exercicios-alga-17-18

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Exercicios ALGA 17-18
Álgebra e Conexões Matemáticas (Instituto Politécnico de Lisboa)
A StuDocu não é patrocinada ou endossada por alguma faculdade ou universidade
Exercicios ALGA 17-18
Álgebra e Conexões Matemáticas (Instituto Politécnico de Lisboa)
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
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https://www.studocu.com/pt?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=exercicios-alga-17-18
https://www.studocu.com/pt/document/instituto-politecnico-de-lisboa/algebra-e-conexoes-matematicas/trabalhos-praticos/exercicios-alga-17-18/2113821/view?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=exercicios-alga-17-18
https://www.studocu.com/pt/course/instituto-politecnico-de-lisboa/algebra-e-conexoes-matematicas/2842348?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=exercicios-alga-17-18
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https://www.studocu.com/pt/document/instituto-politecnico-de-lisboa/algebra-e-conexoes-matematicas/trabalhos-praticos/exercicios-alga-17-18/2113821/view?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=exercicios-alga-17-18
https://www.studocu.com/pt/course/instituto-politecnico-de-lisboa/algebra-e-conexoes-matematicas/2842348?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=exercicios-alga-17-18
➪❧❣❡❜r❛ ▲✐♥❡❛r ❡ ●❡♦♠❡tr✐❛ ❆♥❛❧ít✐❝❛➪❧❣❡❜r❛ ▲✐♥❡❛r ❡ ●❡♦♠❡tr✐❛ ❆♥❛❧ít✐❝❛
❉❡♣❛rt❛♠❡♥t♦ ❞❡ ▼❛t❡♠át✐❝❛ ❋❈❚✲❯◆▲
❊①❡r❝í❝✐♦s ❆▲●❆ ✲ ✶♦ s❡♠❡str❡ ✲ ✷✵✶✼✴✷✵✶✽
✶ ✲ ▼❛tr✐③❡s
✶✳✶ ✲ ❈♦♥s✐❞❡r❡ ❛s ♠❛tr✐③❡s
A =


1 −1 0 1
2 1 1 0
−1 1 3 1

, B =


3 0 0
0 2 0
0 0 1

, C =


1
−1
2

,
D =
[
−3 1 4 1
]
, E =
[
2
]
, F =




0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0




,
G =


1 4
2 5
3 6

, H =


0 0 0
1 0 0
2 4 0

 ❡ I =
[
1 0
0 1
]
.
■♥❞✐q✉❡ q✉❛✐s sã♦ ♠❛tr✐③❡s✿
✭❛✮ ◗✉❛❞r❛❞❛s✳
✭❜✮ ❚r✐❛♥❣✉❧❛r❡s ✐♥❢❡r✐♦r❡s✳
✭❝✮ ❉✐❛❣♦♥❛✐s✳
✭❞✮ ❊s❝❛❧❛r❡s✳
✶✳✸ ✲ ❈♦♥s✐❞❡r❡ ❛s ♠❛tr✐③❡s ❞❡ M2×3(R)
A =
[
3 1 0
1 1 −1
]
, B =
[
1 0 4
−1 2 −1
]
❡ C =
[
0 0 1
−2 −2 1
]
.
❉❡t❡r♠✐♥❡✿
✭❛✮ A+B + C✳
✭❜✮ 2A+ 2C + 2B✳
✭❝✮ A−B✳
✭❞✮ 2A− 3(B + C)✳
✶✳✹ ✲ ❉❛❞❛s ❛s ♠❛tr✐③❡s ❞❡ M3×3(R)
A =


1 0 0
0 1 0
0 0 1

 ❡ B =


1 1 1
1 1 1
1 1 1

,
❞❡t❡r♠✐♥❡ ✉♠❛ ♠❛tr✐③ X ∈ M3×3(R)✱ t❛❧ q✉❡
X +A = 2(X −B).
✶✳✺ ✲ ❙❡❥❛♠ A =
[
1 2 −1
] ∈ M1×3(R) ❡ B =


0
1
3

 ∈ M3×1(R)✳
❉❡t❡r♠✐♥❡✱ s❡ ♣♦ssí✈❡❧✱ AB ❡ BA✳
✶✳✼ ✲ ❈♦♥s✐❞❡r❡ ❛s ♠❛tr✐③❡s
A =
[
1 2
]
, B =
[
2 1
0 2
]
, C =


1 −1
0 1
2 0

 ❡ D =
[
−1 1 1
1 −1 0
]
.
❉❡t❡r♠✐♥❡✱ s❡ ♣♦ssí✈❡❧✱ ❝❛❞❛ ✉♠ ❞♦s s❡❣✉✐♥t❡s ♣r♦❞✉t♦s✿
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✭❛✮ AB✳
✭❜✮ BA✳
✭❝✮ CD✳
✭❞✮ DC✳
✶✳✽ ✲ ❈♦♥s✐❞❡r❡ ❛s ♠❛tr✐③❡s
A =
[
4 2
2 1
]
, B =
[
−1 −1
2 2
]
, C =
[
0 −3
3 0
]
∈ M2×2(R).
❱❡r✐✜q✉❡ q✉❡✿
✭❛✮ AB 6= BA✳
✭❜✮ AB = 0 ❝♦♠ A 6= 0 ❡ B 6= 0✳
✭❝✮ BA = CA ❡ A 6= 0 ♠❛s B 6= C✳
✶✳✾ ✲ ❙❡❥❛♠ D,D′ ∈ Mn×n(K) ♠❛tr✐③❡s ❞✐❛❣♦♥❛✐s✳ ▼♦str❡ q✉❡ DD′ é ✉♠❛ ♠❛tr✐③ ❞✐❛❣♦♥❛❧ ❝♦♠
(DD′)ii = diid′ii✱ i = 1, . . . , n✱ ❡ q✉❡ DD
′ = D′D✳
✶✳✶✵ ✲ ❙❡❥❛♠ A ∈ Mm×n(K) ❡ B ∈ Mn×p(K)✳ ❏✉st✐✜q✉❡ q✉❡✿
✭❛✮ ❙❡ A t❡♠ ❛ ❧✐♥❤❛ i ♥✉❧❛ ❡♥tã♦ ❛ ♠❛tr✐③ AB t❡♠ ❛ ❧✐♥❤❛ i ♥✉❧❛✳
✭❜✮ ❙❡ B t❡♠ ❛ ❝♦❧✉♥❛ k ♥✉❧❛ ❡♥tã♦ ❛ ♠❛tr✐③ AB t❡♠ ❛ ❝♦❧✉♥❛ k ♥✉❧❛✳
✭❝✮ ❙❡ A t❡♠ ❛s ❧✐♥❤❛s i ❡ j ✐❣✉❛✐s✱ ❝♦♠❡❡ i 6= j✱ ❡♥tã♦ ❛ ♠❛tr✐③ AB t❡♠ ❛s ❧✐♥❤❛s i ❡ j ✐❣✉❛✐s✳
✭❞✮ ❙❡ B t❡♠ ❛s ❝♦❧✉♥❛s k ❡ l ✐❣✉❛✐s✱ ❝♦♠ k 6= l✱ ❡♥tã♦ ❛ ♠❛tr✐③ AB t❡♠ ❛s ❝♦❧✉♥❛s k ❡ l ✐❣✉❛✐s✳
✶✳✶✹ ✲ ❙❡♠ ❝❛❧❝✉❧❛r (A + B)2✱ (A − B)2 ❡ A2 − B2✱ ✈❡r✐✜q✉❡ q✉❡ ♣❛r❛ ❛s ♠❛tr✐③❡s A =
[
0 1
0 1
]
, B =
[
−1 −1
0 0
]
∈ M2×2(R) s❡ t❡♠✿
✭❛✮ (A+B)2 6= A2 + 2AB +B2✳
✭❜✮ (A−B)2 6= A2 − 2AB +B2✳
✭❝✮ A2 −B2 6= (A−B)(A+B)✳
✶✳✶✺ ✲ ❙❡❥❛ D ∈ Mn×n(K) ✉♠❛ ♠❛tr✐③ ❞✐❛❣♦♥❛❧✳ ❉❡t❡r♠✐♥❡ Dk✱ ❝♦♠ k ∈ N✳
✶✳✶✾ ✲ ❙❡❥❛ A ∈ Mn×n(K)✳ ❆t❡♥❞❡♥❞♦ ❛♦ ❊①❡r❝í❝✐♦ ✶✳✶✵✱ ❥✉st✐✜q✉❡ q✉❡✿
✭❛✮ ❙❡ A t❡♠ ✉♠❛ ❝♦❧✉♥❛ ♥✉❧❛ ❡♥tã♦ A ♥ã♦ é ✐♥✈❡rtí✈❡❧✳
✭❜✮ ❙❡ A t❡♠ ❛s ❝♦❧✉♥❛s i ❡ j ✐❣✉❛✐s✱ ❝♦♠ i 6= j✱ ❡♥tã♦ A ♥ã♦ é ✐♥✈❡rtí✈❡❧✳
✶✳✷✻ ✲ ❙❡❥❛ A ∈ M3×3(R) ✐♥✈❡rtí✈❡❧ ❝♦♠ A−1 =


1 1 2
0 1 3
4 2 1

.
✭❛✮ ❉❡t❡r♠✐♥❡ ✉♠❛ ♠❛tr✐③ B t❛❧ q✉❡ AB =


1 2
0 1
4 1

 ∈ M3×2(R) ❡ ❥✉st✐✜q✉❡ q✉❡ t❛❧ ♠❛tr✐③ é
ú♥✐❝❛✳
✭❜✮ ❉❡t❡r♠✐♥❡ ✉♠❛ ♠❛tr✐③ C t❛❧ q✉❡ AC = A+ 2I3 ❡ ❥✉st✐✜q✉❡ q✉❡ t❛❧ ♠❛tr✐③ é ú♥✐❝❛✳
✶✳✸✹ ✲ ■♥❞✐q✉❡ q✉❛✐s ❞❛s ♠❛tr✐③❡s
A =
[
0 0
0 0
]
, B =


1 2
2 3
0 0

, C =


1 2 3
2 0 4
3 4 5

,
D =


1 2 −3
−2 0 4
3 −4 −1

, E =


0 2 3
−2 0 4
−3 −4 0

 ❡ F =
[
0 0 0
0 0 0
]
✷
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
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✭❛✮ sã♦ s✐♠étr✐❝❛s✳
✭❜✮ sã♦ ❤❡♠✐✕s✐♠étr✐❝❛s✳
✶✳✹✵ ✲ ❈♦♥s✐❞❡r❡ ❛s ♠❛tr✐③❡s
A = In, B =
[
0 2− 3i
−2− 3i 0
]
, C =
[
1 i
i 2
]
,
D =


0 i 2
−i 0 0
−2 0 0

 ❡ E =


0 i 2
−i 0 1 + 2i
2 1− 2i 1

.
■♥❞✐q✉❡ q✉❛✐s sã♦ ♠❛tr✐③❡s✿
✭❛✮ ❍❡r♠ít✐❝❛s✳
✭❜✮ ❍❡♠✐✕❤❡r♠ít✐❝❛s✳
✶✳✹✷ ✲ ■♥❞✐q✉❡ s❡ ❝❛❞❛ ✉♠❛ ❞❛s s❡❣✉✐♥t❡s ♠❛tr✐③❡s é ✉♠❛ ♠❛tr✐③ ❡❧❡♠❡♥t❛r ❡✱ ❡♠ ❝❛s♦ ❛✜r♠❛t✐✈♦✱ s❡ é
❞♦ t✐♣♦ ■✱ ■■ ♦✉ ■■■✳
✭❛✮


1 0 0
0 1 −1
0 0 1

✳
✭❜✮


2 0 0
0 1 0
0 0 1

✳
✭❝✮


0 1 0
0 0 1
1 0 0

✳
✭❞✮


1 0 0
0 1 0
0 0 0

✳
✭❡✮ In✳
✶✳✹✸ ✲ ❙❡❥❛ A ∈ M3×5(K)✳ ❉❡t❡r♠✐♥❡ ❛s ♠❛tr✐③❡s ❡❧❡♠❡♥t❛r❡s q✉❡✱ ♠✉❧t✐♣❧✐❝❛❞❛s à ❡sq✉❡r❞❛ ❞❡ A✱
❡❢❡❝t✉❛♠ ❡♠ A ❝❛❞❛ ✉♠❛ ❞❛s s❡❣✉✐♥t❡s tr❛♥s❢♦r♠❛çõ❡s✿
✭❛✮ ❚r♦❝❛ ❞❛s ❧✐♥❤❛s 1 ❡ 3✳
✭❜✮ ▼✉❧t✐♣❧✐❝❛çã♦ ❞❛ ❧✐♥❤❛ 1 ♣♦r 6✳
✭❝✮ ❆❞✐çã♦✱ à ❧✐♥❤❛ 3✱ ❞❛ ❧✐♥❤❛ 2 ♠✉❧t✐♣❧✐❝❛❞❛ ♣♦r 15 ✳
✶✳✹✹ ✲ ❙❡♠ ❡❢❡❝t✉❛r ♠✉❧t✐♣❧✐❝❛çõ❡s ❞❡ ♠❛tr✐③❡s✱ ✐♥❞✐q✉❡ ♦ r❡s✉❧t❛❞♦ ❞❡✿
✭❛✮


0 1 0
1 0 0
0 0 1




a b c d
e f g h
i j k l

✳
✭❜✮


5 0 0
0 1 0
0 0 1




0 1 0
1 0 0
0 0 1




a b c d
e f g h
i j k l

✳
✭❝✮


a b c d
e f g h
i j k l






1 0 0 0
0 1 0 0
0 0 1 3
0 0 0 1




✳
✭❞✮
[
2 0
0 1
][
a b c
d e f
]


1 0 0
0 1 −5
0 0 1

✳
✶✳✹✻ ✲ ❉❡t❡r♠✐♥❡ ❛ ✐♥✈❡rs❛ ❞❡ ❝❛❞❛ ✉♠❛ ❞❛s s❡❣✉✐♥t❡s ♠❛tr✐③❡s ❡❧❡♠❡♥t❛r❡s✿
✭❛✮


1 0 0
0 5 0
0 0 1


✸
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✭❜✮


0 0 1
0 1 0
1 0 0


✭❝✮


1 0 0
0 1 0
−3 0 1


✶✳✹✽ ✲ ■♥❞✐q✉❡ s❡ ❡stã♦ ❡♠ ❢♦r♠❛ ❞❡ ❡s❝❛❞❛ ❝❛❞❛ ✉♠❛ ❞❛s s❡❣✉✐♥t❡s ♠❛tr✐③❡s✿
✭❛✮ In✳
✭❜✮




0 0 0
5 1 4
0 1 3
0 0 2




✳
✭❝✮
[
0 5 0 0
]
✳
✭❞✮


0 1 0
0 0 1
0 0 1

✳
✶✳✹✾ ✲ ■♥❞✐q✉❡ ✉♠❛ ♠❛tr✐③ ❡♠ ❢♦r♠❛ ❞❡ ❡s❝❛❞❛ ❡ ❡q✉✐✈❛❧❡♥t❡ ♣♦r ❧✐♥❤❛s ❛ ❝❛❞❛ ✉♠❛ ❞❛s ♠❛tr✐③❡s✿
✭❛✮


1 2 1
2 1 0
−1 0 1

✳
✭❜✮


2 4 −2 6 0
4 8 −4 7 5
−2 −4 2 −1 −5

✳
✭❝✮


2 2 1
−2 −2 1
1 1 2

✳
✶✳✺✶ ✲ ■♥❞✐q✉❡ s❡ ❡stã♦ ❡♠ ❢♦r♠❛ ❞❡ ❡s❝❛❞❛ r❡❞✉③✐❞❛ ❝❛❞❛ ✉♠❛ ❞❛s s❡❣✉✐♥t❡s ♠❛tr✐③❡s ❡♠ ❢♦r♠❛ ❞❡
❡s❝❛❞❛✿
✭❛✮
[
0 0 0 1 5
]
✳
✭❜✮


0 1 0 1 1
0 0 1 1 1
0 0 0 0 0

✳
✭❝✮


0 1 2 5 0
0 0 0 1 1
0 0 0 0 0

✳
✭❞✮
[
0 1 2 5
]
✳
✭❡✮


1
0
0

✳
✶✳✺✷ ✲ ■♥❞✐q✉❡ ❛ ❢♦r♠❛ ❞❡ ❡s❝❛❞❛ r❡❞✉③✐❞❛ ❞❡ ❝❛❞❛ ✉♠❛ ❞❛s s❡❣✉✐♥t❡s ♠❛tr✐③❡s✿
✭❛✮


1 2 1
2 1 0
−1 0 1

✳
✭❜✮


2 4 −2 6 0
4 8 −4 7 5
−2 −4 2 −1 −5

✳
✭❝✮


2 2 1
−2 −2 1
1 1 2

✳
❖❜s❡r✈❛çã♦ ✕ ❈❛s♦ t❡♥❤❛ r❡s♦❧✈✐❞♦ ♦ ❊①❡r❝í❝✐♦ ✶✳✹✾ ❥á ❞❡t❡r♠✐♥♦✉ ✉♠❛ ♠❛tr✐③ ❡♠ ❢♦r♠❛ ❞❡
❡s❝❛❞❛ ❡ ❡q✉✐✈❛❧❡♥t❡ ♣♦r ❧✐♥❤❛s ❛ ❝❛❞❛ ✉♠❛ ❞❡st❛s ♠❛tr✐③❡s✳
✶✳✺✺ ✲ ❉❡t❡r♠✐♥❡ s❡ sã♦ ❡q✉✐✈❛❧❡♥t❡s ♣♦r ❧✐♥❤❛s ❛s ♠❛tr✐③❡s✿
✹
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
✭❛✮
[
1 1 1
0 0 2
]
❡
[
0 1 0
1 −1 0
]
✳
✭❜✮
[
1 2 3
−2 −4 −5
]
❡
[
1 21
1 2 0
]
✳
✶✳✺✻ ✲ ▼♦str❡ q✉❡ ❛s ♠❛tr✐③❡s
A =
[
2 0 0
1 1 0
]
❡ B =
[
1 2 0
−1 2 0
]
sã♦ ❡q✉✐✈❛❧❡♥t❡s ♣♦r ❧✐♥❤❛s ❡ ✐♥❞✐q✉❡ ✉♠❛ s❡q✉ê♥❝✐❛ ❞❡ tr❛♥s❢♦r♠❛çõ❡s ❡❧❡♠❡♥t❛r❡s s♦❜r❡ ❧✐♥❤❛s
t❛❧ q✉❡✿
✭❛✮ A−−−−−−−→
(linhas)
B✳
✭❜✮ B−−−−−−−→
(linhas)
A✳
✶✳✺✼ ✲ ❈♦♥s✐❞❡r❡ ❛s ♠❛tr✐③❡s
A1 =


0 1 0 1 1
2 3 0 2 1
−2 −1 0 1 −1

, A2 =




2 1 0
1 1 2
3 −1 1
−1 1 0




,
A3 =


1 4 2
2 3 1
−1 1 1

 ❡ A4 =


2 1 1
0 0 1
1 −1 2

.
❉❡t❡r♠✐♥❡ ❛ ❝❛r❛❝t❡ríst✐❝❛ ❞❡ Ai✱ ❝♦♠ i = 1, 2, 3, 4✳
✶✳✺✽ ✲ ❉✐s❝✉t❛✱ s❡❣✉♥❞♦ ♦s ✈❛❧♦r❡s ❞❡ α ❡ ❞❡ β✱ ❛ ❝❛r❛❝t❡ríst✐❝❛ ❞❛s ♠❛tr✐③❡s ❞❡ ❡❧❡♠❡♥t♦s r❡❛✐s
Aα =


