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# p2-081g

DisciplinaÁlgebra Linear II1.032 materiais8.203 seguidores
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P2 de Álgebra Linear II
15/05/08

GabaritoGabaritoGabaritoGabarito

1ª questão.
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
=
2222
1110
1000
2200
A
Fatoração QR

:= v1 [ ], , ,0 0 0 2 := v2 [ ], , ,0 0 1 2 := v3 [ ], , ,2 0 1 2 := v4 [ ], , ,2 1 1 2
u1:= v1/norm(v1,2);r11:=innerprod(u1,v1);u1:= v1/norm(v1,2);r11:=innerprod(u1,v1);u1:= v1/norm(v1,2);r11:=innerprod(u1,v1);u1:= v1/norm(v1,2);r11:=innerprod(u1,v1);

:= u1 [ ], , ,0 0 0 1 := r11 2
Encontrando um vetor w
2
, que é ortogonal a u
1

> r12:=innerprod(v2,u1); w2:= v2r12:=innerprod(v2,u1); w2:= v2r12:=innerprod(v2,u1); w2:= v2r12:=innerprod(v2,u1); w2:= v2---- r12*u1; r12*u1; r12*u1; r12*u1;
:= r12 2 := w2 [ ], , ,0 0 1 0
Normalize o vetor w
2
para obter o segundo vetor ortonormal u
2

> u2:= w2/norm(w2,2);r22:=innerprod(u2,w2) ; u2:= w2/norm(w2,2);r22:=innerprod(u2,w2) ; u2:= w2/norm(w2,2);r22:=innerprod(u2,w2) ; u2:= w2/norm(w2,2);r22:=innerprod(u2,w2) ;
:= u2 [ ], , ,0 0 1 0 := r22 1
Calcule o terceiro vetor ortonormal
> r13:=innerprod(v3,u1); r23:=innerprod(v3,u2); w3:= v3 r13:=innerprod(v3,u1); r23:=innerprod(v3,u2); w3:= v3 r13:=innerprod(v3,u1); r23:=innerprod(v3,u2); w3:= v3 r13:=innerprod(v3,u1); r23:=innerprod(v3,u2); w3:= v3 ---- r13*u1 r13*u1 r13*u1 r13*u1 ---- r23*u2 r23*u2 r23*u2 r23*u2
:= r13 2 := r23 1 := w3 [ ], , ,2 0 0 0
> u3:= w3/norm(w3,2);r33:=innerprod(u3,wu3:= w3/norm(w3,2);r33:=innerprod(u3,wu3:= w3/norm(w3,2);r33:=innerprod(u3,wu3:= w3/norm(w3,2);r33:=innerprod(u3,w3);3);3);3);
:= u3 [ ], , ,1 0 0 0 := r33 2
Calcule o quarto vetor ortonormal
> r14:=innerprod(v4,u1);r24:=innerprod(v4,u2);r34:=innerprod(v4,u3);r14:=innerprod(v4,u1);r24:=innerprod(v4,u2);r34:=innerprod(v4,u3);r14:=innerprod(v4,u1);r24:=innerprod(v4,u2);r34:=innerprod(v4,u3);r14:=innerprod(v4,u1);r24:=innerprod(v4,u2);r34:=innerprod(v4,u3);
w4:= v4w4:= v4w4:= v4w4:= v4---- r14*u1 r14*u1 r14*u1 r14*u1---- r24*u2 r24*u2 r24*u2 r24*u2 ---- r34*u3 r34*u3 r34*u3 r34*u3
:= r14 2 := r24 1 := r34 2
:= w4 [ ], , ,0 1 0 0
> u4:=evalm(w4/norm(w4,2));u4:=evalm(w4/norm(w4,2));u4:=evalm(w4/norm(w4,2));u4:=evalm(w4/norm(w4,2));
:= u4 [ ], , ,0 1 0 0

:= Q
\uf8ee
\uf8f0
\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef
\uf8f9
\uf8fb
\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa
0 0 1 0
0 0 0 1
0 1 0 0
1 0 0 0
:= R
\uf8ee
\uf8f0
\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef
\uf8f9
\uf8fb
\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa
2 2 2 2
0 1 1 1
0 0 2 2
0 0 0 1

Note que para transformar A em uma matriz triangular superior basta o uso de permutações.
2
2ª questão.
Considere a matriz abaixo:
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
=
1311
1312
3111
2111
A

A fatoração precisa de permutação logo, logo teremos PA=LU
LUPA =
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
\u2212
\u2212\u2212
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
=
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
=
1000
1200
3110
2111
1001
0101
0012
0001
1311
1312
3111
2111
0010
1000
0100
0001

Para resolver o sistema o sistema Ax=(1,2,0,1), precisamos fazer PAx=PB, logo vamos resolver:

\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
=
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
\u2212
\u2212\u2212
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
1
0
2
1
0010
1000
0100
0001
1000
1200
3110
2111
1001
0101
0012
0001
4
3
2
1
x
x
x
x

Fazemos então
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
=
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
\u2212
\u2212\u2212
4
3
2
1
4
3
2
1
1000
1200
3110
2111
y
y
y
y
x
x
x
x
e resolvemos primeiro
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
=
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
2
1
0
1
1001
0101
0012
0001
4
3
2
1
y
y
y
y

temos então que 11 =y , 22 \u2212=y , 03 =y e 14 =y . Finalmente, resolvendo
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
\u2212
=
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
\u2212
\u2212\u2212
1
0
2
1
1000
1200
3110
2111
4
3
2
1
x
x
x
x
, obtemos 11 \u2212=x ,
2
1
2 \u2212=x ,
2
1
3 =x e 14 =x .
Conferindo:
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
=
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
\u2212
\u2212
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
=
1
0
2
1
1
21
21
1
1311
1312
3111
2111
Ax
3
3ª Questão
Seja W o espaço gerado pelos vetores ( ) ( ){ }1,0,2,1,1,0,1,1 .

a. Encontre as equações para W.
O espaço W é gerado pelos vetores:
\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb
\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb
0010
1001
~
1021
1011
Logo as equações de W são
\uf8f3
\uf8f2
\uf8f1
=
=\u2212
0
0
3
41
x
xx

b. Encontre uma base para \u22a5W .
As equações de \u22a5W são
\uf8f3
\uf8f2
\uf8f1
=
=+
0
0
2
41
x
xx
logo uma base é ( ) ( ){ }1,0,0,1,0,1,0,0 \u2212 .

c. Calcule a projeção ortogonal do vetor v= (1,1,1,1) em \u22a5W
Seja a matriz
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb\u2212
=
01
10
00
01
A , como o vetor não pertence ao espaço coluna da matriz A ( \u22a5W ),
usaremos a solução de mínimos quadrados que dará a projeção ortogonal de v em \u22a5W .
Fazendo vAAxA TT = , logo a projeção será: ( ) vAAAAproj TT 1\u2212=

\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb
=
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb\u2212
\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb\u2212
10
02
01
10
00
01
0100
1001
cuja inversa é \uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb
10
021
logo:

\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
=\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb\u2212
=\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb
\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb\u2212
=
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb
\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb\u2212
\uf8f7\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ed
\uf8eb
\uf8f7\uf8f7
\uf8f7
\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec
\uf8ec
\uf8ec
\uf8ec
\uf8ed
\uf8eb\u2212
0
1
0
0
1
0
01
10
00
01
1
0
10
021
01
10
00
01
1
1
1
1
0100
1001
10
021
01
10
00
01