicial da populac¸a\u2dco entre os tre\u2c6s estados e e´ chamada vetor de
estado. Apo´s uma unidade de tempo a populac¸a\u2dco estara´ dividida entre os tre\u2c6s estados da seguinte
forma
\ud4431 =
\u23a1
\u23a3 \ud46111\ud45d1 + \ud46112\ud45d2 + \ud46113\ud45d3\ud46121\ud45d1 + \ud46122\ud45d2 + \ud46123\ud45d3
\ud46131\ud45d1 + \ud46132\ud45d2 + \ud46133\ud45d3
\u23a4
\u23a6 estara´ no estado 1estara´ no estado 2
estara´ no estado 3
Matrizes Vetores e Geometria Anal\u131´tica Marc¸o 2010
1.1 Matrizes 17
Lembre-se que \ud461\ud456\ud457 e´ a probabilidade de mudanc¸a do estado \ud457 para o estado \ud456. Assim o vetor de estado
apo´s uma unidade de tempo e´ dada pelo produto de matrizes:
\ud4431 = \ud447\ud4430.
Exemplo 1.9. Vamos considerar a matriz de transic¸a\u2dco
\ud447 =
1\u20dd 2\u20dd 3\u20dd\u23a1
\u23a2\u23a3
1
2
1
4
0
1
2
1
2
1
2
0 1
4
1
2
\u23a4
\u23a5\u23a6 1\u20dd2\u20dd
3\u20dd
(1.5)
e o vetor de estados inicial
\ud4430 =
\u23a1
\u23a3 131
3
1
3
\u23a4
\u23a6 esta´ no estado 1esta´ no estado 2
esta´ no estado 3
(1.6)
que representa uma populac¸a\u2dco dividida de forma que 1/3 da populac¸a\u2dco esta´ em cada estado.
Apo´s uma unidade de tempo a matriz de estado sera´ dada por
\ud4431 = \ud447\ud4430 =
\u23a1
\u23a2\u23a3
1
2
1
4
0
1
2
1
2
1
2
0 1
4
1
2
\u23a4
\u23a5\u23a6
\u23a1
\u23a2\u23a3
1
3
1
3
1
3
\u23a4
\u23a5\u23a6 =
\u23a1
\u23a2\u23a3
1
4
1
2
1
4
\u23a4
\u23a5\u23a6
Como estamos assumindo que em cada unidade de tempo a matriz de transic¸a\u2dco e´ a mesma,
enta\u2dco apo´s \ud458 unidades de tempo a populac¸a\u2dco estara´ dividida entre os tre\u2c6s estados segundo a matriz
de estado
\ud443\ud458 = \ud447\ud443\ud458\u22121 = \ud447 2\ud443\ud458\u22122 = \u22c5 \u22c5 \u22c5 = \ud447 \ud458\ud4430
Marc¸o 2010 Reginaldo J. Santos
18 Matrizes e Sistemas Lineares
Assim a matriz \ud447 \ud458 da´ a transic¸a\u2dco entre \ud458 unidades de tempo.
Matrizes Vetores e Geometria Anal\u131´tica Marc¸o 2010
1.1 Matrizes 19
Exerc\u131´cios Nume´ricos (respostas na pa´gina 540)
1.1.1. Considere as seguintes matrizes
\ud434 =
[
2 0
6 7
]
, \ud435 =
[
0 4
2 \u22128
]
, \ud436 =
[ \u22126 9 \u22127
7 \u22123 \u22122
]
\ud437 =
\u23a1
\u23a3 \u22126 4 01 1 4
\u22126 0 6
\u23a4
\u23a6 , \ud438 =
\u23a1
\u23a3 6 9 \u22129\u22121 0 \u22124
\u22126 0 \u22121
\u23a4
\u23a6
Se for poss\u131´vel calcule:
(a) \ud434\ud435 \u2212\ud435\ud434,
(b) 2\ud436 \u2212\ud437,
(c) (2\ud437\ud461 \u2212 3\ud438\ud461)\ud461,
(d) \ud4372 \u2212\ud437\ud438.
1.1.2. Conhecendo-se somente os produtos \ud434\ud435 e \ud434\ud436, como podemos calcular \ud434(\ud435 + \ud436), \ud435\ud461\ud434\ud461,
\ud436\ud461\ud434\ud461 e (\ud434\ud435\ud434)\ud436?
1.1.3. Considere as seguintes matrizes
\ud434 =
[ \u22123 2 1
1 2 \u22121
]
, \ud435 =
\u23a1
\u23a3 2 \u221212 0
0 3
\u23a4
\u23a6
Marc¸o 2010 Reginaldo J. Santos
20 Matrizes e Sistemas Lineares
\ud436 =
\u23a1
\u23a3 \u22122 1 \u221210 1 1
\u22121 0 1
\u23a4
\u23a6 , \ud437 =
\u23a1
\u23a3 \ud4511 0 00 \ud4512 0
0 0 \ud4513
\u23a4
\u23a6
\ud4381 =
\u23a1
\u23a3 10
0
\u23a4
\u23a6 , \ud4382 =
\u23a1
\u23a3 01
0
\u23a4
\u23a6 , \ud4383 =
\u23a1
\u23a3 00
1
\u23a4
\u23a6
Verifique que:
(a) \ud434\ud435 e´ diferente de \ud435\ud434.
(b) \ud434\ud438\ud457 e´ a \ud457-e´sima coluna de \ud434, para \ud457 = 1, 2, 3 e \ud438\ud461\ud456\ud435 e´ a \ud456-e´sima linha de