1 0 −1 1
1 1 0 1
α 1 −1 2

, Bα =




1 −1 0 1
1 1 0 −1
α 1 1 0
0 1 α 1




,
Cα,β =


0 0 α
0 β 2
3 0 1

 ❡ Dα,β =




α 0 −1 β
1 0 β 0
1 1 1 1
1 1 0 1




.
✶✳✺✾ ✲ ❈❛❧❝✉❧❛♥❞♦ ❛s ❝❛r❛❝t❡ríst✐❝❛s✱ ❥✉st✐✜q✉❡ q✉❡ ❛s ♠❛tr✐③❡s
[
1 2
4 8
]
❡
[
0 1
1 2
]
♥ã♦ sã♦ ❡q✉✐✈❛❧❡♥t❡s ♣♦r ❧✐♥❤❛s✳
✶✳✻✷ ✲
✭❛✮ ❉❡t❡r♠✐♥❡ ♦ ❝♦♥❥✉♥t♦ ❞♦s ✈❛❧♦r❡s ❞❡ α ∈ R ♣❛r❛ ♦s q✉❛✐s ❛ ♠❛tr✐③


1 2 1
1 4 2
2 4 7 + α

 ∈ M3×3(R)
é ✐♥✈❡rtí✈❡❧✳
✭❜✮ ❉❡t❡r♠✐♥❡ ♦ ❝♦♥❥✉♥t♦ ❞♦s ✈❛❧♦r❡s ❞❡ α ❡ ♦ ❝♦♥❥✉♥t♦ ❞♦s ✈❛❧♦r❡s ❞❡ β✱ ❝♦♠ α, β ∈ R✱ ♣❛r❛
♦s q✉❛✐s ❛ ♠❛tr✐③


1 2 1
1 α+ 3 2
2 4 β

 ∈ M3×3(R)
é ✐♥✈❡rtí✈❡❧✳
✺
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✶✳✻✺ ✲ ❙❡❥❛ A =
[
1 −1
2 0
]
∈ M2×2(R)✳
✭❛✮ ▼♦str❡ q✉❡ A é ✐♥✈❡rtí✈❡❧ ❡ ❞❡t❡r♠✐♥❡ A−1✳
✭❜✮ ❊①♣r✐♠❛ A−1 ❡ A ❝♦♠♦ ♣r♦❞✉t♦ ❞❡ ♠❛tr✐③❡s ❡❧❡♠❡♥t❛r❡s✳
✶✳✻✻ ✲ ❈♦♥s✐❞❡r❡ ❛s ♠❛tr✐③❡s
A =


3 1 0
1 2 1
2 −1 −1

 ∈ M3×3(R), B =


1 −1 0
2 1 2
0 1 −1

 ∈ M3×3(R),
C =
[
1 1 + i
−i 1
]
∈ M2×2(C) ❡ D =




1 −1 1 2
2 −2 1 1
1 −1 0 1
−2 0 2 −2




∈ M4×4(R).
■♥❞✐q✉❡ q✉❛✐s sã♦ ✐♥✈❡rtí✈❡✐s ❡✱ ❡♠ ❝❛s♦ ❛✜r♠❛t✐✈♦✱ ❞❡t❡r♠✐♥❡ ❛ r❡s♣❡❝t✐✈❛ ✐♥✈❡rs❛✳
✷ ✲ ❙✐st❡♠❛s ❞❡ ❊q✉❛çõ❡s ▲✐♥❡❛r❡s
✷✳✷ ✲ ❙❡❥❛♠
A =


1 1 2 −1
2 2 −2 2
0 0 6 −4

 ∈ M3×4(R), B =


−1
4
−6

 ∈ M3×1(R)
❡ (S) ♦ s✐st❡♠❛ ❞❡ ❡q✉❛çõ❡s ❧✐♥❡❛r❡s AX = B✳ ❙❡♠ r❡s♦❧✈❡r ♦ s✐st❡♠❛✱ ♠♦str❡ q✉❡✿
✭❛✮ (−1, 1, 1, 3) é s♦❧✉çã♦ ❞❡ (S)✳
✭❜✮ (1, 0, 1, 0) ♥ã♦ é s♦❧✉çã♦ ❞❡ (S)✳
✷✳✸ ✲ ❏✉st✐✜q✉❡ q✉❡ ❡①✐st❡ ✉♠ s✐st❡♠❛ (S) ❞❡ ❡q✉❛çõ❡s ❧✐♥❡❛r❡s✱ AX = B✱ ❝♦♠
A =




1 0 −1
2 4 3
−1 0 2
3 4 2




∈ M4×3(R)
❡ t❛❧ q✉❡ (1, 2, 3) é s♦❧✉çã♦ ❞❡ (S)✳ ■♥❞✐q✉❡ ❛s ❡q✉❛çõ❡s ❞❡ ✉♠ s✐st❡♠❛ ♥❡ss❛s ❝♦♥❞✐çõ❡s✳
✷✳✼ ✲ ❉✐s❝✉t❛ ❝❛❞❛ ✉♠ ❞♦s s❡❣✉✐♥t❡s s✐st❡♠❛s ❞❡ ❡q✉❛çõ❡s ❧✐♥❡❛r❡s✱ ♥❛s ✐♥❝ó❣♥✐t❛s x1, x2, x3✱ s♦❜r❡ R✱
❡ r❡s♦❧✈❛✲♦s ♥♦s ❝❛s♦s ❡♠ q✉❡ sã♦ ♣♦ssí✈❡✐s✳
(S1)



x1 + x2 − x3 = 0
2x1 + x2 = 1
x1 − x3 = 1
2x1 + x2 − 2x3 = 1
(S2)



2x1 + x2 = 1
−x1 + 3x2 + x3 = 2
x1 + 4x2 + x3 = 3
(S3)
{
x1 + 2x2 + x3 = −1
2x1 + 4x2 + 2x3 = 3
✷✳✽ ✲ ❈♦♠ ❛ ✐♥❢♦r♠❛çã♦ ❞❛❞❛ ♥♦ q✉❛❞r♦ s❡❣✉✐❞❛♠❡♥t❡ ❛♣r❡s❡♥t❛❞♦ ❞❡t❡r♠✐♥❡✱ ❝❛s♦ s❡❥❛ ♣♦ssí✈❡❧✱ s❡
❝❛❞❛ ✉♠ ❞♦s s✐st❡♠❛s ❞❡ ❡q✉❛çõ❡s ❧✐♥❡❛r❡s AX = B é ♣♦ssí✈❡❧ ✭❞❡t❡r♠✐♥❛❞♦ ♦✉ ✐♥❞❡t❡r♠✐♥❛❞♦✮
♦✉ ✐♠♣♦ssí✈❡❧ ❡✱ ♣❛r❛ ♦s s✐st❡♠❛s ✐♥❞❡t❡r♠✐♥❛❞♦s✱ ✐♥❞✐q✉❡ ♦ r❡s♣❡❝t✐✈♦ ❣r❛✉ ❞❡ ✐♥❞❡t❡r♠✐♥❛çã♦✳
✻
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
▼❛tr✐③ A r(A) r([A | B])
✭❛✮ 3×3 ✸ ✸
✭❜✮ 3×3 ✷ ✸
✭❝✮ 3×3 ✶ ✶
✭❞✮ 5×7 ✸ ✸
✭❡✮ 5×7 ✷ ✸
✭❢✮ 6×2 ✷ ✷
✭❣✮ 4×4 ✵ ✵
✷✳✾ ✲ ■♥❞✐q✉❡ ✉♠ s✐st❡♠❛ ❞❡ ❡q✉❛çõ❡s ❧✐♥❡❛r❡s ❝♦♠ 3 ✐♥❝ó❣♥✐t❛s q✉❡ s❡❥❛ ♣♦ssí✈❡❧ ✐♥❞❡t❡r♠✐♥❛❞♦✱ ❝♦♠
❣r❛✉ ❞❡ ✐♥❞❡t❡r♠✐♥❛çã♦
✭❛✮ 1✳
✭❜✮ 2✳
❖ ❣r❛✉ ❞❡ ✐♥❞❡t❡r♠✐♥❛çã♦ ♣♦❞❡ s❡r 3❄
✷✳✶✶ ✲ ■♥❞✐q✉❡ ♦ ❝♦♥❥✉♥t♦ C ❞♦s ✈❛❧♦r❡s r❡❛✐s ❞❡ k ♣❛r❛ ♦s q✉❛✐s ♦ s✐st❡♠❛ ❞❡ ❡q✉❛çõ❡s ❧✐♥❡❛r❡s
{
x− y = 1
3x− 3y = k
♥❛s ✐♥❝ó❣♥✐t❛s x, y✱ s♦❜r❡ R✱ é
✭❛✮ ✐♠♣♦ssí✈❡❧✳
✭❜✮ ♣♦ssí✈❡❧ ❞❡t❡r♠✐♥❛❞♦✳
✭❝✮ ♣♦ssí✈❡❧ ✐♥❞❡t❡r♠✐♥❛❞♦✳
✷✳✶✺ ✲ ▼♦str❡ q✉❡ ❛ ♠❛tr✐③ A =


−3 2 −1
2 0 −2
−1 1 1

 ∈ M3×3(R) é ✐♥✈❡rtí✈❡❧ ❡ ✉t✐❧✐③❡ A−1 ♣❛r❛ r❡s♦❧✈❡r ♦
s✐st❡♠❛ ❞❡ ❡q✉❛çõ❡s ❧✐♥❡❛r❡s✱ ♥❛s ✐♥❝ó❣♥✐t❛s x, y, z✱ s♦❜r❡ R✱



−3x+ 2y − z = α
2x− 2z = β
−x+ y + z = γ
, ❝♦♠ α, β, γ ∈ R.
✷✳✷✹ ✲ ❉✐s❝✉t❛ ❝❛❞❛ ✉♠ ❞♦s s❡❣✉✐♥t❡s s✐st❡♠❛s ❞❡ ❡q✉❛çõ❡s ❧✐♥❡❛r❡s✱ ♥❛s ✐♥❝ó❣♥✐t❛s x1, x2, x3✱ s♦❜r❡ R✱
✼
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
https://www.studocu.com/pt?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=exercicios-alga-17-18
❡ r❡s♦❧✈❛✲♦s ♥♦s ❝❛s♦s ❡♠ q✉❡ sã♦ ♣♦ssí✈❡✐s✳
(S1)



−5x1 − 2x2 + x3 = −1
6x1 + 2x2 + x3 = 0
−4x1 − 2x2 + 3x3 = −2
2x1 + 4x3 = −2
−6x1 − 3x2 + 2x3 = −1
(S2)



−x1 + 2x3 = 1
x1 + 2x2 = −1
2x2 + 2x3 = 0
x1 − 2x3 = −1
(S3)



x1 + x2 + 2x3 = 1
2x1 − x2 + x3 = 1
3x2 + 3x3 = 0
(S4)



2x1 − x2 + x3 = −1
x1 + 2x2 + x3 = 0
x1 − 3x2 = −1
4x1 − 2x2 + 2x3 = −2
−2x1 + x2 − x3 = 1
(S5)



x1 + 2x2 = 1
x1 + x2 = 1
−x1 + x2 = −1
(S6)



x1 + x2 + x3 = −1
2x1 + x2 = 0
x2 + x3 = 2
x1 − x3 = −1
(S7)



x1 + x2 − x3 = 0
2x1 + x2 = 1
−x1 − x3 = −1
✷✳✸✺ ✲ P❛r❛ ❝❛❞❛ α ∈ R ❡ ❝❛❞❛ β ∈ R✱ ❝♦♥s✐❞❡r❡ ♦ s✐st❡♠❛ ❞❡ ❡q✉❛çõ❡s ❧✐♥❡❛r❡s✱ ♥❛s ✐♥❝ó❣♥✐t❛s x, y, z✱
s♦❜r❡ R✱ 


x+ y − z = 1
−x− αy + z = −1
−x− y + (α+ 1)z = β − 2
.
✭❛✮ ❉✐s❝✉t❛ ♦ s✐st❡♠❛✱ ❡♠ ❢✉♥çã♦ ❞❡ α ❡ β✳
✭❜✮ P❛r❛ α = 0 ❡ β = 1 ✐♥❞✐q✉❡ ♦ ❝♦♥❥✉♥t♦ ❞❛s s♦❧✉çõ❡s ❞♦ s✐st❡♠❛✳
✷✳✸✼ ✲ P❛r❛ ❝❛❞❛ α ∈ R ❡ ❝❛❞❛ β ∈ R✱ ❝♦♥s✐❞❡r❡ ♦ s✐st❡♠❛ ❞❡ ❡q✉❛çõ❡s ❧✐♥❡❛r❡s✱ ♥❛s ✐♥❝ó❣♥✐t❛s x, y, z✱
s♦❜r❡ R✱
(Sα,β)



x+ αy + βz = 1
α(β − 1)y = α
x+ αy + z = β2
.
✭❛✮ ❉✐s❝✉t❛ ♦ s✐st❡♠❛✱ ❡♠ ❢✉♥çã♦ ❞❡ α ❡ β✳
✭❜✮ ✐✳ ❏✉st✐✜q✉❡ q✉❡ S2,2 t❡♠ ✉♠❛ ❡ ✉♠❛ só s♦❧✉çã♦✳
✐✐✳ ❏✉st✐✜q✉❡ q✉❡ ❛ ♠❛tr✐③ s✐♠♣❧❡s ❞❡ S2,2 é ✐♥✈❡rtí✈❡❧✳
✐✐✐✳ ❉❡t❡r♠✐♥❡ ❛ s♦❧✉çã♦ ❞❡ S2,2✱ ✉t✐❧✐③❛♥❞♦ ❛ ✐♥✈❡rs❛ ❞❛ ♠❛tr✐③ s✐♠♣❧❡s ❞♦ s✐st❡♠❛✳
✸ ✲ ❉❡t❡r♠✐♥❛♥t❡s
✸✳✶ ✲ ❈❛❧❝✉❧❡ ♦ ❞❡t❡r♠✐♥❛♥t❡ ❞❛s s❡❣✉✐♥t❡s ♠❛tr✐③❡s✿
✭❛✮ A =
[
1 1
1 2
]
∈ M2×2(R)✳
✽
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
✭❜✮ B =
[
1 2
2 1
]
∈ M2×2(R)✳
✭❝✮ C =
[
1 i
i −1
]
∈ M2×2(C)✳
✸✳✷ ✲ ❙❡❥❛ A =


0 a a2
a−1 0 a
a−2 a−1 0

 ∈ M3×3(R)✱ ❝♦♠ a 6= 0✳ ❈❛❧❝✉❧❡ ♦ ❞❡t❡r♠✐♥❛♥t❡ ❞❡ A ♣❡❧❛ ❘❡❣r❛ ❞❡
❙❛rr✉s✳
✸✳✸ ✲ ❙❡❥❛ A =


1 0 3
−1 2 4
3 1 2

 ∈ M3×3(R)✳ ❉❡t❡r♠✐♥❡✿
✭❛✮ â11✳
✭❜✮ â32✳
✭❝✮ â23✳
✸✳✹ ✲ ❈❛❧❝✉❧❡✱ ❞❡ ❞✉❛s ❢♦r♠❛s ❞✐❢❡r❡♥t❡s✱ ♦ ❞❡t❡r♠✐♥❛♥t❡ ❞❡ ❝❛❞❛ ✉♠❛ ❞❛s s❡❣✉✐♥t❡s ♠❛tr✐③❡s✿
✭❛✮ A =


1 1 0
2 1 1
1 1 1

 ∈ M3×3(R)✳
✭❜✮ B =


1 0 i
0 0 2
−i 2 1

 ∈ M3×3(C)✳
✭❝✮ C =




1 0 −1 0
−2 0 2 −1
1 1 −1 1
3 3 −6 6




∈ M4×4(R)✳
✸✳✻ ✲ P❛r❛ ❝❛❞❛ λ ∈ R✱ ❝♦♥s✐❞❡r❡
Aλ =


3− λ −3 2
0 −2− λ 2
0 −3 3− λ

.
❉❡t❡r♠✐♥❡ ♦ ❝♦♥❥✉♥t♦ ❞♦s ✈❛❧♦r❡s ❞❡ λ ♣❛r❛ ♦s q✉❛✐s detAλ = 0✳
✸✳✶✵ ✲ ❙❡❥❛ A =


a b c
d e f
g h i

 ∈ M3×3(R) t❛❧ q✉❡ detA = γ✳
■♥❞✐q✉❡✱ ❡♠ ❢✉♥çã♦ ❞❡ γ✱ ♦ ✈❛❧♦r ❞❡ ❝❛❞❛ ✉♠ ❞♦s s❡❣✉✐♥t❡s ❞❡t❡r♠✐♥❛♥t❡s✿
✭❛✮
∣
∣
∣
∣
∣
∣
d e f
g h i
a b c
∣
∣
∣
∣
∣
∣
✳
✭❜✮
∣
∣
∣
∣
∣
∣
3a 3b 3c
−d −e −f
4g 4h 4i
∣
∣
∣
∣
∣
∣
✳
✭❝✮
∣
∣
∣
∣
∣
∣
a+ g b+ h c+ i
d e f
g h i
∣
∣
∣
∣
∣
∣
✳
✭❞✮
∣
∣
∣
∣
∣
∣
−3a −3b −3c
d e f
g − 4d h− 4e i− 4f
∣
∣
∣
∣
∣
∣
✳
✭❡✮
∣
∣
∣
∣
∣
∣
b e h
a d g
c f i
∣
∣
∣
∣
∣
∣
✳
✸✳✶✺ ✲ P❛r❛ ❝❛❞❛ k ∈ R✱ ❝♦♥s✐❞❡r❡ ❛ ♠❛tr✐③
Bk =




1 0 −1 0
2 −1 −1 k
0 k −k k
−1 1 1 2




∈ M4×4(R).
❉❡t❡r♠✐♥❡ ♦ ❝♦♥❥✉♥t♦ ❞♦s ✈❛❧♦r❡s ❞❡ k ♣❛r❛ ♦s q✉❛✐s s❡ t❡♠ detBk = 2✳
✾
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
https://www.studocu.com/pt?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=exercicios-alga-17-18
✸✳✶✾ ✲ P❛r❛ ❝❛❞❛ t ∈ R✱ s❡❥❛
At =


1 t −1
2 4 −2
−3 −7 t+ 3

 ∈ M3×3(R).
❉❡t❡r♠✐♥❡ ♦ ❝♦♥❥✉♥t♦ ❞♦s ✈❛❧♦r❡s ❞❡ t ♣❛r❛ ♦s q✉❛✐s At é ✐♥✈❡rtí✈❡❧✳
✸✳✷✵ ✲ ❙❡❥❛♠ A,B,C ∈ Mn×n(R) t❛✐s q✉❡ detA = 2✱ detB = −5 ❡ detC = 4✳ ❈❛❧❝✉❧❡ det (AB⊤C)✱
det (3B) ❡ det
(
B2C
)
✳
✸✳✷✶ ✲ ▼♦str❡ q✉❡✱ ♣❛r❛ q✉❛✐sq✉❡r A,B ∈ Mn×n(K)✱ s❡ t❡♠✿
✭❛✮ det (AB) = det (BA)✳
✭❜✮ ❙❡ AB é ✐♥✈❡rtí✈❡❧ ❡♥tã♦ ♦ ♠❡s♠♦ s✉❝❡❞❡ ❛ A ❡ ❛ B✳
✸✳✷✷ ✲ ❙❡❥❛♠ A =
[
−1 2
] ∈ M1×2(R) ❡ B =
[
0
1
]
∈ M2×1(R)✳
✭❛✮ ▼♦str❡ q✉❡ det(AB) 6= det (BA)✳
✭❜✮ ❈♦♠❡♥t❡ ❛ ❛❧í♥❡❛ ❛♥t❡r✐♦r✱ ❛t❡♥❞❡♥❞♦ à ❛❧í♥❡❛ ✭❛✮ ❞♦ ❊①❡r❝í❝✐♦ ✸✳✷✶✳
✸✳✷✸ ✲ ❙❡❥❛ A ∈ Mn×n(K) ✉♠❛ ♠❛tr✐③ ✐❞❡♠♣♦t❡♥t❡ ✭✐st♦ é✱ A2 = A✮✳ ▼♦str❡ q✉❡ detA ∈ {0, 1}.
✸✳✷✺ ✲ ❈♦♥s✐❞❡r❡ ❛s ♠❛tr✐③❡s
A =




1 −1 1 1
0 2 4 4
1 3 1 1
0 0 −2 0




∈ M4×4(R) ❡ B =


2 1 −1
1 1 1
−1 0 2

 ∈ M3×3(R).
✭❛✮ ❈❛❧❝✉❧❡ detA ❡ detB✳
✭❜✮ ❉❡t❡r♠✐♥❡ s❡ A ❡ B sã♦ ✐♥✈❡rtí✈❡✐s ❡✱ ❡♠ ❝❛s♦ ❛✜r♠❛t✐✈♦✱ ✐♥❞✐q✉❡ ♦ ❞❡t❡r♠✐♥❛♥t❡ ❞❛ r❡s✲
♣❡❝t✐✈❛ ✐♥✈❡rs❛✳
✭❝✮ ■♥❞✐q✉❡ s❡ sã♦ ❞❡t❡r♠✐♥❛❞♦s ♦s s✐st❡♠❛s
✐✳ AX = 0✳
✐✐✳ BX = 0✳
✸✳✷✻ ✲ ❈♦♥s✐❞❡r❡ ❛ ♠❛tr✐③ A =


−4 −3 −3
1 0 1
4 4 3

 ∈ M3×3(R)✳ ❱❡r✐✜q✉❡ q✉❡ adjA = A✳
✸✳✷✽ ✲ ▼♦str❡ q✉❡ ❝❛❞❛ ✉♠❛ ❞❛s ♠❛tr✐③❡s s❡❣✉✐♥t❡s é ✐♥✈❡rtí✈❡❧ ❡ ❞❡t❡r♠✐♥❡ ❛ s✉❛ ✐♥✈❡rs❛ ❛ ♣❛rt✐r ❞❛
s✉❛ ❛❞❥✉♥t❛✳
✭❛✮ A =


3 1 2
1 2 1
2 2 2

 ∈ M3×3(R)✳
✭❜✮ Vα =
[
cosα − senα
senα cosα
]
, ❝♦♠ α ∈ R✳
✭❝✮ A =
[
z w
−w z
]
∈ M2×2(C)✱ ❝♦♠ z 6= 0 ♦✉ w 6= 0✳
✸✳✸✷ ✲ ❙❡❥❛♠
A =


1 2 3
0 2 1
1 1 1

 ∈ M3×3(R), B =


14
7
6

 ∈ M3×1(R)
❡ ❝♦♥s✐❞❡r❡ ♦ s✐st❡♠❛ (S) ❞❡ ❡q✉❛çõ❡s ❧✐♥❡❛r❡s AX = B✳
✭❛✮ ❈❛❧❝✉❧❡ detA ❡ ❥✉st✐✜q✉❡ q✉❡ ♦ s✐st❡♠❛ (S) é ✉♠ s✐st❡♠❛ ❞❡ ❈r❛♠❡r✳
✭❜✮ ❯t✐❧✐③❛♥❞♦ ❛ ❘❡❣r❛ ❞❡ ❈r❛♠❡r✱ ❞❡t❡r♠✐♥❡ ❛ s♦❧✉çã♦ ❞♦ s✐st❡♠❛ (S)✳
✶✵
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
✸✳✸✸ ✲ P❛r❛ ❝❛❞❛ k ∈ R✱ ❝♦♥s✐❞❡r❡ ❛ ♠❛tr✐③
Ak =


1 −k 1
0 k k
k k −k

 ∈ M3×3(R).
✭❛✮ ❯s❛♥❞♦ ❞❡t❡r♠✐♥❛♥t❡s✱ ✐♥❞✐q✉❡ ♣❛r❛ q✉❡ ✈❛❧♦r❡s ❞❡ k ❛ ♠❛tr✐③ Ak é ✐♥✈❡rtí✈❡❧✳
✭❜✮ P❛r❛ k = −1 ❥✉st✐✜q✉❡ q✉❡ ♦ s✐st❡♠❛ ❞❡ ❡q✉❛çõ❡s ❧✐♥❡❛r❡s
AX =


1
0
0


é ✉♠ s✐st❡♠❛ ❞❡ ❈r❛♠❡r ❡ ❞❡t❡r♠✐♥❡ ❛ s✉❛ s♦❧✉çã♦✳
✹ ✲ ❊s♣❛ç♦s ❱❡❝t♦r✐❛✐s
✹✳✸ ✲ ❈♦♥s✐❞❡r❡ R2 ❝♦♠ ✉♠❛ ❛❞✐çã♦ ❡ ✉♠❛ ♠✉❧t✐♣❧✐❝❛çã♦ ❡①t❡r♥❛ ❞❡✜♥✐❞❛s✱ r❡s♣❡❝t✐✈❛♠❡♥t❡✱ ♣♦r
(a1, a2) + (b1, b2) = (a1 + b1, a2 + b2)
α(a1, a2) = (αa1, 0),
♣❛r❛ q✉❛✐sq✉❡r (a1, a2), (b1, b2) ∈ R2 ❡ q✉❛❧q✉❡r α ∈ R✳ ▼♦str❡ q✉❡ (R2,+, ·) ♥ã♦ é ✉♠ ❡s♣❛ç♦
✈❡❝t♦r✐❛❧ s♦❜r❡ R✳
✹✳✻ ✲ ■♥❞✐q✉❡ 0E ♥♦s s❡❣✉✐♥t❡s ❝❛s♦s✿
✭❛✮ E = R4✳
✭❜✮ E = M2×3(R)✳
✭❝✮ E = R3[x]✳
✹✳✽ ✲ ❙❡❥❛ E ✉♠ ❡s♣❛ç♦ ✈❡❝t♦r✐❛❧ s♦❜r❡ K✳ ❙❡❥❛♠ α, β ∈ K ❡ s❡❥❛♠ u, v ∈ E✳ ❏✉st✐✜q✉❡ q✉❡✿
✭❛✮ ❙❡ αu = αv ❡ α 6= 0 ❡♥tã♦ u = v✳
✭❜✮ ❙❡ αu = βu ❡ u 6= 0E ❡♥tã♦ α = β✳
✹✳✶✸ ✲ ❉❡t❡r♠✐♥❡ q✉❛✐s ❞♦s s❡❣✉✐♥t❡s ❝♦♥❥✉♥t♦s sã♦ s✉❜❡s♣❛ç♦s ❞♦ ❡s♣❛ç♦ ✈❡❝t♦r✐❛❧ ✐♥❞✐❝❛❞♦✳
✭❛✮ F1 =
{
(a, b) ∈ R2 : a ≥ 0
}
❡♠ R2✳
✭❜✮ F2 = {(0, 0, 0), (0, 1, 0), (0,−1, 0)} ❡♠ R3✳
✭❝✮ F3 =
{
(a, b, c) ∈ R3 : 2a = b ∧ c = 0
}
❡♠ R3✳
✭❞✮ F4 =
{
(a, b, c) ∈ R3 : 2a = b
}
❡♠ R3✳
✹✳✶✺ ✲ ▼♦str❡ q✉❡ é ✉♠ s✉❜❡s♣❛ç♦ ❞❡ Mn×n(K) ♦ ❝♦♥❥✉♥t♦ ❞❛s ♠❛tr✐③❡s ❞❡ Mn×n(K)✿
✭❛✮ ❈♦♠ ❛ ❞✐❛❣♦♥❛❧ ♣r✐♥❝✐♣❛❧ ♥✉❧❛✳
✭❜✮ ❚r✐❛♥❣✉❧❛r❡s s✉♣❡r✐♦r❡s✳
✭❝✮ ❉✐❛❣♦♥❛✐s✳
✭❞✮ ❊s❝❛❧❛r❡s✳
✭❡✮ ❙✐♠étr✐❝❛s✳
✶✶
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
https://www.studocu.com/pt?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=exercicios-alga-17-18
✭❢✮ ❍❡♠✐✲s✐♠étr✐❝❛s✳
✹✳✶✻ ✲ ❏✉st✐✜q✉❡ q✉❡ ♥ã♦ é ✉♠ s✉❜❡s♣❛ç♦ ❞❡ Mn×n(K) ♦ ❝♦♥❥✉♥t♦ ❞❛s ♠❛tr✐③❡s ❞❡ Mn×n(K)✿
✭❛✮ ❈♦♠ ❛ ❞✐❛❣♦♥❛❧ ♣r✐♥❝✐♣❛❧ ♥ã♦ ♥✉❧❛✳
✭❜✮ ■♥✈❡rtí✈❡✐s✳
✭❝✮ ◆ã♦ ✐♥✈❡rtí✈❡✐s✳
✹✳✷✵ ✲ ▼♦str❡ q✉❡ ♦s s❡❣✉✐♥t❡s ❝♦♥❥✉♥t♦s sã♦ s✉❜❡s♣❛ç♦s ❞♦ ❡s♣❛ç♦ ✈❡❝t♦r✐❛❧ ✐♥❞✐❝❛❞♦✿
✭❛✮ F =
{
(a, b, c, d) ∈ R4 : a− 2b = 0 ∧ b+ c = 0
}
❡♠ R4✳
✭❜✮ G =
{
[
a b
c d
]
∈ M2×2(R) : a− 2b = 0 ∧ b+ c = 0
}
❡♠ M2×2(R)✳
✭❝✮ H =
{
ax3 + bx2 + cx+ d ∈ R3[x] : a− 2b = 0 ∧ b+ c = 0
}
❡♠ R3[x]✳
✹✳✷✷ ✲ ❙❡❥❛ G =
{
[
a a+ b
−b 0
]
: a, b ∈ R
}
✳ ▼♦str❡ q✉❡ G é ✉♠ s✉❜❡s♣❛ç♦ ❞❡ M2×2(R) ✐♥❞✐❝❛♥❞♦ ✉♠❛
s❡q✉ê♥❝✐❛ ❣❡r❛❞♦r❛ ❞❡ G✳
✹✳✷✸ ✲ ❆♣r❡s❡♥t❛♥❞♦ ✉♠❛ s❡q✉ê♥❝✐❛ ❣❡r❛❞♦r❛✱ ❥✉st✐✜q✉❡ q✉❡ ♦s s❡❣✉✐♥t❡s ❝♦♥❥✉♥t♦s sã♦ s✉❜❡s♣❛ç♦s ❞♦
❡s♣❛ç♦ ✈❡❝t♦r✐❛❧ ✐♥❞✐❝❛❞♦✳
✭❛✮
{
(a, b, c) ∈ R3 : a− c = 0
}
❡♠ R3✳
✭❜✮
{
[
a b
c d
]
∈ M2×2(R) : a+ d = 0
}
❡♠ M2×2(R)✳
✭❝✮
{
ax3 + bx2 + cx+ d ∈ R3[x] : a− 2c+ d = 0
}
❡♠ R3[x]✳
✹✳✸✶ ✲ ❙❡❥❛ E ✉♠ ❡s♣❛ç♦ ✈❡❝t♦r✐❛❧ s♦❜r❡ K ❡ s❡❥❛♠ u1, u2, u3 ∈ E✳ ❏✉st✐✜q✉❡ ❛s ❛✜r♠❛çõ❡s✿
✭❛✮ S = (u1, u2, u3) é ❧✐♥❡❛r♠❡♥t❡ ✐♥❞❡♣❡♥❞❡♥t❡ s❡✱ ❡ só s❡✱
S′ = (u1, u1 + u2, u1 + u2 + u3)
é ❧✐♥❡❛r♠❡♥t❡ ✐♥❞❡♣❡♥❞❡♥t❡✳
✭❜✮ S = (u1, u2, u3) é ❧✐♥❡❛r♠❡♥t❡ ✐♥❞❡♣❡♥❞❡♥t❡ s❡✱ ❡ só s❡✱
S′′ = (u1 − u2, u2 − u3, u1 + u3)
é ❧✐♥❡❛r♠❡♥t❡ ✐♥❞❡♣❡♥❞❡♥t❡✳
✭❝✮ ❆ s❡q✉ê♥❝✐❛
S′′′ = (u1 − u2, u2 − u3, u1 − u3)
é ❧✐♥❡❛r♠❡♥t❡ ❞❡♣❡♥❞❡♥t❡✳
✹✳✸✸ ✲ ❊♠ R3✱ ❝♦♥s✐❞❡r❡ ♦ s✉❜❡s♣❛ç♦ F =
〈
(2, 3, 3)
〉
✳ ■♥❞✐q✉❡✱ ♣❛r❛ F ✱ ❞✉❛s ❜❛s❡s ❞✐st✐♥t❛s✳
✹✳✸✺ ✲ ❙❡❥❛ G =
{
[
a a+ b
−b 0
]
: a, b ∈ R
}
♦ s✉❜❡s♣❛ç♦ ❞❡ M2×2(R) r❡❢❡r✐❞♦ ♥♦ ❊①❡r❝í❝✐♦ ✹✳✷✷✳ ❉❡t❡r✲
♠✐♥❡ ✉♠❛ ❜❛s❡ ❞❡ G✳
✹✳✹✶ ✲ ❊♠ R3✱ ❝♦♥s✐❞❡r❡ ♦ s✉❜❡s♣❛ç♦
F =
〈
(1, 2, 1), (2,−1,−3), (0, 1, 1)
〉
.
✭❛✮ ❱❡r✐✜q✉❡ q✉❡
(
(1, 2, 1), (2,−1,−3), (0, 1, 1)
)
♥ã♦ é ✉♠❛ ❜❛s❡ ❞❡ F ✳
✶✷
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
✭❜✮ ❉❡t❡r♠✐♥❡ ✉♠❛ ❜❛s❡ ❞❡ F ❝♦♥st✐t✉í❞❛ ♣♦r ✈❡❝t♦r❡s ❞❛ s❡q✉ê♥❝✐❛ ✐♥❞✐❝❛❞❛ ♥❛ ❛❧í♥❡❛ ❛♥t❡r✐♦r✳
✹✳✹✹ ✲ ❊♠ M2×2(R)✱ ❝♦♥s✐❞❡r❡ ❛s ❜❛s❡s
B =
(
[
1 0
0 0
]
,
[
1 1
0 0
]
,
[
1 1
1 0
]
,
[
1 1
1 1
]
)
❡
B′ =
(
[
1 0
0 0
]
,
[
0 1
0 0
]
,
[
0 0
1 0
]
,
[
0 0
0 1
]
)
.
✭❛✮ ❉❡t❡r♠✐♥❡ ❛ s❡q✉ê♥❝✐❛ ❞❛s ❝♦♦r❞❡♥❛❞❛s ❞♦ ✈❡❝t♦r
[
4 3
2 1
]
❡♠ ❝❛❞❛ ✉♠❛ ❞❛s ❜❛s❡s B ❡ B′✳
✭❜✮ ❉❡t❡r♠✐♥❡ ❛ s❡q✉ê♥❝✐❛ ❞❛s ❝♦♦r❞❡♥❛❞❛s ❞❡
[
a b
c d
]
∈ M2×2(R) ❡♠ ❝❛❞❛ ✉♠❛ ❞❛s ❜❛s❡s B
❡ B′✳
✹✳✹✽ ✲ ❙❡❥❛♠
F =
{
(a, b, c, d) ∈ R4 : a− c = 0 ∧ a− b+ d = 0
}
❡
G =
〈
(1, 1, 0, 1), (2, 1, 2,−1)
〉
.
❉❡t❡r♠✐♥❡ ✉♠❛ ❜❛s❡ ❞❡ F ∩G✳
✹✳✺✷ ✲ ❊♠ R4✱ ❝♦♥s✐❞❡r❡ ♦s s✉❜❡s♣❛ç♦s
F =
{
(a, b, c, d) ∈ R4 : a− b = 0 ∧ a = b+ d
}
,
G =
{
(a, b, c, d) ∈ R4 : b− c = 0 ∧ d = 0
}
❡
H =
〈
(1, 0, 0, 3), (2, 0, 0, 1)
〉
.
❉❡t❡r♠✐♥❡ ✉♠❛ ❜❛s❡ ❞❡
✭❛✮ F ✳
✭❜✮ G✳
✭❝✮ F +G✳
✭❞✮ F +H✳
✹✳✺✻ ✲ ❊♠ M2×2(R)✱ ❝♦♥s✐❞❡r❡ ♦s s✉❜❡s♣❛ç♦s
F =
{
[
a b
0 0
]
: a, b ∈ R
}
❡ G =
{
[
0 0
c d
]
: c, d ∈ R
}
.
✭❛✮ ▼♦str❡ q✉❡ M2×2(R) = F ⊕G✳
✭❜✮ ❈♦♥s✐❞❡r❛♥❞♦ A =
[
4 5
0 6
]
❞❡t❡r♠✐♥❡ ❛ ♣r♦❥❡❝çã♦ ❞❡ A s♦❜r❡ F ✱ s❡❣✉♥❞♦ G✱ ❡ ❛ ♣r♦❥❡❝çã♦
❞❡ A s♦❜r❡ G✱ s❡❣✉♥❞♦ F ✳
✹✳✻✷ ✲ ❊♠ R3✱ ❝♦♥s✐❞❡r❡ ♦s s✉❜❡s♣❛ç♦s
F =
〈
(1, 0, 1), (1,−1, 2)
〉
❡ G =
〈
(1, α, 3)
〉
.
✭❛✮ ❉❡t❡r♠✐♥❡ ♦ ❝♦♥❥✉♥t♦ ❞♦s ✈❛❧♦r❡s ❞❡ α ♣❛r❛ ♦s q✉❛✐s s❡ t❡♠
dim(F +G) = 3.
✶✸
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
https://www.studocu.com/pt?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=exercicios-alga-17-18
✭❜✮ ❈♦♥❝❧✉❛ q✉❡ R3 = F ⊕G s❡✱ ❡ só s❡✱ α ∈ R \ {−2}✳
✹✳✻✺ ✲ ❊♠ R3✱ ❝♦♥s✐❞❡r❡ ❛ s❡q✉ê♥❝✐❛ ❞❡ ✈❡❝t♦r❡s
Sk =
(
(1, 0, 2), (−1, 2,−3), (−1, 4, k)
)
.
❉❡t❡r♠✐♥❡ ♦ ❝♦♥❥✉♥t♦ ❞♦s ✈❛❧♦r❡s ❞❡ k ♣❛r❛ ♦s q✉❛✐s Sk é ✉♠❛ ❜❛s❡ ❞❡ R3✳
✹✳✻✾ ✲ ❊♠ R4✱ ❝♦♥s✐❞❡r❡ ♦ s✉❜❡s♣❛ç♦ F =
〈
(1, 0, 1, 0), (−1, 1, 0, 1), (1, 1, 2, 1)
〉
.
✭❛✮ ■♥❞✐q✉❡ ✉♠❛ ❜❛s❡ ❞❡ F ✳
✭❜✮ ❱❡r✐✜q✉❡ q✉❡ (1, 2, 3, 2) ∈ F ✳
✭❝✮ ❉❡t❡r♠✐♥❡ ✉♠❛ ❜❛s❡ ❞❡ R4 q✉❡ ✐♥❝❧✉❛ ♦s ✈❡❝t♦r❡s ❞❛ ❜❛s❡ ❞❡ F ✐♥❞✐❝❛❞❛ ❡♠ ✭❛✮✳
✹✳✼✶ ✲ ❊♠ M3×1(R)✱ ❝♦♥s✐❞❡r❡ ❛s s❡q✉ê♥❝✐❛s
S1 =
(


1
−1
1

,


1
1
0


)
❡ S2 =
(


1
−1
1

,


1
1
0

,


2
0
1


)
.
❉❡t❡r♠✐♥❡ s❡ Si✱ i = 1, 2✱ é ✉♠❛ s❡q✉ê♥❝✐❛ ❧✐♥❡❛r♠❡♥t❡ ❞❡♣❡♥❞❡♥t❡ ❡✱ ❡♠ ❝❛s♦ ❛✜r♠❛t✐✈♦✱ ✐♥❞✐q✉❡
✉♠ ✈❡❝t♦r ❞❛ s❡q✉ê♥❝✐❛ q✉❡ s❡❥❛ ❝♦♠❜✐♥❛çã♦ ❧✐♥❡❛r ❞♦s r❡st❛♥t❡s✳
✹✳✼✹ ✲ ■♥❞✐q✉❡ ❛ ❞✐♠❡♥sã♦ ❡ ✉♠❛ ❜❛s❡ ❞♦ s✉❜❡s♣❛ç♦
✭❛✮ F =
〈
2x3 + 2x2 − 2x, x3 + 2x2 − x− 1, x3 + x+ 5, x3 + 3, 2x3 + 2x2 − x+ 2
〉
❞❡ R3[x]✳
✭❜✮ G =
〈
[
1 1
1 1
]
,
[
1 1
1 0
]
,
[
2 −3
1 1
]
,
[
4 −1
3 2
]
〉
❞❡ M2×2(R)✳
✹✳✶✺✺ ✲ ❊♠ R4[x]✱ ❝♦♥s✐❞❡r❡ ♦ s✉❜❡s♣❛ç♦
F =
{
a0 + a1x+ a2x
2 + a3x
3 + a4x
4 ∈ R4[x] : −2a0 + 2a1 + a4 = 0 ∧ −a0 + a1 + 5a4 = 0
}
.
✭❛✮ ❉❡t❡r♠✐♥❡ ✉♠❛ ❜❛s❡ ❞❡ F ✳
✭❜✮ ❉❡t❡r♠✐♥❡ ✉♠❛ ❜❛s❡ ❞❡ R4[x] q✉❡ ✐♥❝❧✉❛ ❛ ❜❛s❡ ❞❡ F ✐♥❞✐❝❛❞❛ ❡♠ ✭❛✮✳
✭❝✮ ■♥❞✐q✉❡✱ ❝❛s♦ ❡①✐st❛✱ ✉♠ s✉❜❡s♣❛ç♦ G ❞❡ R4[x] t❛❧ q✉❡
dim(F +G) = 4 ❡ dim(F ∩G) = 1.
✹✳✶✺✻ ✲ ❊♠ R4✱ ❝♦♥s✐❞❡r❡ ♦s s✉❜❡s♣❛ç♦s
F =
〈
(0, 1, 0, 1), (1, 2, 1, 2), (1, 1, 1, 1), (1, 2, 3, 4)
〉
❡
Gt =
〈
(1, 1, 0, 1), (1, 0, 1, 1), (t, 2,−1, 1)
〉
.
✭❛✮ ▼♦str❡ q✉❡ ❡①✐st❡ ✉♠✱ ❡ ✉♠ só✱ ✈❛❧♦r ❞❡ t ♣❛r❛ ♦ q✉❛❧ dimGt = 2✳
✭❜✮ P❛r❛ ♦ ✈❛❧♦r ❞❡ t ❞❡t❡r♠✐♥❛❞♦ ❡♠ ✭❛✮✱ ❞❡t❡r♠✐♥❡ ✉♠❛ ❜❛s❡ ❞❡ F + Gt ❡ ❛ ❞✐♠❡♥sã♦ ❞❡
F ∩Gt✳
✺ ✲ ❆♣❧✐❝❛çõ❡s ▲✐♥❡❛r❡s
✶✹
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)lOMoARcPSD|6016133
✺✳✷ ✲ ❉❡t❡r♠✐♥❡ s❡ é ❧✐♥❡❛r ❛ ❛♣❧✐❝❛çã♦ f : R3 −→ R2 t❛❧ q✉❡✱ ♣❛r❛ q✉❛❧q✉❡r (x, y, z) ∈ R3✱ s❡ t❡♠
✭❛✮ f(x, y, z) = (y, 0)✳
✭❜✮ f(x, y, z) = (x− 1, y)✳
✭❝✮ f(x, y, z) = (xy, 0)✳
✭❞✮ f(x, y, z) = (x, |z|)✳
✺✳✹ ✲ ❙❡❥❛ g : Mm×n(K) −→ Mn×m(K) t❛❧ q✉❡
g(A) = A⊤,
♣❛r❛ q✉❛❧q✉❡r A ∈ Mm×n(K)✳ ❏✉st✐✜q✉❡ q✉❡ g é ✉♠❛ ❛♣❧✐❝❛çã♦ ❧✐♥❡❛r✳
✺✳✻ ✲ ❙❡❥❛ f : R3 −→ M2×2(R) ✉♠❛ ❛♣❧✐❝❛çã♦ ❧✐♥❡❛r✳ ❏✉st✐✜q✉❡ q✉❡✿
✭❛✮ f(0, 0, 0) =
[
0 0
0 0
]
✳
✭❜✮ f(2, 4,−2) = 2f(1, 2,−1)✳
✭❝✮ f(−3, 1, 2) = f(−2, 0, 1) + f(−1, 1, 1)✳
✺✳✼ ✲ ❉❡t❡r♠✐♥❡ s❡ é ❧✐♥❡❛r ❝❛❞❛ ✉♠❛ ❞❛s ❛♣❧✐❝❛çõ❡s s❡❣✉✐♥t❡s✿
✭❛✮ f : R3 −→ R2 t❛❧ q✉❡
f(a, b, c) = (2a, b+ 1),
♣❛r❛ q✉❛❧q✉❡r (a, b, c) ∈ R3✳
✭❜✮ g : R2[x] −→ M2×2(R) t❛❧ q✉❡
g(ax2 + bx+ c) =
[
c b
a+ b 2
]
,
♣❛r❛ q✉❛❧q✉❡r ax2 + bx+ c ∈ R2[x]✳
✺✳✶✵ ✲ ❉❡t❡r♠✐♥❡ ♦ ♥ú❝❧❡♦✱ ✉♠❛ ❜❛s❡ ❞♦ ♥ú❝❧❡♦ ❡ ✉♠❛ ❜❛s❡ ❞❛ ✐♠❛❣❡♠ ❞❡ ❝❛❞❛ ✉♠❛ ❞❛s ❛♣❧✐❝❛çõ❡s
❧✐♥❡❛r❡s s❡❣✉✐♥t❡s✿
✭❛✮ f : R3 −→ R2 t❛❧ q✉❡
f(x, y, z) = (y, z),
♣❛r❛ q✉❛❧q✉❡r (x, y, z) ∈ R3✳
✭❜✮ g : M2×2(R) −→ R3 t❛❧ q✉❡
g
(
[
a b
c d
]
)
= (2a, c+ d, 0),
♣❛r❛ q✉❛❧q✉❡r
[
a b
c d
]
∈ M2×2(R)✳
✭❝✮ h : R3 −→ R2[x] t❛❧ q✉❡
h(a, b, c) = (a+ b)x2 + c,
♣❛r❛ q✉❛❧q✉❡r (a, b, c) ∈ R3✳
✭❞✮ t : R3[x] −→ M2×2(R) t❛❧ q✉❡
t(ax3 + bx2 + cx+ d) =
[
a− c 0
0 b+ d
]
,
♣❛r❛ q✉❛❧q✉❡r ax3 + bx2 + cx+ d ∈ R3[x]✳
✶✺
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
https://www.studocu.com/pt?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=exercicios-alga-17-18
✺✳✶✹ ✲ ■♥❞✐q✉❡ s❡ ❝❛❞❛ ✉♠❛ ❞❛s ❛♣❧✐❝❛çõ❡s ❧✐♥❡❛r❡s s❡❣✉✐♥t❡s é ✐♥❥❡❝t✐✈❛✱ ❞❡t❡r♠✐♥❛♥❞♦ ♦ s❡✉ ♥ú❝❧❡♦✳
✭❛✮ f : R3 −→ R3 t❛❧ q✉❡ f(a, b, c) = (2a, b+ c, b− c),
♣❛r❛ q✉❛❧q✉❡r (a, b, c) ∈ R3✳
✭❜✮ g : R2[x] −→ M2×2(R) t❛❧ q✉❡ g(ax2 + bx+ c) =
[
2a b+ c
0 a+ b− c
]
,
♣❛r❛ q✉❛❧q✉❡r ax2 + bx+ c ∈ R2[x]✳
✺✳✶✼ ✲ ❉❡t❡r♠✐♥❡ ❛ ♥✉❧✐❞❛❞❡ ❞❡ ❝❛❞❛ ✉♠❛ ❞❛s ❛♣❧✐❝❛çõ❡s ❧✐♥❡❛r❡s s❡❣✉✐♥t❡s✿
✭❛✮ f : R5 −→ R8 ❝♦♠ dim Im f = 4✳
✭❜✮ g : R3[x] −→ R3[x] ❝♦♠ dim Im g = 1✳
✭❝✮ h : R6 −→ R3 ❝♦♠ h s♦❜r❡❥❡❝t✐✈❛✳
✭❞✮ t : M3×3(R) −→ M3×3(R) ❝♦♠ t s♦❜r❡❥❡❝t✐✈❛✳
✺✳✶✽ ✲ ❏✉st✐✜q✉❡ q✉❡ ♥ã♦ ❡①✐st❡ ♥❡♥❤✉♠❛ ❛♣❧✐❝❛çã♦ ❧✐♥❡❛r f : R7 −→ R3 ❝✉❥♦ ♥ú❝❧❡♦ t❡♥❤❛ ❞✐♠❡♥sã♦
✐♥❢❡r✐♦r ♦✉ ✐❣✉❛❧ ❛ 3✳
✺✳✶✾ ✲ ❙❡❥❛ f : R5 −→ R3 ✉♠❛ ❛♣❧✐❝❛çã♦ ❧✐♥❡❛r ❝♦♠ ♥✉❧✐❞❛❞❡ n(f) ❡ ❝❛r❛❝t❡ríst✐❝❛ r(f)✳ ■♥❞✐q✉❡ t♦❞♦s
♦s ♣❛r❡s ♣♦ssí✈❡✐s (n(f), r(f))✳
✺✳✷✷ ✲ ❯t✐❧✐③❛♥❞♦ ❛ Pr♦♣♦s✐çã♦ ❛❞❡q✉❛❞❛✱ ❞❡t❡r♠✐♥❡ s❡ é ❜✐❥❡❝t✐✈❛ ❝❛❞❛ ✉♠❛ ❞❛s s❡❣✉✐♥t❡s ❛♣❧✐❝❛çõ❡s
❧✐♥❡❛r❡s✳
✭❛✮ ❆ ❛♣❧✐❝❛çã♦ ❞❛ ❛❧í♥❡❛ ✭❛✮ ❞♦ ❊①❡r❝í❝✐♦ ✺✳✶✹✱ f : R3 −→ R3 t❛❧ q✉❡
f(a, b, c) = (2a, b+ c, b− c),
♣❛r❛ q✉❛❧q✉❡r (a, b, c) ∈ R3✳
✭❜✮ g : M2×2(R) −→ R3[x] t❛❧ q✉❡
g
(
[
a b
c d
]
)
= (a+ d)x3 + 2ax2 + (b− c)x+ (a+ c),
♣❛r❛ q✉❛❧q✉❡r
[
a b
c d
]
∈ M2×2(R)✳
✺✳✷✻ ✲ ■♥❞✐q✉❡ s❡ ❡①✐st❡ ❛❧❣✉♠❛ ❛♣❧✐❝❛çã♦ ❧✐♥❡❛r ♥❛s ❝♦♥❞✐çõ❡s r❡❢❡r✐❞❛s ❡✱ ❡♠ ❝❛s♦ ❛✜r♠❛t✐✈♦✱ ❞ê ✉♠
❡①❡♠♣❧♦✳
✭❛✮ f : R4 −→ R4 t❛❧ q✉❡ Im f =
〈
(1, 0, 0, 1), (0, 1, 1, 0), (0, 1, 2, 0)
〉
❡ dimKer f = 2.
✭❜✮ g : R4 −→ R3 t❛❧ q✉❡ Ker g =
〈
(0, 1, 1, 0), (1, 1, 1, 1)
〉
❡ (1, 1, 1) ∈ Im g.
✭❝✮ h : R3 −→ R4 t❛❧ q✉❡ Imh =
〈
(1, 2, 0,−4), (2, 0,−1,−3)
〉
.
✺✳✸✻ ✲ ❈♦♥s✐❞❡r❡ ❛ ❛♣❧✐❝❛çã♦ ❧✐♥❡❛r f : R3 −→ R2 t❛❧ q✉❡
f(x, y, z) = (y, z),
♣❛r❛ q✉❛❧q✉❡r (x, y, z) ∈ R3 ❡ ❛ ❛♣❧✐❝❛çã♦ ❧✐♥❡❛r g : M2×2(R) −→ R3 t❛❧ q✉❡
g
(
[
a b
c d
]
)
= (2a, c+ d, 0),
♣❛r❛ q✉❛❧q✉❡r
[
a b
c d
]
∈ M2×2(R)✳
✶✻
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
✭❛✮ ❉❡t❡r♠✐♥❡ M(f ; B,B′) s❡♥❞♦
B =
(
(1, 2, 3), (0,−2, 1), (0, 0, 3)
)
❡ B′ =
(
(0,−2), (−1, 0)
)
.
✭❜✮ ❉❡t❡r♠✐♥❡ M(g; B′′, b. c.R3) s❡♥❞♦
B′′ =
(
[
1 1
0 1
]
,
[
0 2
3 1
]
,
[
0 0
2 −1
]
,
[
0 0
1 0
]
)
.
✺✳✸✾ ✲ ❙❡❥❛♠ B ❡ B′ ❜❛s❡s ❞❡ R4 ❡ R3✱ r❡s♣❡❝t✐✈❛♠❡♥t❡✱ ❡ s❡❥❛ f : R4 −→ R3 ❛ ❛♣❧✐❝❛çã♦ ❧✐♥❡❛r t❛❧ q✉❡
M(f ;B,B′) =


−1 2 1 1
0 1 1 0
1 0 1 −1

.
❈❛❧❝✉❧❛♥❞♦ ❛ ❝❛r❛❝t❡ríst✐❝❛ ❞❛ ♠❛tr✐③ ❛♥t❡r✐♦r✱ ❞❡t❡r♠✐♥❡ s❡ f é s♦❜r❡❥❡❝t✐✈❛✳
✺✳✹✷ ✲ ❙❡❥❛ f : R3 −→ R2 ❛ ❛♣❧✐❝❛çã♦ ❧✐♥❡❛r t❛❧ q✉❡
f(a, b, c) = (a+ b, b+ c),
♣❛r❛ q✉❛❧q✉❡r (a, b, c) ∈ R3✳ ❊♠ R3✱ ❝♦♥s✐❞❡r❡ ❛s ❜❛s❡s
B1 = b. c.R3 , B2 =
(
(0, 1, 0), (1, 0, 1), (1, 0, 0)
)
❡✱ ❡♠ R2✱ ❝♦♥s✐❞❡r❡ ❛s ❜❛s❡s
B′1 = b. c.R2 , B′2 =
(
(1, 1), (1, 0)
)
.
✭❛✮ ❈❛❧❝✉❧❡ f(1, 2, 3)✳
✭❜✮ ❉❡t❡r♠✐♥❡ M (f ;B1,B′1) ❡ ❝❛❧❝✉❧❡ f(1, 2, 3) ✉t✐❧✐③❛♥❞♦ ❡st❛ ♠❛tr✐③✳
✭❝✮ ❉❡t❡r♠✐♥❡ M (f ;B2,B′1) ❡ ❝❛❧❝✉❧❡ f(1, 2, 3) ✉t✐❧✐③❛♥❞♦ ❡st❛ ♠❛tr✐③✳
✭❞✮ ❉❡t❡r♠✐♥❡ M (f ;B1,B′2) ❡ ❝❛❧❝✉❧❡ f(1, 2, 3) ✉t✐❧✐③❛♥❞♦ ❡st❛ ♠❛tr✐③✳
✭❡✮ ❉❡t❡r♠✐♥❡ M (f ;B2,B′2) ❡ ❝❛❧❝✉❧❡ f(1, 2, 3) ✉t✐❧✐③❛♥❞♦ ❡st❛ ♠❛tr✐③✳
✺✳✹✸ ✲ ❊♠ R3✱ ❝♦♥s✐❞❡r❡ ❛s ❜❛s❡s
B1 =
(
(1,−1, 0), (−1, 1,−1), (0, 1, 0)
)
, B2 = b. c.R3
❡
B3 =
(
(1, 1, 1), (0, 1, 1), (0, 0, 1)
)
.
❉❡t❡r♠✐♥❡ ❛ ♠❛tr✐③ ❞❡ ♠✉❞❛♥ç❛ ❞❡ ❜❛s❡ ❞❡
✭❛✮ B1 ♣❛r❛ B2✳
✭❜✮ B2 ♣❛r❛ B1✳
✭❝✮ B1 ♣❛r❛ B3✳
✺✳✹✻ ✲ ❙❡❥❛ f : R3 −→ R2 ❛ ❛♣❧✐❝❛çã♦ ❧✐♥❡❛r t❛❧ q✉❡
M (f ; b. c.R3 , b. c.R2) =
[
1 1 0
0 1 1
]
.
❈♦♥s✐❞❡r❡ ❛s ❜❛s❡s
B =
(
(0, 1, 0), (1, 0, 1), (1, 0, 0)
)
❡ B′ =
(
(1, 1), (1, 0)
)
❞❡ R3 ❡ ❞❡ R2✱ r❡s♣❡❝t✐✈❛♠❡♥t❡✳ ❯t✐❧✐③❛♥❞♦ ♠❛tr✐③❡s ❞❡ ♠✉❞❛♥ç❛ ❞❡ ❜❛s❡✱ ❞❡t❡r♠✐♥❡✿
✶✼
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
https://www.studocu.com/pt?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=exercicios-alga-17-18
✭❛✮ M (f ; B, b. c.R2)✳
✭❜✮ M (f ; b. c.R3 ,B′)✳
✭❝✮ M (f ; B,B′)✳
✺✳✶✶✵ ✲ ❙❡❥❛ E ✉♠ ❡s♣❛ç♦ ✈❡❝t♦r✐❛❧ r❡❛❧ ❡ s❡❥❛ B′ = (u1, u2, u3) ✉♠❛ ❜❛s❡ ❞❡ E✳ ❈♦♥s✐❞❡r❡ ❛ ❛♣❧✐❝❛çã♦
❧✐♥❡❛r f : R5 −→ E t❛❧ q✉❡
f(a, b, c, d, e) = (−b− c+ d)u1 + (2a+ b+ 3c− 3d)u2 + (b+ c− d)u3,
♣❛r❛ q✉❛✐sq✉❡r a, b, c, d, e ∈ R✳
✭❛✮ ❉❡t❡r♠✐♥❡ M(f ; b. c.R5 ,B′)✳
✭❜✮ ❉❡t❡r♠✐♥❡ ✉♠❛ ❜❛s❡ ❞❡ Im f ✳
✭❝✮ ❊♠ R5✱ ❝♦♥s✐❞❡r❡ ♦s ✈❡❝t♦r❡s
v1 = (2, 2, 0, 2, 2), v2 = (−1,−1, 1, 0, 1) ❡ v3 = (0, 0, 0, 0, 1).
▼♦str❡ q✉❡ (v1, v2, v3) é ✉♠❛ ❜❛s❡ ❞❡ Ker f ✳
✭❞✮ ❉❡t❡r♠✐♥❡ ✉♠❛ ❜❛s❡ ❞❡ R5 q✉❡ ✐♥❝❧✉❛ ♦s ✈❡❝t♦r❡s v1✱ v2 ❡ v3✳
✭❡✮ ❙❡♥❞♦ B ❛ ❜❛s❡ ♦❜t✐❞❛ ❡♠ ✭❞✮✱ ❞❡t❡r♠✐♥❡ M(f ;B,B′)✳
✻ ✲ ❱❛❧♦r❡s ❡ ❱❡❝t♦r❡s Pró♣r✐♦s
✻✳✶ ✲ ❙❡❥❛ f : R3 −→ R3 ❛ ❛♣❧✐❝❛çã♦ ❧✐♥❡❛r t❛❧ q✉❡
f(a, b, c) = (a+ b, b, 2c),
♣❛r❛ q✉❛❧q✉❡r (a, b, c) ∈ R3✳ ❈♦♥s✐❞❡r❡ ♦s ✈❡❝t♦r❡s u1 = (2, 0, 0)✱ u2 = (0, 0, 7) ❡ u3 = (0, 0, 0)✳
❱❡r✐✜q✉❡ s❡ ❝❛❞❛ ✉♠ ❞♦s ✈❡❝t♦r❡s u1✱ u2✱ u3 é ✉♠ ✈❡❝t♦r ♣ró♣r✐♦ ❞❡ f ❡✱ ❡♠ ❝❛s♦ ❛✜r♠❛t✐✈♦✱
✐♥❞✐q✉❡ ♦ ✈❛❧♦r ♣ró♣r✐♦ ❛ss♦❝✐❛❞♦✳
✻✳✷ ✲ ❙❡❥❛ A =
[
1 0
1 2
]
∈ M2×2(R).
✭❛✮ ▼♦str❡ q✉❡
[
1
−1
]
❡
[
0
2
]
sã♦ ✈❡❝t♦r❡s ♣ró♣r✐♦s ❞❡ A ❡ ✐♥❞✐q✉❡ ♦s ✈❛❧♦r❡s ♣ró♣r✐♦s ❝♦rr❡s✲
♣♦♥❞❡♥t❡s✳
✭❜✮ ◗✉❡stã♦ ❛♥á❧♦❣❛ à ❞❡ ✭❛✮ ♣❛r❛
[
α
−α
]
❡
[
0
α
]
✱ ❝♦♠ α ∈ R \ {0}✳
✻✳✸ ✲ ❙❡❥❛ α ✉♠ ✈❛❧♦r ♣ró♣r✐♦ ❞❡ A ∈ Mn×n(C)✳ ❏✉st✐✜q✉❡ q✉❡ α é ✈❛❧♦r ♣ró♣r✐♦ ❞❡ A✳
✻✳✼ ✲ ❙❡❥❛ A ∈ Mn×n(K) ✉♠❛ ♠❛tr✐③ ✐❞❡♠♣♦t❡♥t❡ ✭✐st♦ é✱ A2 = A✮✳
✭❛✮ ▼♦str❡ q✉❡ t♦❞♦ ♦ ✈❛❧♦r ♣ró♣r✐♦ ❞❡ A ♣❡rt❡♥❝❡ ❛♦ ❝♦♥❥✉♥t♦ {0, 1}✳
✭❜✮ ■♥❞✐q✉❡ ✉♠❛ ♠❛tr✐③ q✉❡ t❡♥❤❛ t♦❞♦s ♦s ✈❛❧♦r❡s ♣ró♣r✐♦s ♥♦ ❝♦♥❥✉♥t♦ {0, 1} ❡ q✉❡ ♥ã♦ s❡❥❛
✐❞❡♠♣♦t❡♥t❡✳
✻✳✶✷ ✲ ❉❡t❡r♠✐♥❡ ♦s ✈❛❧♦r❡s ♣ró♣r✐♦s ❞❛ ♠❛tr✐③
A =


2 −i 0
i 2 0
0 0 3

 ∈ M3×3(C)
❡ ❛s r❡s♣❡❝t✐✈❛s ♠✉❧t✐♣❧✐❝✐❞❛❞❡s ❛❧❣é❜r✐❝❛s✳
✶✽
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
✻✳✶✸ ✲ ❙❡❥❛♠ A =
[
0 1
−1 0
]
, B =
[
−2 −1
5 2
]
∈ M2×2(K). ▼♦str❡ q✉❡✿
✭❛✮ ❙❡ K = R ❡♥tã♦ A ♥ã♦ t❡♠ ✈❛❧♦r❡s ♣ró♣r✐♦s✳
✭❜✮ ❙❡ K = C ❡♥tã♦ A t❡♠ ❞♦✐s ✈❛❧♦r❡s ♣ró♣r✐♦s ❞✐st✐♥t♦s✳
✭❝✮ ❆s ♠❛tr✐③❡s A ❡ B tê♠ ♦ ♠❡s♠♦ ♣♦❧✐♥ó♠✐♦ ❝❛r❛❝t❡ríst✐❝♦✳
✻✳✶✹ ✲ ❙❡❥❛ A =


0 2 0
−2 0 0
0 0 3

 ∈ M3×3(R).
✭❛✮ ❉❡t❡r♠✐♥❡ ♦s ✈❛❧♦r❡s ♣ró♣r✐♦s ❞❡ A ❡ ❛s r❡s♣❡❝t✐✈❛s ♠✉❧t✐♣❧✐❝✐❞❛❞❡s ❛❧❣é❜r✐❝❛s✳
✭❜✮ ❈❛❧❝✉❧❡ ♦ ❞❡t❡r♠✐♥❛♥t❡ ❞❡ A✳
✻✳✶✾ ✲ ❙❡❥❛ A ∈ Mn×n(K) ✉♠❛ ♠❛tr✐③ ✐♥✈❡rtí✈❡❧✳ ▼♦str❡ q✉❡✿
✭❛✮ ❙❡ α é ✈❛❧♦r ♣ró♣r✐♦ ❞❡ A ❡♥tã♦ α 6= 0 ❡ α−1 é ✈❛❧♦r ♣ró♣r✐♦ ❞❡ A−1✳
✭❜✮ ❙❡ X ∈ Mn×1(K) é ✈❡❝t♦r ♣ró♣r✐♦ ❞❡ A ❛ss♦❝✐❛❞♦ ❛♦ ✈❛❧♦r ♣ró♣r✐♦ α ❡♥tã♦ X é ✈❡❝t♦r
♣ró♣r✐♦ ❞❡ A−1 ❛ss♦❝✐❛❞♦ ❛♦ ✈❛❧♦r ♣ró♣r✐♦ α−1✳
✻✳✷✽ ✲ ❙❡❥❛ f : R3 −→ R3 ❛ ❛♣❧✐❝❛çã♦ ❧✐♥❡❛r ❞❡✜♥✐❞❛ ♣♦r
f(a, b, c) = (−b− c,−2a+ b− c, 4a+ 2b+ 4c),
♣❛r❛ q✉❛❧q✉❡r (a, b, c) ∈ R3✳ ❉❡t❡r♠✐♥❡ ♦s ✈❛❧♦r❡s ♣ró♣r✐♦s ❡ ♦s s✉❜❡s♣❛ç♦s ♣ró♣r✐♦s ❞❡ f ✳
✻✳✸✺ ✲ ❈♦♥s✐❞❡r❡ ❛s ♠❛tr✐③❡s tr✐❛♥❣✉❧❛r❡s
A =
[
−2 1
0 2
]
, B =
[
5 0
4 1
]
∈ M2×2(R).
❙❡♠ ❡❢❡❝t✉❛r ❝á❧❝✉❧♦s✱ ❥✉st✐✜q✉❡ q✉❡ A ❡ B sã♦ ❛♠❜❛s ❞✐❛❣♦♥❛❧✐③á✈❡✐s❡ ✐♥❞✐q✉❡ ✉♠❛ ♠❛tr✐③
❞✐❛❣♦♥❛❧ DA s❡♠❡❧❤❛♥t❡ ❛ A ❡ ✉♠❛ ♠❛tr✐③ ❞✐❛❣♦♥❛❧ DB s❡♠❡❧❤❛♥t❡ ❛ B✳
✻✳✸✻ ✲ ❙❡❥❛ A =


2 5 2
0 3 0
2 −1 2

 ∈ M3×3(R)✳ ❈❛❧❝✉❧❡ ♦s ✈❛❧♦r❡s ♣ró♣r✐♦s ❞❡ A ❡✱ s❡♠ ❞❡t❡r♠✐♥❛r ♦s
s✉❜❡s♣❛ç♦s ♣ró♣r✐♦s ❞❡ A✱ ❝♦♥❝❧✉❛ q✉❡ A é ❞✐❛❣♦♥❛❧✐③á✈❡❧✳
✻✳✸✼ ✲ ❈♦♥s✐❞❡r❡ ❛ ♠❛tr✐③
A =


3 2 0
−4 −3 0
4 2 −1

 ∈ M3×3(R).
✭❛✮ ❈❛❧❝✉❧❡ ♦s ✈❛❧♦r❡s ♣ró♣r✐♦s ❞❡ A ❡ ✐♥❞✐q✉❡ ❛s r❡s♣❡❝t✐✈❛s ♠✉❧t✐♣❧✐❝✐❞❛❞❡s ❛❧❣é❜r✐❝❛s✳
✭❜✮ ❉❡t❡r♠✐♥❡ ✉♠❛ ❜❛s❡ ♣❛r❛ ❝❛❞❛ ✉♠ ❞♦s s✉❜❡s♣❛ç♦s ♣ró♣r✐♦s ❞❡ A✳
✭❝✮ ▼♦str❡ q✉❡ A é ❞✐❛❣♦♥❛❧✐③á✈❡❧ ❡ ✐♥❞✐q✉❡ ✉♠❛ ♠❛tr✐③ ✐♥✈❡rtí✈❡❧
P ∈ M3×3(R) ❡ ✉♠❛ ♠❛tr✐③ ❞✐❛❣♦♥❛❧ D t❛✐s q✉❡
P−1AP = D.
✻✳✹✶ ✲ ❙❡❥❛ f ✉♠ ❡♥❞♦♠♦r✜s♠♦ ❞❡ R3 ❡ s❡❥❛ B = (e1, e2, e3) ✉♠❛ ❜❛s❡ ❞❡ R3✳ ❙❛❜❡♥❞♦ q✉❡
M (f ;B,B) =


3 2 0
−4 −3 0
4 2 −1

,
❞❡t❡r♠✐♥❡✿
✶✾
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
https://www.studocu.com/pt?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=exercicios-alga-17-18
✭❛✮ ❖s ✈❛❧♦r❡s ♣ró♣r✐♦s ❞❡ f ✳
✭❜✮ ❯♠❛ ❜❛s❡ B′✱ ❞❡ R3✱ ❝♦♥st✐t✉í❞❛ ♣♦r ✈❡❝t♦r❡s ♣ró♣r✐♦s ❞❡ f ✳
✭❝✮ M (f ;B′,B′) s❡♥❞♦ B′ ❛ ❜❛s❡ ✐♥❞✐❝❛❞❛ ❡♠ ✭❜✮✳
❖❜s❡r✈❛çã♦✿ ❈♦♠♣❛r❡ ♦s r❡s✉❧t❛❞♦s ❝♦♠ ♦s ♦❜t✐❞♦s ♥♦ ❊①❡r❝í❝✐♦ ✻✳✸✼✳
✻✳✽✾ ✲ ❈♦♥s✐❞❡r❡ ❛s ♠❛tr✐③❡s
A =


1 0 −1
1 2 1
2 2 3

, B =


0 1 0
0 0 1
1 −3 3

 ∈ M3×3(R).
✭❛✮ ❉❡t❡r♠✐♥❡ ♦s ✈❛❧♦r❡s ♣ró♣r✐♦s ❞❡ ❝❛❞❛ ✉♠❛ ❞❛s ♠❛tr✐③❡s ❡ ✐♥❞✐q✉❡ ❛s r❡s♣❡❝t✐✈❛s ♠✉❧t✐♣❧✐✲
❝✐❞❛❞❡s ❛❧❣é❜r✐❝❛s✳
✭❜✮ ✐✳ ▼♦str❡ q✉❡ A é ❞✐❛❣♦♥❛❧✐③á✈❡❧✳
✐✐✳ ■♥❞✐q✉❡ s❡ B é ❞✐❛❣♦♥❛❧✐③á✈❡❧✳
✭❝✮ ❉❡t❡r♠✐♥❡ ✉♠❛ ♠❛tr✐③ ✐♥✈❡rtí✈❡❧ P ∈ M3×3(R) t❛❧ q✉❡ P−1AP s❡❥❛ ✉♠❛ ♠❛tr✐③ ❞✐❛❣♦♥❛❧ ❡
♦s ❡❧❡♠❡♥t♦s ❞❛ ❞✐❛❣♦♥❛❧ ♣r✐♥❝✐♣❛❧ ❞❡ P−1AP ❡st❡❥❛♠ ♦r❞❡♥❛❞♦s ♣♦r ♦r❞❡♠ ❝r❡s❝❡♥t❡✳
✻✳✾✸ ✲ ❙❡❥❛ A ∈ M3×3(R) t❛❧ q✉❡
A


1
2
3

 =


2
4
6

, A


0
1
2

 =


0
0
0

 ❡ A


0
0
1

 =


0
0
2

.
✭❛✮ ■♥❞✐q✉❡ ♦s ✈❛❧♦r❡s ♣ró♣r✐♦s ❞❡ A ❡ ❛s r❡s♣❡❝t✐✈❛s ♠✉❧t✐♣❧✐❝✐❞❛❞❡s ❣❡♦♠étr✐❝❛s✳
✭❜✮ ■♥❞✐q✉❡✱ s❡ ❡①✐st✐r✱ ✉♠❛ ♠❛tr✐③ ❞✐❛❣♦♥❛❧ s❡♠❡❧❤❛♥t❡ ❛ A✳
✭❝✮ ❉❡t❡r♠✐♥❡ ✉♠❛ ♠❛tr✐③ A ♥❛s ❝♦♥❞✐çõ❡s ❞♦ ❡♥✉♥❝✐❛❞♦✳
✻✳✾✺ ✲ ❊♠ R3✱ ❝♦♥s✐❞❡r❡ ♦ s✉❜❡s♣❛ç♦
F =
{
(x, y, z) ∈ R3 : x+ 2y + z = 0
}
.
❙❡❥❛ f : R3 −→ R3 ❛ ❛♣❧✐❝❛çã♦ ❧✐♥❡❛r t❛❧ q✉❡ (1,−1, 0) é ✈❡❝t♦r ♣ró♣r✐♦ ❞❡ f ❛ss♦❝✐❛❞♦ ❛♦ ✈❛❧♦r
♣ró♣r✐♦ 2 ❡
f(a, b, c) = (0, 0, 0),
♣❛r❛ q✉❛❧q✉❡r (a, b, c) ∈ F ✳
✭❛✮ ❏✉st✐✜q✉❡ q✉❡ B =
(
(1,−1, 0), (1, 1,−3), (1, 0,−1)
)
é ✉♠❛ ❜❛s❡ ❞❡ R3 ❝♦♥st✐t✉í❞❛ ♣♦r ✈❡❝✲
t♦r❡s ♣ró♣r✐♦s ❞❡ f ✳
✭❜✮ ▼♦str❡ q✉❡ 0 é ✈❛❧♦r ♣ró♣r✐♦ ❞❡ f ❡ q✉❡ mg(0) = ma(0)✳
✭❝✮ ❉❡t❡r♠✐♥❡ M(f ;B, b. c.R3)✳
✼ ✲ Pr♦❞✉t♦ ■♥t❡r♥♦✱ ❊①t❡r♥♦ ❡ ▼✐st♦
✼✳✶ ✲ ◆♦ r❡❢❡r❡♥❝✐❛❧ ♦rt♦♥♦r♠❛❞♦ (O; e1, e2, e3) ❝♦♥s✐❞❡r❡ ♦s s❡❣✉✐♥t❡s ✈❡❝t♦r❡s✿
a = 2e1 + αe2 + e3 ❡ b = 4e1 − 2e2 − 2e3.
❉❡t❡r♠✐♥❡ ♣❛r❛ q✉❡ ✈❛❧♦r ❞❡ α ♦s ✈❡❝t♦r❡s sã♦ ♣❡r♣❡♥❞✐❝✉❧❛r❡s✳
✷✵
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
✼✳✷ ✲ ◆♦ r❡❢❡r❡♥❝✐❛❧ ♦rt♦♥♦r♠❛❞♦ (O; e1, e2, e3) ❝♦♥s✐❞❡r❡ ♦s s❡❣✉✐♥t❡s ✈❡❝t♦r❡s✿
a = 2e1 + e2 − e3 ❡ b = 6e1 − 3e2 + e3.
❉❡t❡r♠✐♥❡ ♦ â♥❣✉❧♦ ❢♦r♠❛❞♦ ♣❡❧♦s ✈❡❝t♦r❡s a ❡ b✳
✼✳✸ ✲ ◆♦ r❡❢❡r❡♥❝✐❛❧ ♦rt♦♥♦r♠❛❞♦ (O; e1, e2, e3) ❝♦♥s✐❞❡r❡ ♦s s❡❣✉✐♥t❡s ✈❡❝t♦r❡s✿
a = e1 + e2 + e3 ❡ b = e1 − 2e2 + 3e3.
❉❡t❡r♠✐♥❡ ♦ s❡♥♦ ❡ ♦ ❝♦✲s❡♥♦ ❞♦ â♥❣✉❧♦ ❢♦r♠❛❞♦ ♣❡❧♦s ✈❡❝t♦r❡s a ❡ b✳
✼✳✹ ✲ ❈♦♥s✐❞❡r❡✱ ♥♦ r❡❢❡r❡♥❝✐❛❧ ♦rt♦♥♦r♠❛❞♦ ❡ ❞✐r❡❝t♦ (O; e1, e2, e3) ♦s s❡❣✉✐♥t❡s ✈❡❝t♦r❡s✿
u = e1 − e2 + e3, v = e2 + 2e3 ❡ w = e1 + e2.
❉❡t❡r♠✐♥❡
✭❛✮ u× v❀
✭❜✮ v × w❀
✭❝✮ (u× v)× w❀
✭❞✮ u× (v × w)❀
✭❡✮ ❯♠ ✈❡❝t♦r ♣❡r♣❡♥❞✐❝✉❧❛r ❛ u ❡ ❛ v ❞❡ ♥♦r♠❛ ✐❣✉❛❧ ❛
√
15✳
✼✳✺ ✲ ❉❡t❡r♠✐♥❡✱ ❝♦♥s✐❞❡r❛♥❞♦ ♦ r❡❢❡r❡♥❝✐❛❧ ♦rt♦♥♦r♠❛❞♦ ❡ ❞✐r❡❝t♦ (O; e1, e2, e3)✱ ❛ ár❡❛ ❞♦ tr✐â♥❣✉❧♦
q✉❡ t❡♠ ♣♦r ✈ért✐❝❡s ♦s ♣♦♥t♦s A(1, 2, 3)✱ B(2,−1, 1) ❡ C(−1, 2, 3)✳
✼✳✻ ✲ ◆♦ r❡❢❡r❡♥❝✐❛❧ ♦rt♦♥♦r♠❛❞♦ (O; e1, e2, e3) ❡♠ q✉❡ O = (0, 0, 0)✱ ❝♦♥s✐❞❡r❡ ♦s s❡❣✉✐♥t❡s ✈❡❝t♦r❡s✿
−→
OA = e1 − 3e2 + 2e3 ❡
−−→
OB = 2e1 + e2 − e3.
❏✉st✐✜q✉❡ q✉❡ ♦s ♣♦♥t♦s O✱ A ❡ B ❞❡✜♥❡♠ ✉♠ ♣❧❛♥♦ ❡ ❞❡t❡r♠✐♥❡ ✉♠ ✈❡❝t♦r ♣❡r♣❡♥❞✐❝✉❧❛r ❛ ❡ss❡
♣❧❛♥♦✳
✼✳✼ ✲ ◆♦ r❡❢❡r❡♥❝✐❛❧ ♦rt♦♥♦r♠❛❞♦ (O; e1, e2, e3)✱ ❝♦♥s✐❞❡r❡ ♦s ✈❡❝t♦r❡s
a = e1 + e2, b = e2 + e3 ❡ c = e1 + e3.
❈❛❧❝✉❧❡ ♦ ✈♦❧✉♠❡ ❞♦ ♣❛r❛❧❡❧❡♣í♣❡❞♦ ❞❡t❡r♠✐♥❛❞♦ ♣❡❧♦s ✈❡❝t♦r❡s a✱ b ❡ c✳
✼✳✽ ✲ ◆♦ r❡❢❡r❡♥❝✐❛❧ ♦rt♦♥♦r♠❛❞♦ ❞✐r❡❝t♦ (O; e1, e2, e3) ❡♠ q✉❡ O = (0, 0, 0)✱ ❝♦♥s✐❞❡r❡ ♦s s❡❣✉✐♥t❡s
✈❡❝t♦r❡s✿ −−→
OP = 2e1 − e2 + e3 ❡
−−→
OQ = e1 + e2 + e3.
❉❡t❡r♠✐♥❡✿
✭❛✮ ❆ ❡①♣r❡ssã♦ ❞♦ ✈❡❝t♦r
−−→
PQ ✉t✐❧✐③❛♥❞♦ ❛s s✉❛s ❝♦♠♣♦♥❡♥t❡s ❡ ♦s s❡✉s ❝♦✲s❡♥♦s ❞✐r❡❝t♦r❡s✳
✭❜✮ ❯♠ ✈❡❝t♦r ❞♦ ♣❧❛♥♦ OXY ✱ ❝♦♠ ♥♦r♠❛ 4 ✱ q✉❡ ❢❛③ ❝♦♠ ♦s ❡✐①♦s OX ❡ OY ✱ r❡s♣❡❝t✐✈❛♠❡♥t❡✱
♦s â♥❣✉❧♦s ❞❡ 30o ❡ 60o.
✭❝✮ ❆ ár❡❛ ❞♦ tr✐â♥❣✉❧♦ [OPQ]✳
✼✳✾ ✲ ❈♦♥s✐❞❡r❡ ♦ r❡❢❡r❡♥❝✐❛❧ ♦rt♦♥♦r♠❛❞♦ ❡ ❞✐r❡❝t♦ (O; e1, e2, e3)✳
✭❛✮ ❉❡t❡r♠✐♥❡ λ ❞❡ ♠♦❞♦ q✉❡ s❡❥❛♠ ♣❡r♣❡♥❞✐❝✉❧❛r❡s ♦s ✈❡❝t♦r❡s u(2,−1, 15) ❡ v(λ,−3, λ2)✳
✷✶
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
https://www.studocu.com/pt?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=exercicios-alga-17-18
✭❜✮ ❈♦♥s✐❞❡r❡ ♦s ✈❡❝t♦r❡s a(3,−1, 0) ❡ b(1, 1, 1)✳
✐✳ ❉❡t❡r♠✐♥❡ ❛ ❛♠♣❧✐t✉❞❡ ❞♦ â♥❣✉❧♦ ❢♦r♠❛❞♦ ♣❡❧♦s ✈❡❝t♦r❡s a ❡ b✳
✐✐✳ ❉❡t❡r♠✐♥❡ ❛ ❡①♣r❡ssã♦ ❣❡r❛❧ ❞♦s ✈❡❝t♦r❡s ♣❡r♣❡♥❞✐❝✉❧❛r❡s ❛♦ ✈❡❝t♦r b✳
✐✐✐✳ ❉❡t❡r♠✐♥❡ ✉♠ ✈❡❝t♦r ✉♥✐tár✐♦ ♣❡r♣❡♥❞✐❝✉❧❛r ❛♦s ✈❡❝t♦r❡s a ❡ b✳
✭❝✮ ❉❛❞♦s ♦s ♣♦♥t♦s A(6, 0, 0)✱ ❇(2, 3, 0) ❡ C(1, 5, k)✱ ❞❡t❡r♠✐♥❡ k ❞❡ ♠♦❞♦ ❛ q✉❡ ♦s ✈❡❝t♦r❡s−−→
AB ❡
−→
AC ❢♦r♠❡♠ ✉♠ â♥❣✉❧♦ ❞❡ 60o✳
✼✳✶✵ ✲ ❈♦♥s✐❞❡r❡ ♦ r❡❢❡r❡♥❝✐❛❧ ♦rt♦♥♦r♠❛❞♦ ❡ ❞✐r❡❝t♦ (O; e1, e2, e3) ❡ ♦s ♣♦♥t♦s
A(0, k,−3), B(−1, 2, 2) ❡ C(2,−3, 0).
❉❡t❡r♠✐♥❡ k ❞❡ ♠♦❞♦ q✉❡✿
✭❛✮ ❖ ✈♦❧✉♠❡ ❞♦ ♣❛r❛❧❡❧❡♣í♣❡❞♦ ❞❡ ❛r❡st❛s [OA], [OB] ❡ [OC] s❡❥❛ ✐❣✉❛❧ ❛ 10✳
✭❜✮ ❖s ♣♦♥t♦s O,A,B ❡ C s❡❥❛♠ ❝♦♠♣❧❛♥❛r❡s✳
✽ ✲ ❘❡❝t❛ ❡ P❧❛♥♦
◆♦t❛✿ ◆♦s ❡①❡r❝í❝✐♦s q✉❡ s❡ s❡❣✉❡♠✱ ❝♦♥s✐❞❡r❛♠♦s s❡♠♣r❡ ♦ ❡s♣❛ç♦ R3 ❡ ✉♠ r❡❢❡r❡♥❝✐❛❧ ♦rt♦♥♦r♠❛❞♦ ❡
❞✐r❡❝t♦ (O; e1, e2, e3)✳
✽✳✶ ✲ ❉❡t❡r♠✐♥❡ ✉♠❛ ❡q✉❛çã♦ ✈❡❝t♦r✐❛❧ ❡ ✉♠ s✐st❡♠❛ ❞❡ ❡q✉❛çõ❡s ❝❛rt❡s✐❛♥❛s ♣❛r❛ ❝❛❞❛ ✉♠❛ ❞❛s
s❡❣✉✐♥t❡s r❡❝t❛s✿
✭❛✮ ❘❡❝t❛ q✉❡ ♣❛ss❛ ♣❡❧♦ ♣♦♥t♦ A(3, 2,−1) ❡ t❡♠ ✈❡❝t♦r ❞✐r❡❝t♦r u(−2, 2, 3)✳
✭❜✮ ❘❡❝t❛ q✉❡ ♣❛ss❛ ♣❡❧♦s ♣♦♥t♦s A(3, 2,−1) ❡ B(2, 1,−1)✳
✭❝✮ ❘❡❝t❛ q✉❡ ♣❛ss❛ ♣❡❧♦ ♣♦♥t♦ C(1, 0, 2) ❡ é ♦rt♦❣♦♥❛❧ ❛♦ ♣❧❛♥♦ ❞❡ ❡q✉❛çã♦ ❣❡r❛❧ 2x−y+z = 0✳
✭❞✮ ❘❡❝t❛ q✉❡ ♣❛ss❛ ♣❡❧♦ ♣♦♥t♦ C(1, 0, 2) ❡ t❡♠ ✈❡❝t♦r ❞✐r❡❝t♦r u = (2e1 − 3e2)× (e1+ e2 + e3)✳
✭❡✮ ❘❡❝t❛ q✉❡ ♣❛ss❛ ♣❡❧♦ ♣♦♥t♦ O ❡ t❡♠ ✈❡❝t♦r ❞✐r❡❝t♦r e1 ✭❡st❛ r❡❝t❛ é ❤❛❜✐t✉❛❧♠❡♥t❡ ❞❡s✐❣♥❛❞❛
♣♦r ❡✐①♦ ❞♦s xx✮✳
✭❢✮ ❘❡❝t❛ q✉❡ ♣❛ss❛ ♣❡❧♦ ♣♦♥t♦ O ❡ t❡♠ ✈❡❝t♦r ❞✐r❡❝t♦r e2 ✭❡st❛ r❡❝t❛ é ❤❛❜✐t✉❛❧♠❡♥t❡ ❞❡s✐❣♥❛❞❛
♣♦r ❡✐①♦ ❞♦s yy✮✳
✭❣✮ ❘❡❝t❛ q✉❡ ♣❛ss❛ ♣❡❧♦ ♣♦♥t♦ O ❡ t❡♠ ✈❡❝t♦r ❞✐r❡❝t♦r e3 ✭❡st❛ r❡❝t❛ é ❤❛❜✐t✉❛❧♠❡♥t❡ ❞❡s✐❣♥❛❞❛
♣♦r ❡✐①♦ ❞♦s zz✮✳
➚s r❡❝t❛s ❝♦♥s✐❞❡r❛❞❛s ❡♠ ✭❡✮✱ ✭❢✮ ❡ ✭❣✮ ❞❛♠♦s ♦ ♥♦♠❡ ❞❡ ❡✐①♦s ❝♦♦r❞❡♥❛❞♦s ✳
✽✳✷ ✲ ❉❡t❡r♠✐♥❡ ✉♠❛ ❡q✉❛çã♦ ✈❡❝t♦r✐❛❧ ❞❛ r❡❝t❛ r❡♣r❡s❡♥t❛❞❛ ❡♠ ❝❛❞❛ ✉♠❛ ❞❛s ❛❧í♥❡❛s s❡❣✉✐♥t❡s✿
✭❛✮
{
x = 0
y = 3z
❀
✭❜✮
{
2x+ y + z = 1
x− y − z − 2 = 0 ❀
✭❝✮
{
y = 2
z = −4 ❀
✭❞✮ x−1
3
=
y+2
2
=
z
6
.
✽✳✸ ✲ P❛r❛ ❝❛❞❛ ✉♠ ❞♦s s❡❣✉✐♥t❡s ♣❧❛♥♦s ❞❡t❡r♠✐♥❡ ✉♠❛ ❡q✉❛çã♦ ✈❡❝t♦r✐❛❧ ❡ ✉♠❛ ❡q✉❛çã♦ ❣❡r❛❧✳
✷✷
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
✭❛✮ P❧❛♥♦ q✉❡ ♣❛ss❛ ♣❡❧♦ ♣♦♥t♦ A(1, 0, 2) ❡ é ♣❛r❛❧❡❧♦ ❛♦s ✈❡❝t♦r❡s u(−2,−1, 0) ❡ v(3, 0, 2)✳
✭❜✮ P❧❛♥♦ q✉❡ ♣❛ss❛ ♣❡❧♦s ♣♦♥t♦s A(2,−1, 4)✱ B(0, 0, 1) ❡ C(0, 3,−5)✳
✭❝✮ P❧❛♥♦ q✉❡ ♣❛ss❛ ♣❡❧♦ ♣♦♥t♦ ❞❡ ❝♦♦r❞❡♥❛❞❛s (3, 0, 0) ❡ é ♦rt♦❣♦♥❛❧ ❛♦ ✈❡❝t♦r v(1, 2, 3)✳
✭❞✮ P❧❛♥♦ q✉❡ ♣❛ss❛ ♣❡❧♦ ♣♦♥t♦ O ❡ é ♣❛r❛❧❡❧♦ ❛♦s ✈❡❝t♦r❡s e1 ❡ e2 ✭❡st❡ ♣❧❛♥♦ é ❤❛❜✐t✉❛❧♠❡♥t❡
❞❡s✐❣♥❛❞♦ ♣♦r ♣❧❛♥♦ xOy✮✳
✭❡✮ P❧❛♥♦ q✉❡ ♣❛ss❛ ♣❡❧♦ ♣♦♥t♦ O ❡ é ♣❛r❛❧❡❧♦ ❛♦s ✈❡❝t♦r❡s e1 ❡ e3 ✭❡st❡ ♣❧❛♥♦ é ❤❛❜✐t✉❛❧♠❡♥t❡
❞❡s✐❣♥❛❞♦ ♣♦r ♣❧❛♥♦ xOz✮✳
✭❢✮ P❧❛♥♦ q✉❡ ♣❛ss❛ ♣❡❧♦ ♣♦♥t♦ O ❡ é ♣❛r❛❧❡❧♦ ❛♦s ✈❡❝t♦r❡s e2 ❡ e3 ✭❡st❡ ♣❧❛♥♦ é ❤❛❜✐t✉❛❧♠❡♥t❡
❞❡s✐❣♥❛❞♦ ♣♦r ♣❧❛♥♦ yOz✮✳
❆♦s ♣❧❛♥♦s ❞❛s ❛❧í♥❡❛s ✭❞✮✱ ✭❡✮ ❡ ✭❢✮ ❞❛♠♦s ♦ ♥♦♠❡ ❞❡ ♣❧❛♥♦s ❝♦♦r❞❡♥❛❞♦s ✳
✽✳✹ ✲ ❊s❝r❡✈❛ ✉♠❛ ❡q✉❛çã♦ ✈❡❝t♦r✐❛❧ ❞❡ ❝❛❞❛ ✉♠ ❞♦s ♣❧❛♥♦s ❞❡ ❡q✉❛çã♦ ❣❡r❛❧✿
✭❛✮ x+ 5z = 0❀
✭❜✮ z = x− 2y❀
✭❝✮ x+ 3 = 0❀
✭❞✮ 3x− 2y + 4z − 6 = 0.
✽✳✺ ✲ ❊s❝r❡✈❛ ✉♠❛ ❡q✉❛çã♦ ❣❡r❛❧ ❞♦ ♣❧❛♥♦ ❞❡✜♥✐❞♦ ♣❡❧♦ ♣♦♥t♦ A ❡ ♣❡❧♦s ✈❡❝t♦r❡s u ❡ v✱ ❝♦♠✿
✭❛✮ A(0, 1, 2);u(2, 0,−1) ❡ v(0,−1, 3)❀
✭❜✮ A(2, 0, 0);u(5, 1,−1) ❡ v(−10, 1, 2)✳
✽✳✻ ✲ ❉❡t❡r♠✐♥❡ ♦s ♣♦♥t♦s ❞❡ ✐♥t❡rs❡❝çã♦❞♦ ♣❧❛♥♦ ❞❡ ❡q✉❛çã♦ 3x − 2y + 5z − 6 = 0 ❝♦♠ ♦s ❡✐①♦s
❝♦♦r❞❡♥❛❞♦s✳
✽✳✼ ✲ ❈♦♥s✐❞❡r❡ ❛ r❡❝t❛ r ❞❡✜♥✐❞❛ ♣❡❧♦ s✐st❡♠❛ ❞❡ ❡q✉❛çõ❡s ❝❛rt❡s✐❛♥❛s ✿
{
x− 3 = z
y = z − 1 .
✭❛✮ ❉❡t❡r♠✐♥❡ ✉♠❛ r❡♣r❡s❡♥t❛çã♦ ❞❛ r❡❝t❛ s ♣❛r❛❧❡❧❛ à r❡❝t❛ ❞❛❞❛ ❡ q✉❡ ♣❛ss❛ ♣❡❧♦ ♣♦♥t♦ ❞❡
❝♦♦r❞❡♥❛❞❛s (1,−2, 3)✳
✭❜✮ ❊s❝r❡✈❛ ✉♠❛ ❡q✉❛çã♦ ❣❡r❛❧ ❞♦ ♣❧❛♥♦ q✉❡ ❝♦♥té♠ ❛s r❡❝t❛s r ❡ s✳
✭❝✮ ❉❡t❡r♠✐♥❡ ✉♠ s✐st❡♠❛ ❞❡ ❡q✉❛çõ❡s ❝❛rt❡s✐❛♥❛s ❞❛ r❡❝t❛ q✉❡ ♣❛ss❛ ♣♦r A(1,−2, 3) ❡ é ♦rt♦✲
❣♦♥❛❧ ❛♦ ♣❧❛♥♦ ❞❡ ❡q✉❛çã♦ 3x− z = y − 2✳
✭❞✮ ❉❡t❡r♠✐♥❡ ✉♠❛ ❡q✉❛çã♦ ✈❡❝t♦r✐❛❧ ❡ ✉♠ s✐st❡♠❛ ❞❡ ❡q✉❛çõ❡s ❝❛rt❡s✐❛♥❛s ❞❛ r❡❝t❛ q✉❡ ♣❛ss❛
♣❡❧♦ ♣♦♥t♦ C(1, 0, 2) ❡ é ♣❛r❛❧❡❧❛ ❛♦s ♣❧❛♥♦s ❞❡ ❡q✉❛çõ❡s 2x− y = 5 ❡ x+ y + z = 4✳
✽✳✽ ✲ ❱❡r✐✜q✉❡ s❡ ❛s r❡❝t❛s✱ ❝♦♥s✐❞❡r❛❞❛s ❡♠ ❝❛❞❛ ✉♠❛ ❞❛s s❡❣✉✐♥t❡s ❛❧í♥❡❛s✱ sã♦ ♣❛r❛❧❡❧❛s✳
✭❛✮ r : (x, y, z) = (2, 2, 2) + λ(1, 10, 1), λ ∈ R
s : (x, y, z) = (1, 2, 1) + µ(0, 1, 2), µ ∈ R✳
✭❜✮ r : (x, y, z) = (1, 2, 3) + λ(1, 0, 1), λ ∈ R
s : (x, y, z) = (1, 2, 1) + µ(0, 1, 2), µ ∈ R✳
✽✳✾ ✲ ❙❡❥❛♠ r ❡ s r❡❝t❛s r❡♣r❡s❡♥t❛❞❛s✱ r❡s♣❡❝t✐✈❛♠❡♥t❡✱ ♣❡❧♦s s❡❣✉✐♥t❡s s✐st❡♠❛s ❞❡ ❡q✉❛çõ❡s ❝❛rt❡s✐✲
❛♥❛s✿ {
x = 2z + 1
y = z + 2
❡
{
x = 1
y = z + 5
.
❉❡t❡r♠✐♥❡ ✉♠ s✐st❡♠❛ ❞❡ ❡q✉❛çõ❡s ❝❛rt❡s✐❛♥❛s ❞❡ ✉♠❛ r❡❝t❛ ♦rt♦❣♦♥❛❧ ❛ r ❡ ❛ s✳
✷✸
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✽✳✶✵ ✲ P❛r❛ ❝❛❞❛ k ∈ R ❡ ♣❛r❛ ❝❛❞❛ m ∈ R✱ ❝♦♥s✐❞❡r❡ ♦s ♣❧❛♥♦s ❞❡ ❡q✉❛çõ❡s ❣❡r❛✐s✿
2x+ ky + 3z − 5 = 0 ❡ mx− 6y − 6z + 2 = 0.
❉❡t❡r♠✐♥❡ ♦s ✈❛❧♦r❡s ❞❡ k ❡ ❞❡ m ♣❛r❛ ♦s q✉❛✐s ♦s ♣❧❛♥♦s sã♦ ♣❛r❛❧❡❧♦s✳
✽✳✶✶ ✲
✭❛✮ ❉❡t❡r♠✐♥❡ ✉♠❛ ❡q✉❛çã♦ ❣❡r❛❧ ❞♦ ♣❧❛♥♦ q✉❡ ♣❛ss❛ ♣❡❧♦ ♣♦♥t♦ (0,−8,−2) ❡ é ♦rt♦❣♦♥❛❧ à
r❡❝t❛ ❞❡ ❡q✉❛çã♦ ✈❡❝t♦r✐❛❧
(x, y, z) = (0, 7, 1) + λ(0,−1, 1), λ ∈ R.
✭❜✮ ❉❡t❡r♠✐♥❡ ♦ â♥❣✉❧♦ ❞♦ ♣❧❛♥♦ ❞❛ ❛❧í♥❡❛ ❛♥t❡r✐♦r ❝♦♠ ✉♠❛ r❡❝t❛ ♣❛r❛❧❡❧❛ à r❡❝t❛ ❞❡ ❡q✉❛çã♦
✈❡❝t♦r✐❛❧
(x, y, z) = (3, 1, 2) + µ(1, 1, 0), µ ∈ R.
✽✳✶✷ ✲ ❉❡t❡r♠✐♥❡ ♦ â♥❣✉❧♦ ❢♦r♠❛❞♦ ♣❡❧♦s ♣❧❛♥♦s ❞❡ ❡q✉❛çõ❡s✿
✭❛✮ x+ y = 0 ❡ y + z = 0❀
✭❜✮ (x, y, z) = λ(−3, 3, 0) + µ(1, 4,−10), λ, µ ∈ R ❡ 2x− 2y − z = 0✳
✽✳✶✸ ✲ ❊♠ ❝❛❞❛ ✉♠❛ ❞❛s ❛❧í♥❡❛s s❡❣✉✐♥t❡s✱ ❞❡t❡r♠✐♥❡ ❛ ❞✐stâ♥❝✐❛ ❞♦ ♣♦♥t♦ A ❛♦ ♣❧❛♥♦ π✳
✭❛✮ A(−2,−4, 3) ❡ π : 2x− y + 2z = 0❀
✭❜✮ A(3,−6, 7) ❡ π : 4x− 3z − 1 = 0✳
✽✳✶✹ ✲ ❉❡t❡r♠✐♥❡ ❛ ❞✐stâ♥❝✐❛ ❞♦ ♣♦♥t♦ A(1, 2, 0) á r❡❝t❛ ❞❡ ❡q✉❛çã♦ ✈❡❝t♦r✐❛❧✿
(x, y, z) = (1, 1, 1) + λ(1, 0, 1), λ ∈ R
✽✳✶✺ ✲ ❈♦♥s✐❞❡r❡ ❛s r❡❝t❛s
r : (x, y, z) = (1, 2, 0) + λ(1, 1, 0), λ ∈ R ❡ s : (x, y, z) = (0, 0, 2) + µ(0, 2, 0), µ ∈ R.
❉❡t❡r♠✐♥❡✿
✭❛✮ ❆ ♣♦s✐çã♦ r❡❧❛t✐✈❛ ❞❛s ❞✉❛s r❡❝t❛s✳
✭❜✮ ❯♠❛ r❡❝t❛ ♦rt♦❣♦♥❛❧ ❛ r ❡ ❛ s ❡ q✉❡ ❛s ✐♥t❡rs❡❝t❡✳
✭❝✮ ❆ ❞✐stâ♥❝✐❛ ❡♥tr❡ r ❡ s✳
✽✳✶✻ ✲ ❈♦♥s✐❞❡r❡ ❛s r❡❝t❛s r ❡ s ❞❡✜♥✐❞❛s P♦r
r : (x, y, z) = (1, 2, 3) + λ(4, 0,−3), λ ∈ R ❡ s : x− 2
3
=
y − 2
3
=
z + 4
4
.
✭❛✮ ❱❡r✐✜q✉❡ q✉❡ ❛s r❡❝t❛s ♥ã♦ s❡ ✐♥t❡rs❡❝t❛♠ ❡ q✉❡ ♥ã♦ sã♦ ♣❛r❛❧❡❧❛s✳
✭❜✮ ❊s❝r❡✈❛ ✉♠❛ ❡q✉❛çã♦ ❞♦ ♣❧❛♥♦ q✉❡ ❝♦♥té♠ r ❡ é ♣❛r❛❧❡❧♦ ❛ s✳
✭❝✮ ❉❡t❡r♠✐♥❡ ❛ ❞✐stâ♥❝✐❛ ❡♥tr❡ r ❡ s✳
✽✳✶✼ ✲ ❙❡❥❛♠✱ r ❛ r❡❝t❛ ❞❡✜♥✐❞❛ ♣♦r✿ {
x−1
2 = 1
z = 3
❡ π ♦ ♣❧❛♥♦ ❞❡✜♥✐❞♦ ♣♦r 


x = 1 + 2λ+ µ
y = λ
z = 3 + µ
, λ, µ ∈ R.
❉❡t❡r♠✐♥❡✿
✷✹
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lOMoARcPSD|6016133
✭❛✮ ❯♠❛ ❡q✉❛çã♦ ✈❡❝t♦r✐❛❧ ❞❛ r❡❝t❛ r✳
✭❜✮ ❯♠❛ ❡q✉❛çã♦ ❣❡r❛❧ ❞♦ ♣❧❛♥♦ π✳
✭❝✮ ❆ ❞✐stâ♥❝✐❛ ❞❡ r ❛ π✳
✭❞✮ ❯♠ ♣❧❛♥♦ q✉❡ ❝♦♥t❡♥❤❛ ❛ r❡❝t❛ r ❡ s❡❥❛ ♦rt♦❣♦♥❛❧ ❛ π✳
✽✳✶✽ ✲ ❈♦♥s✐❞❡r❡ ♦s ♣❧❛♥♦s✿
π1 : −2x+ 4y − 2z + 3 = 0; π2 : x+ 2z − 1 = 0; π3 : 2x+ 4y + 6z + 2 = 0.
❉❡t❡r♠✐♥❡✿
✭❛✮ ❯♠❛ ❡q✉❛çã♦ ✈❡❝t♦r✐❛❧ ❞♦ ♣❧❛♥♦ π1✳
✭❜✮ ❖ â♥❣✉❧♦ ❢♦r♠❛❞♦ ♣❡❧♦s ♣❧❛♥♦s π1 ❡ π2✳
✭❝✮ ❯♠❛ ❡q✉❛çã♦ ✈❡❝t♦r✐❛❧ ❞❛ r❡❝t❛ r q✉❡ é ❛ ✐♥t❡rs❡❝çã♦ ❞♦s ♣❧❛♥♦s π1 ❡ π2✳
✭❞✮ ❆ ❞✐stâ♥❝✐❛ ❞❛ r❡❝t❛ r ❛♦ ♣❧❛♥♦ π3✳
❙♦❧✉çõ❡s
✶ ✲ ▼❛tr✐③❡s
✶✳✶ ✲ ✭❛✮ B✱ E✱ F ✱ H✱ I
✭❜✮ B✱ E✱ F ✱ H✱ I
✭❝✮ B✱ E✱ F ✱ I
✭❞✮ E✱ F ✱ I
✶✳✸ ✲ ✭❛✮
[
4 1 5
−2 1 −1
]
✭❜✮
[
8 2 10
−4 2 −2
]
✭❝✮
[
2 1 −4
2 −1 0
]
✭❞✮
[
3 2 −15
11 2 −2
]
✶✳✹ ✲ X =


3 2 2
2 3 2
2 2 3


✶✳✺ ✲ AB = [−1 ]✱ BA =
[
0 0 0
1 2 −1
3 6 −3
]
✶✳✼ ✲ ✭❛✮ [ 2 5 ]
✭❜✮ ◆ã♦ ❡stá ❞❡✜♥✐❞♦
✭❝✮
[−2 2 1
1 −1 0
−2 2 2
]
✭❞✮
[
1 2
1 −2
]
✶✳✶✺ ✲ ❙❡ D = diag (d1, . . . , dn) ❡♥tã♦
Dk = diag
(
d1
k, . . . , dn
k
)
✶✳✷✻ ✲ ✭❛✮
[
9 5
12 4
8 11
]
✭❜✮ C = I3 + 2A−1 =
[
3 2 4
0 3 6
8 4 3
]
✶✳✸✹ ✲ ✭❛✮ A✱ C
✭❜✮ A✱ E
✶✳✹✵ ✲ ✭❛✮ A✱ E
✭❜✮ B
✶✳✹✷ ✲ ✭❛✮ ❙✐♠✱ ❞♦ t✐♣♦ ■■■
✭❜✮ ❙✐♠✱ ❞♦ t✐♣♦ ■■
✭❝✮ ◆ã♦
✭❞✮ ◆ã♦
✭❡✮ ❙✐♠✱ ❞♦ t✐♣♦ ■■ ♦✉ ❞♦ t✐♣♦ ■■■
✶✳✹✸ ✲ ✭❛✮
[
0 0 1
0 1 0
1 0 0
]
✭❜✮
[
6 0 0
0 1 0
0 0 1
]
✭❝✮
[
1 0 0
0 1 0
0 1
5
1
]
✶✳✹✹ ✲ ✭❛✮
[
e f g h
a b c d
i j k l
]
✭❜✮
[
5e 5f 5g 5h
a b c d
i j k l
]
✭❝✮
[
a b c d+3c
e f g h+3g
i j k l+3k
]
✭❞✮
[
2a 2b 2c−10b
d e f−5e
]
✶✳✹✻ ✲ ✭❛✮
[
1 0 0
0 1
5
0
0 0 1
]
✭❜✮
[
0 0 1
0 1 0
1 0 0
]
✷✺
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✭❝✮
[
1 0 0
0 1 0
3 0 1
]
✶✳✹✽ ✲ ✭❛✮ ❙✐♠
✭❜✮ ◆ã♦
✭❝✮ ❙✐♠
✭❞✮ ◆ã♦
✶✳✹✾ ✲ ✭❛✮ P♦r ❡①❡♠♣❧♦✱[
1 2 1
0 1 1
0 0 1
]
✭❜✮ P♦r ❡①❡♠♣❧♦✱[
2 4 −2 6 0
0 0 0 −5 5
0 0 0 0 0
]
✭❝✮ P♦r ❡①❡♠♣❧♦✱[
1 1 2
0 0 5
0 0 0
]
✶✳✺✶ ✲ ✭❛✮ ❙✐♠
✭❜✮ ❙✐♠
✭❝✮ ◆ã♦
✭❞✮ ❙✐♠
✭❡✮ ❙✐♠
✶✳✺✷ ✲ ✭❛✮
[
1 0 0
0 1 0
0 0 1
]
✭❜✮
[
1 2 −1 0 3
0 0 0 1 −1
0 0 0 0 0
]
✭❝✮
[
1 1 0
0 0 1
0 0 0
]
✶✳✺✺ ✲ ✭❛✮ ◆ã♦
✭❜✮ ❙✐♠
✶✳✺✻ ✲ ✭❛✮ P♦r ❡①❡♠♣❧♦✱
(12 l1, l2 + (−1)l1, l1 + 2l2, 4l2,
l2 + (−1)l1)
✭❜✮ P♦r ❡①❡♠♣❧♦✱
(l2 + l1,
1
4 l2, l1 + (−2)l2, l2 + l1, 2l1)
✶✳✺✼ ✲ r(A1) = 3✱ r(A2) = 3✱
r(A3) = 2✱ r(A4) = 3
✶✳✺✽ ✲ r(Aα) =
{
2, s❡ α = 2
3, s❡ α 6= 2
r(Bα) =
{
3, s❡ α = 2
4, s❡ α 6= 2
r(Cα,β) =
{
2, s❡ α = 0 ♦✉ β = 0
3, s❡ α 6= 0 ❡ β 6= 0
r(Dα,β) =
{
3, s❡ β = 0 ❡ α ∈ R
4, s❡ β 6= 0 ❡ α ∈ R
✶✳✺✾ ✲ r ([ 1 24 8 ]) = 1 ❡ r ([
0 1
1 2 ]) = 2
✶✳✻✷ ✲ ✭❛✮ α ∈ R \ {−5}
✭❜✮ α ∈ R \ {−1} ❡ β ∈ R \ {2}
✶✳✻✺ ✲ ✭❛✮
[
0 1
2
−1 1
2
]
✭❜✮ P♦r ❡①❡♠♣❧♦✱
A−1 = [ 1 10 1 ]
[
1 0
0 1
2
] [
1 0
−2 1
]
A = [ 1 02 1 ] [
1 0
0 2 ]
[
1 −1
0 1
]
✶✳✻✻ ✲ B−1 = 15
[
3 1 2
−2 1 2
−2 1 −3
]
C−1 =
[−i −1+i
1 −i
]
D−1 = 12
[ 0 2 −4 −1
1 1 −5 −1
1 1 −3 0
1 −1 1 0
]
✷ ✲ ❙✐st❡♠❛s
❆❜r❡✈✐❛t✉r❛s ✉t✐❧✐③❛❞❛s✿
❙✳P✳❉✳✕ ❙✐st❡♠❛ P♦ssí✈❡❧ ❉❡t❡r♠✐♥❛❞♦
❙✳■✳✕ ❙✐st❡♠❛ ■♠♣♦ssí✈❡❧
❙✳P✳■✳✕ ❙✐st❡♠❛ P♦ssí✈❡❧ ■♥❞❡t❡r♠✐♥❛❞♦
❣✳✐✳ ✕ ❣r❛✉ ❞❡ ✐♥❞❡t❡r♠✐♥❛çã♦
✷✳✸ ✲ ❇❛st❛ t♦♠❛r ❛ ♠❛tr✐③
B = A
[
1
2
3
]
=
[−2
19
5
17
]
r❡s✉❧t❛♥❞♦ ♦ s✐st❡♠❛✱ ♥❛s ✐♥❝ó❣♥✐t❛s x✱ y✱ z✱
s♦❜r❡ R✱


x− z = −2
2x+ 4y + 3z = 19
−x+ 2z = 5
3x+ 4y + 2z = 17
✷✳✼ ✲ (S1) ❙✳P✳❉✳
❈♦♥❥✉♥t♦ ❞❛s s♦❧✉çõ❡s ❞❡ (S1)✿
{(1,−1, 0)}
(S2) ❙✳P✳■✳ ❝♦♠ ❣✳✐✳ ✶
❈♦♥❥✉♥t♦ ❞❛s s♦❧✉çõ❡s ❞❡ (S2)✿{(
1
7 +
1
7α,
5
7 − 27α, α
)
: α ∈ R
}
(S3) ❙✳■✳
✷✳✽ ✲ ✭❛✮ ❙✳P✳❉✳
✭❜✮ ❙✳■✳
✭❝✮ ❙✳P✳■✳ ❝♦♠ ❣✳✐✳ ✷
✭❞✮ ❙✳P✳■✳ ❝♦♠ ❣✳✐✳ ✹
✭❡✮ ❙✳■✳
✷✻
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✭❢✮ ❙✳P✳❉✳
✭❣✮ ❙✳P✳■✳ ❝♦♠ ❣✳✐✳ ✹
✷✳✾ ✲ ✭❛✮ P♦r ❡①❡♠♣❧♦✱
{
x− 2z = 1
y − z = −4
✭❜✮ P♦r ❡①❡♠♣❧♦✱
{
x− y + 2z = 0
2x− 2y + 4z = 0
❙✐♠✱ ❜❛st❛ t♦♠❛r ♦ s✐st❡♠❛
0x+ 0y + 0z = 0
✷✳✶✶ ✲ ✭❛✮ C = R \ {3}
✭❜✮ C = ∅
✭❝✮ C = {3}
✷✳✶✺ ✲ ❈♦♥❥✉♥t♦ ❞❛s s♦❧✉çõ❡s✿{(
−14α+ 38β+ 12γ, 12β+γ,−14α− 18β+ 12γ
)
:
α, β, γ ∈ R}
✷✳✷✹ ✲ (S1) ❙✳P✳❉✳
❈♦♥❥✉♥t♦ ❞❛s s♦❧✉çõ❡s ❞❡ (S1)✿{(
3
5 ,−75 ,−45
)}
(S2) ❙✳P✳■✳ ❝♦♠ ❣✳✐✳ ✶
❈♦♥❥✉♥t♦ ❞❛s s♦❧✉çõ❡s ❞❡ (S2)✿
{(−1 + 2α,−α, α) : α ∈ R}
(S3) ❙✳■✳
(S4) ❙✳P✳■✳ ❝♦♠ ❣✳✐✳ ✶
❈♦♥❥✉♥t♦ ❞❛s s♦❧✉çõ❡s ❞❡ (S4)✿{(
−25 − 35α, 15 − 15α, α
)
: α ∈ R
}
(S5) ❙✳P✳■✳ ❝♦♠ ❣✳✐✳ ✶
❈♦♥❥✉♥t♦ ❞❛s s♦❧✉çõ❡s ❞❡ (S5)✿
{(1, 0, α) : α ∈ R}
(S6) ❙✳■✳
(S7) ❙✳P✳■✳ ❝♦♠ ❣✳✐✳ ✶
❈♦♥❥✉♥t♦ ❞❛s s♦❧✉çõ❡s ❞❡ (S7)✿
{(1− α,−1 + 2α, α) : α ∈ R}
✷✳✸✺ ✲ ✭❛✮ ❙❡ α 6= 0 ❡ α 6= 1 ❡ β ∈ R✱ ❙✳P✳❉✳
❙❡ α = 1 ❡ β ∈ R✱ ❙✳P✳■✳ ❝♦♠ ❣✳✐✳ 1
❙❡ α = 0 ❡ β 6= 1✱ ❙✳■✳
❙❡ α = 0 ❡ β = 1✱ ❙✳P✳■✳ ❝♦♠ ❣✳✐✳ 1
✭❜✮ ❈♦♥❥✉♥t♦ ❞❛s s♦❧✉çõ❡s✿
{(1 + γ, 0, γ) : γ ∈ R}
✷✳✸✼ ✲ ✭❛✮ ❙❡ α 6= 0 ❡ β 6= 1✱ ❙✳P✳❉✳
❙❡ α = 0 ❡ β = 1✱ ❙✳P✳■✳ ❝♦♠ ❣✳✐✳ 2
❙❡ α = 0 ❡ β 6= 1✱ ❙✳P✳■✳ ❝♦♠ ❣✳✐✳ 1
❙❡ α 6= 0 ❡ β = 1✱ ❙✳■✳
✭❜✮ ✭✐✐✐✮ (5, 1,−3)
✸ ✲ ❉❡t❡r♠✐♥❛♥t❡s
✸✳✶ ✲ ✭❛✮ 1
✭❜✮ −3
✭❝✮ 0
✸✳✷ ✲ 2
✸✳✸ ✲ ✭❛✮ 0
✭❜✮ −7
✭❝✮ −1
✸✳✹ ✲ ✭❛✮ −1
✭❜✮ −4
✭❝✮ 3
✸✳✻ ✲ {0, 1, 3}
✸✳✶✵ ✲ ✭❛✮ γ
✭❜✮ −12γ
✭❝✮ γ
✭❞✮−3γ
✭❡✮ −γ
✸✳✶✺ ✲ k ∈ {−2, 1}
✸✳✶✾ ✲ t ∈ R \ {0, 2}
✸✳✷✵ ✲ |AB⊤C| = −40
|3B| = 3n(−5)
|B2C| = (−5)2 · 4
✸✳✷✺ ✲ ✭❛✮ |A| = −32
|B| = 0
✭❜✮ |A−1| = − 132
✭❝✮ ✭✐✮ ❙✐♠
✭✐✐✮ ◆ã♦
✸✳✷✽ ✲ ✭❛✮ A−1 =
[
1 1 − 3
2
0 1 − 1
2
−1 −2 5
2
]
✭❜✮ V −1α = [
cosα senα
− senα cosα ] = V−α
✭❝✮ A−1 = 1|z|2+|w|2
[
z −w
w z
]
✸✳✸✷ ✭❛✮ |A| = −3
✭❜✮ (1, 2, 3)
✸✳✸✸ ✭❛✮ k ∈ R \ {0,−3}
✭❜✮ (1,−12 , 12)
✷✼
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
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✹ ✲ ❊s♣❛ç♦s ❱❡❝t♦r✐❛✐s
✹✳✻ ✲ ✭❛✮ (0, 0, 0, 0)
✭❜✮ [ 0 0 00 0 0 ]
✭❝✮ 0x3 + 0x2 + 0x+ 0
✹✳✶✸ ✲ ✭❛✮ ◆ã♦
✭❜✮ ◆ã♦
✭❝✮ ❙✐♠
✭❞✮ ❙✐♠
✹✳✷✷ ✲ ✭❛✮ P♦r ❡①❡♠♣❧♦✱ G =
〈
[ 1 10 0 ] ,
[
0 1
−1 0
]〉
✹✳✷✸ ✲ ✭❛✮ P♦r ❡①❡♠♣❧♦✱
(
(1, 0, 1), (0, 1, 0)
)
✭❜✮ P♦r ❡①❡♠♣❧♦✱([−1 0
0 1
]
, [ 0 10 0 ] , [
0 0
1 0 ]
)
✭❝✮ P♦r ❡①❡♠♣❧♦✱(
x2, 2x3 + x,−x3 + 1
)
✹✳✸✸ ✲ P♦r ❡①❡♠♣❧♦✱
(
(2, 3, 3)
)
❡
(
(−4,−6,−6)
)
✹✳✸✺ ✲ P♦r ❡①❡♠♣❧♦✱
(
[ 1 10 0 ] ,
[
0 1
−1 0
])
✹✳✹✶ ✲ ✭❜✮ P♦r ❡①❡♠♣❧♦✱
(
(1, 2, 1), (0, 1, 1)
)
✹✳✹✹ ✲ ✭❛✮ (1, 1, 1, 1)✱ ♥❛ ❜❛s❡ B
(4, 3, 2, 1)✱ ♥❛ ❜❛s❡ b. c.R4
✭❜✮ (a− b, b− c, c− d, d)✱ ♥❛ ❜❛s❡ B
(a, b, c, d)✱ ♥❛ ❜❛s❡ b. c.R4
✹✳✹✽ ✲ P♦r ❡①❡♠♣❧♦✱
(
(2, 1, 2,−1)
)
✹✳✺✷ ✲ ✭❛✮ P♦r ❡①❡♠♣❧♦✱(
(1, 1, 0, 0), (0, 0, 1, 0)
)
✭❜✮ P♦r ❡①❡♠♣❧♦✱(
(1, 0, 0, 0), (0, 1, 1, 0)
)
✭❝✮ P♦r ❡①❡♠♣❧♦✱(
(1, 1, 0, 0), (0, 0, 1, 0), (1, 0, 0, 0)
)
✭❞✮
(
(1, 1, 0, 0), (0, 0, 1, 0), (1, 0, 0, 3),
(2, 0, 0, 1)
)
✹✳✺✻ ✲ ✭❜✮ Pr♦❥❡❝çã♦ ❞❡ A s♦❜r❡ F ✱ s❡❣✉♥❞♦ G✿
[ 4 50 0 ]
Pr♦❥❡❝çã♦ ❞❡ A s♦❜r❡ G✱ s❡❣✉♥❞♦ F ✿
[ 0 00 6 ]
✹✳✻✷ ✲ ✭❛✮ R \ {−2}
✹✳✻✺ ✲ R \ {−4}
✹✳✻✾ ✲ ✭❛✮ P♦r ❡①❡♠♣❧♦✱(
(1, 0, 1, 0), (0, 1, 1, 1)
)
✭❝✮ P♦r ❡①❡♠♣❧♦✱(
(1, 0, 1, 0), (0, 1, 1, 1), (0, 0, 1, 0),
(0, 0, 0, 1)
)
✹✳✼✶ ✲ ✭❛✮ S1 é ❧✐♥❡❛r♠❡♥t❡ ✐♥❞❡♣❡♥❞❡♥t❡
S2 é ❧✐♥❡❛r♠❡♥t❡ ❞❡♣❡♥❞❡♥t❡
✹✳✼✹ ✲ ✭❛✮ dimF = 3
P♦r ❡①❡♠♣❧♦✱(
x3 + x2 − x, x2 − 1, x+ 2
)
✭❜✮ dimG = 3
P♦r ❡①❡♠♣❧♦✱(
[ 1 11 1 ] , [
1 1
1 0 ] ,
[
2 −3
1 1
])
✹✳✶✺✺ ✲ ✭❛✮ P♦r ❡①❡♠♣❧♦✱
(
1 + x, x2, x3
)
✭❜✮ P♦r ❡①❡♠♣❧♦✱(
1, 1 + x, x2, x3, x4
)
✭❝✮ P♦r ❡①❡♠♣❧♦✱ G =
〈
x3, x4
〉
✹✳✶✺✻ ✲ ✭❛✮ t = 1
✭❜✮ P♦r ❡①❡♠♣❧♦✱(
(1, 2, 1, 2), (1, 1, 1, 1),
(1, 2, 3, 4), (1, 1, 0, 1)
)
dim(F ∩Gt) = 1
✺ ✲ ❆♣❧✐❝❛çõ❡s ▲✐♥❡❛r❡s
✺✳✷ ✲ ✭❛✮ ❙✐♠
✭❜✮ ◆ã♦
✭❝✮ ◆ã♦
✭❞✮ ◆ã♦
✺✳✼ ✲ ✭❛✮ ◆ã♦
✭❜✮ ◆ã♦
✺✳✶✵ ✲
✭❛✮ Ker f = {(a, 0, 0) : a ∈ R}
P♦r ❡①❡♠♣❧♦✱
❇❛s❡ ❞❡ Ker f ✿
(
(1, 0, 0)
)
❇❛s❡ ❞❡ Im f ✿
(
(1, 0), (0, 1)
)
✭❜✮ Ker g =
{[
0 b
−d d
]
: b, d ∈ R
}
P♦r ❡①❡♠♣❧♦✱
❇❛s❡ ❞❡ Ker g✿
(
[ 0 10 0 ] ,
[
0 0
−1 1
])
❇❛s❡ ❞❡ Im g✿
(
(2, 0, 0), (0, 1, 0)
)
✭❝✮ Kerh = {(−b, b, 0) : b ∈ R}
✷✽
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
P♦r ❡①❡♠♣❧♦✱
❇❛s❡ ❞❡ Kerh✿
(
(−1, 1, 0)
)
❇❛s❡ ❞❡ Imh✿
(
x2, 1
)
✭❞✮ Ker t ={
ax3 + bx2 + ax− b : a, b ∈ R
}
P♦r ❡①❡♠♣❧♦✱
❇❛s❡ ❞❡ Ker t✿
(
x3 + x, x2 − 1
)
❇❛s❡ ❞❡ Im t✿ ([ 1 00 0 ] , [
0 0
0 1 ])
✺✳✶✹ ✲ ✭❛✮ Ker f = {(0, 0, 0)}
■♥❥❡❝t✐✈❛
✭❜✮ Ker g =
{
0x2 + 0x+ 0
}
■♥❥❡❝t✐✈❛
✺✳✶✼ ✲ ✭❛✮ n(f) = 1
✭❜✮ n(g) = 3
✭❝✮ n(h) = 3
✭❞✮ n(t) = 0
✺✳✶✾ ✲ (2, 3), (3, 2), (4, 1), (5, 0)
✺✳✷✷ ✲ ✭❛✮ ❙✐♠
✭❜✮ ❙✐♠
✺✳✷✻ ✲ ✭❛✮ ◆ã♦
✭❜✮ ❙✐♠✱ ♣♦r ❡①❡♠♣❧♦✱
f(0, 1, 1, 0) = (0, 0, 0)✱
f(1, 1, 1, 1) = (0, 0, 0)✱
f(0, 0, 1, 0) = (1, 1, 1)✱
f(0, 0, 0, 1) = (1, 1, 0)
✭❝✮ ❙✐♠✱ ♣♦r ❡①❡♠♣❧♦✱
f(1, 0, 0) = (1, 2, 0,−4)✱
f(0, 1, 0) = (2, 0,−1,−3)✱
f(0, 0, 1) = (0, 0, 0, 0)
✺✳✸✻ ✲ ✭❛✮
[
− 3
2
− 1
2
− 3
2
−2 2 0
]
✭❜✮
[
2 0 0 0
1 4 1 1
0 0 0 0
]
✺✳✸✾ ✲ r(M(f ;B,B′)) = 2 = dim Im f 6= dimR3
f ♥ã♦ é s♦❜r❡❥❡❝t✐✈❛✳
✺✳✹✷ ✲ ✭❛✮ (3, 5)
✭❜✮ [ 1 1 00 1 1 ]
✭❝✮ [ 1 1 11 1 0 ]
✭❞✮
[
0 1 1
1 0 −1
]
✭❡✮ [ 1 1 00 0 1 ]
✺✳✹✸ ✲ ✭❛✮
[
1 −1 0
−1 1 1
0 −1 0
]
✭❜✮
[
1 0 −1
0 0 −1
1 1 0
]
✭❝✮
[
1 −1 0
−2 2 1
1 −2 −1
]
✺✳✹✻ ✲ ✭❛✮ [ 1 1 11 1 0 ]
✭❜✮
[
0 1 1
1 0 −1
]
✭❝✮ [ 1 1 00 0 1 ]
✺✳✶✶✵ ✲ ✭❛✮
[ 0 −1 −1 1 0
2 1 3 −3 0
0 1 1 −1 0
]
✭❜✮ P♦r ❡①❡♠♣❧♦✱ (u1 − u3, u2)
✭❞✮ P♦r ❡①❡♠♣❧♦✱(
(2, 2, 0, 2, 2), (−1,−1, 1, 0, 1) ,
(0, 0, 0, 0, 1), (0, 1, 0, 0, 0),
(0, 0, 0, 1, 0)
)
✭❡✮
[
0 0 0 −1 1
0 0 0 1 −3
0 0 0 1 −1
]
✻ ✲ ❱❛❧♦r❡s ❡ ❱❡❝t♦r❡s Pró♣r✐♦s
✻✳✶ ✲ u1 ✈❡❝t♦r ♣ró♣r✐♦ ❛ss♦❝✐❛❞♦ ❛♦ ✈❛❧♦r ♣ró♣r✐♦
1
u2 ✈❡❝t♦r ♣ró♣r✐♦ ❛ss♦❝✐❛❞♦ ❛♦ ✈❛❧♦r ♣ró♣r✐♦
2
u3 ♥ã♦ é ✈❡❝t♦r ♣ró♣r✐♦
✻✳✷ ✲ ✭❛✮
[
1
−1
]
✈❡❝t♦r ♣ró♣r✐♦ ❛ss♦❝✐❛❞♦ ❛♦ ✈❛❧♦r
♣ró♣r✐♦ 1
[ 02 ] ✈❡❝t♦r ♣ró♣r✐♦ ❛ss♦❝✐❛❞♦ ❛♦ ✈❛❧♦r
♣ró♣r✐♦ 2
✭❜✮ [ α−α ] ✈❡❝t♦r ♣ró♣r✐♦ ❛ss♦❝✐❛❞♦ ❛♦ ✈❛❧♦r
♣ró♣r✐♦ 1
[ 0α ] ✈❡❝t♦r ♣ró♣r✐♦ ❛ss♦❝✐❛❞♦ ❛♦ ✈❛❧♦r
♣ró♣r✐♦ 2
✻✳✼ ✲ ✭❜✮ P♦r ❡①❡♠♣❧♦✱
[
0 2 1
0 1 2
0 0 1
]
✻✳✶✷ ✲ ❱❛❧♦r❡s ♣ró♣r✐♦s ❞❡ A✿ 3 ❡ 1
ma(3) = 2 ❡ ma(1) = 1
✻✳✶✹ ✲ ✭❛✮ ❱❛❧♦r❡s ♣ró♣r✐♦s ❞❡ A✿ ✸
ma(3) = 1
✻✳✷✽ ✲ ❱❛❧♦r❡s ♣ró♣r✐♦s ❞❡ f ✿ 1 ❡ 2
E1 =
〈
(−1,−1, 2)
〉
E2 =
〈
(−1, 2, 0), (−1, 0, 2)
〉
✻✳✸✺ ✲ DA =
[−2 0
0 2
]
✱ DB = [ 5 00 1 ]
✻✳✸✼ ✲ ✭❛✮ ❱❛❧♦r❡s ♣ró♣r✐♦s ❞❡ A✿ −1 ❡ 1
ma(−1) = 2 ❡ ma(1) = 1
✷✾
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
https://www.studocu.com/pt?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=exercicios-alga-17-18
✭❜✮ P♦r ❡①❡♠♣❧♦✱
❇❛s❡ ❞❡ M−1✿
([
1
−2
0
]
,
[
0
0
1
])
❇❛s❡ ❞❡ M1✿
([
1
−1
1
])
✭❝✮ P♦r ❡①❡♠♣❧♦✱ P =
[
1 0 1
−2 0 −1
0 1 1
]
❡ D =
[−1 0 0
0 −1 0
0 0 1
]
✻✳✹✶✲ ✭❛✮ ❱❛❧♦r❡s ♣ró♣r✐♦s ❞❡ f ✿ −1 ❡ 1
✭❜✮ P♦r ❡①❡♠♣❧♦✱
B = (e1 − 2e2, e3, e1 − e2 + e3)
✭❝✮ M (f ; B,B) =
[−1 0 0
0 −1 0
0 0 1
]
✻✳✽✾ ✲
✭❛✮ ❱❛❧♦r❡s ♣ró♣r✐♦s ❞❡ A✿ 1✱ 2 ❡ 3
ma(1) = ma(2) = ma(3) = 1
❱❛❧♦r❡s ♣ró♣r✐♦s ❞❡ B✿ 1
ma(1) = 3
✭❜✮ ✭✐✐✮ ◆ã♦
✭❝✮
[−1 −2 −1
1 1 1
0 2 2
]
✻✳✾✸ ✲ ✭❛✮ ❱❛❧♦r❡s ♣ró♣r✐♦s ❞❡ A✿ 0 ❡ 2
mg(0) = 1 ❡ mg(2) = 2
✭❜✮
[
0 0 0
0 2 0
0 0 2
]
✭❝✮
[
2 0 0
4 0 0
8 −4 2
]
✻✳✾✺ ✲ ✭❝✮
[
2 0 0
−2 0 0
0 0 0
]
✼ ✲ Pr♦❞✉t♦ ■♥t❡r♥♦✱ ❊①t❡r♥♦ ❡ ▼✐st♦
✼✳✶ ✲ α = 3
✼✳✷ ✲ arccos
(
4√
69
)
✼✳✸ ✲ sen(α) =
√
19
21
cos(α) =
√
2
21
✼✳✹ ✲ ✭❛✮ −3e1 − 2e2 + e3
✭❜✮ −2e1 + 2e2 − e3
✭❝✮ −e1 + e2 − e3
✭❞✮ −e1 − e2
✭❡✮
(
−3
√
15√
14
,−2
√
15√
14
,
√
15√
14
)
✼✳✺ ✲
√
52
2 =
√
13
✼✳✻ ✲ e1 + 5e2 + 7e3
✼✳✼ ✲ 2
✼✳✽ ✲ ✭❛✮
−−→
PQ = −e1 + 2e2√
5(cos(
−−→
PQ, e1)e1 + cos(
−−→
PQ, e2)e2 +
cos(
−−→
PQ, e3)e3) =
√
5
(
− 1√
5
e1 +
2√
5
e2 +
0√
5
e3
)
✭❜✮ 2
√
3e1 + 2e2
✭❝✮
√
7
2
✼✳✾ ✲ ✭❛✮ −λ = 5±
√
10
✭❜✮ ✐✳ arccos
(√
2
15
)
✐✐✳ {(−y − z, y, z) : (y, z) ∈ R2 \ {(0, 0)}}
✐✐✐✳
(
−1√
26
, −3√
26
, 4√
26
)
♦✉
(
1√
26
, 3√
26
, −4√
26
)
✭❝✮ k = ±
√
146
✼✳✶✵ ✲ ✭❛✮ k = 74 ∨ k = −134
✭❜✮ k = −34
✽ ✲ ❘❡❝t❛ ❡ P❧❛♥♦
✽✳✶ ✲ ✭❛✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (3, 2,−1) + λ(−2, 2, 3), λ ∈ R


x = 3− 2λ
y = 2 + 2λ
z = −1 + 3λ
⇔ x−3
−2
= y−2
2
= z+1
3
✭❜✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (3, 2,−1) + λ(−1,−1, 0), λ ∈ R


x = 3− λ
y = 2− λ
z = −1
⇔ x−3
−1
= y−2
−1
∧ z = −1
✭❝✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (1, 0, 2) + λ(2,−1, 1), λ ∈ R


x = 1 + 2λ
y = −λ
z = 2 + λ
⇔ x−1
2
= −y = z − 2
✭❞✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (1, 0, 2) + λ(−3,−2, 5), λ ∈ R


x = 1− 3λ
y = −2λ
z = 2 + 5λ
⇔ x−1
−3
= y
−2
= z−2
5
✭❡✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (0, 0, 0) + λ(1, 0, 0), λ ∈ R
✸✵
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133



x = λ
y = 0
z = 0
⇔ y = 0 ∧ z = 0
✭❢✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (0, 0, 0) + λ(0, 1, 0), λ ∈ R


x = 0
y = λ
z = 0
⇔ x = 0 ∧ z = 0
✭❣✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (0, 0, 0) + λ(1, 0, 0), λ ∈ R


x = 0
y = 0
z = λ
⇔ x = 0 ∧ y = 0
✽✳✷ ✲ ✭❛✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (0, 3, 1) + λ(0, 3, 1), λ ∈ R
✭❜✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (1,−1, 0) + λ(0, 1,−1), λ ∈ R
✭❝✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (0, 2,−4) + λ(1, 0, 0), λ ∈ R
✭❞✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (1,−2, 0) + λ(3, 2, 6), λ ∈ R
✽✳✸ ✲ ✭❛✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (1, 0, 2)+λ(−2,−1, 0)+µ(3, 0, 2) λ, µ ∈
R
2x− 4y − 3z + 4 = 0
✭❜✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (0, 0, 1) + λ(2,−1, 3) + µ(0,−3, 6), λ, µ ∈ R
x− 4y − 2z + 2 = 0
✭❝✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (0, 0, 1) + λ(1, 1,−1) + µ(1,−2, 1), λ, µ ∈ R
x+ 2y + 3z − 3 = 0
✭❞✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (0, 0, 0) + λ(1, 0, 0) + µ(0, 1, 0), , λ, µ ∈ R
z = 0
✭❡✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (0, 0, 0) + λ(1, 0, 0) + µ(0, 0, 1), , λ, µ ∈ R
y = 0
✭❢✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (0, 0, 0) + λ(0, 1, 0) + µ(0, 0,1), λ, µ ∈ R
x = 0
✽✳✹ ✲ ✭❛✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (0, 0, 0) + λ(0, 5, 0) + µ(5, 0,−1), λ, µ ∈ R
✭❜✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (0, 0, 0) + λ(1, 0, 1) + µ(1, 1,−1), λ, µ ∈ R
✭❝✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (−3, 0, 0) + λ(0, 1, 0) + µ(0, 0, 1), λ, µ ∈ R
✭❞✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (2, 0, 0) + λ(2, 3, 0) + µ(2, 1,−1), λ, µ ∈ R
✽✳✺ ✲ ✭❛✮ x+ 6y + 2z − 10 = 0
✭❜✮ x+ 5z − 2 = 0
✽✳✻ ✲ π ∩ xx = (2, 0, 0)
π ∩ yy = (0,−3, 0)
π ∩ zz = (0, 0, 65)
✽✳✼ ✲ ✭❛✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (1,−2, 3) + λ(1, 1, 1), λ ∈ R
✭❜✮ −4x+ 5y − z + 17
✭❝✮ x−1
3
= −2− y = 3− z
✭❞✮ 1− x = −y
2
= z−2
3
✽✳✽ ✲ ✭❛✮ ◆ã♦
✭❜✮ ◆ã♦
✽✳✾ ✲ x = 1 ∧ 7 + 2y = 3− 2z
✽✳✶✵ ✲ m = −4 ∧ k = 3
✽✳✶✶ ✲ ✭❛✮ −y + z − 6 = 0
✭❜✮ π/6
✽✳✶✷ ✲ ✭❛✮ π/3
✭❜✮ arccos(1/9)
✽✳✶✸ ✲ ✭❛✮ 2
✭❜✮ 2
✽✳✶✹ ✲
√
3/2
✽✳✶✺ ✲ ✭❛✮ ❡♥✈✐❡s❛❞❛s
✭❜✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (0, 1, 2) + λ(0, 0, 2), λ ∈ R
✭❝✮ 2
✽✳✶✻ ✲ ✭❜✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (1, 2, 3) + λ(4, 0,−3) + µ(3, 3, 4), λ, µ ∈ R
9x− 25y + 12z + 5 = 0
✭❝✮ 15√
34
✽✳✶✼ ✲ ✭❛✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (3, 0, 3) + λ(0, 1, 0), λ ∈ R
✭❜✮ x− 2y − z + 2 = 0
✭❝✮ 0
✭❞✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (3, 0, 3) + λ(0, 1, 0) + µ(1,−2,−1), λ, µ ∈ R
x+ z − 6 = 0
✽✳✶✽ ✲ ✭❛✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (1, 1
4
, 1) + λ(1, 1, 1) + µ(1, 3
2
, 2), λ, µ ∈ R
✭❜✮ arccos
(
3√
30
)
✭❝✮ P♦r ❡①❡♠♣❧♦✱
(x, y, z) = (1,− 1
4
, 0) + λ(−2,− 1
2
, 1), λ ∈ R
✭❞✮ 3
2
√
14
✸✶
Descarregado por Carolina Direito (carolinasimoesdireito@gmail.com)
lOMoARcPSD|6016133
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