P1 de Algebra com GABARITO

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FOLHA DE QUESTÕES 
COORDENAÇÃO DO CICLO BÁSICO ENGENHARIAS 
DISCIPLINA: 
Álgebra Linear 
Curso: Engenharia Professor: Marcelo Torraca 
ASS.: NOME: 
Resolução da P1 Resolução da P1 
DATA: PROVA: TURMA MATRÍCULA: 
17/04/2013 
GRAU: 
P1 1TEC31G 
 
 
Questão 1 (1,5 pontos) 
Seja 





−−−−====++++++++
====++++−−−−
====++++++++
1zy2x
2zyx
0zyx
m
m , determine o valor de m para o sistema ser possível indeterminado e impossível. 





−−−−====++++++++
====++++−−−−
====++++++++
1zy2x
2zyx
0zyx
m
m ⇒⇒⇒⇒ 










−−−−
−−−−
112m
2m11
0111
 ⇒⇒⇒⇒ 
212 LLL
112m
2m120
0111
−−−−====′′′′










−−−−
−−−−−−−−
 (0,25) ⇒⇒⇒⇒ 
313 LLmL
11m2m0
2m120
0111
−−−−⋅⋅⋅⋅====′′′′










−−−−−−−−
−−−−−−−−
 (0,5) ⇒⇒⇒⇒ 
323
2
L2L)2m(L
4m2mm00
2m120
0111
⋅⋅⋅⋅−−−−⋅⋅⋅⋅−−−−====′′′′′′′′










++++−−−−−−−−
−−−−−−−−
 (0,75) ⇒⇒⇒⇒ 










−−−−−−−−−−−−
−−−−−−−−
)1m(2)m1(m00
2m120
0111
 (1,0) 
Para 1m ==== o sistema será possível indeterminado, pois 










−−−−
0000
2020
0111
. (1,25) 
Para 0m ==== o sistema será impossível, pois 










−−−−
2000
2120
0111
.(1,5) 
 
Questão 2 (1,5 pontos) 
Um comerciante deseja totalizar a quantia de R$500,00 utilizando cédulas de um, cinco e dez reais, num 
total de 92 cédulas, de modo que as quantidades de cédulas de um e de dez reais sejam iguais. Neste caso, 
qual a quantidade de cédulas de cinco reais o comerciante precisará? 
 
Resolução da P1 de álgebra linear 
Marcelo Torraca 
 
Página 2 de 5 
Considere “x”, “y” e “z” respectivamente as quantidades de cédulas de R$1,00; R$5,00 e R$10,00, 



====++++++++
====++++++++
500z10y5x
92zyx
 (0,5), lembrando que zx ==== , então 



====++++
====++++
500y5x11
92yx2
 (0,75). Escalonando, o sistema: 
212 L.2L.11L
500y5x11
92yx2
−−−−====′′′′



====++++
====++++
(1,0) ⇒⇒⇒⇒ 



====
====++++
12y
92yx2 (1,25). Logo são necessárias 12 cédulas de R$5,00. (1,5) 
 
Questão 3 (2,0 pontos) 
Resolva o sistema linear 





====++++++++
====−−−−++++
====++++++++
8zy23x
6zyx2
3zyx
111
111
111
 e 







====++++++++++++
====++++++++++++
====++++++++++++
====++++++++++++
4w2zyx
3wz2yx
2wzy2x
1wzyx2
2222
2222
2222
2222
, determine 
2
2212121 )w(zzyyxx −−−−⋅⋅⋅⋅++++⋅⋅⋅⋅−−−−⋅⋅⋅⋅ . 





====++++++++
====−−−−++++
====++++++++
8zy23x
6zyx2
3zyx
111
111
111
 ⇒⇒⇒⇒ 










−−−−
8123
6112
3111
 ⇒⇒⇒⇒ 
313
212
LL3L
LL2L
1210
0310
0111
−−−−⋅⋅⋅⋅====′′′′
−−−−⋅⋅⋅⋅====










 (0,25) ⇒⇒⇒⇒ 
323 LLL1100
0310
0111
−−−−====′′′′′′′′









−−−−
 (0,5) ⇒⇒⇒⇒ 





−−−−====
====++++
====++++++++
1z
03zy
3zyx
1
11
111
 (0,75) ⇒⇒⇒⇒ 





====
====→→→→====−−−−⋅⋅⋅⋅++++
====→→→→====−−−−++++
1z
3y0)1(3y
1x313x
1
11
11
 (1,0) 
⇒⇒⇒⇒ })1,3,1({S1 −−−−==== 
 







====++++++++++++
====++++++++++++
====++++++++++++
====++++++++++++
4w2zyx
3wz2yx
2wzy2x
1wzyx2
2222
2222
2222
2222
 ⇒⇒⇒⇒ 












42111
31211
21121
11112
 ⇒⇒⇒⇒ 
14
41
LL
LL
11112
31211
21121
42111
→→→→
→→→→












 (1,05) ⇒⇒⇒⇒ 
414
313
212
LL2L
LLL
LLL
73110
11100
21010
42111
−−−−⋅⋅⋅⋅====′′′′
−−−−====′′′′
−−−−====′′′′












−−−−
−−−−
 ⇒⇒⇒⇒ 
424 LLL94100
11100
21010
42111
′′′′++++====′′′′′′′′












−−−−
−−−−
 (1,15) ⇒⇒⇒⇒ 
434 LLL105000
11100
21010
42111
′′′′′′′′++++====′′′′′′′′′′′′












−−−−
−−−−
 ⇒⇒⇒⇒ 







====
====++++−−−−
====++++−−−−
====++++++++++++
10w5
1wz
2wy
4w2zyx
2
22
22
2222
 (1,25) ⇒⇒⇒⇒ 







====
====++++−−−−
====++++−−−−
====++++++++++++
2w
1wz
2wy
4w2zyx
2
22
22
2222
 
Resolução da P1 de álgebra linear 
Marcelo Torraca 
 
Página 3 de 5 
(1,5) ⇒⇒⇒⇒ 







====
====→→→→====++++−−−−
====→→→→====++++−−−−
−−−−====→→→→====⋅⋅⋅⋅++++++++++++
2w
1z12z
0y22y
1x42210x
2
22
22
22
(1,75) ⇒⇒⇒⇒ })2,1,0,1({S2 −−−−==== 
2
2212121 )w(zzyyxx −−−−⋅⋅⋅⋅++++⋅⋅⋅⋅−−−−⋅⋅⋅⋅ ⇒⇒⇒⇒ 2)2(1)1(03)1(1 −−−−⋅⋅⋅⋅−−−−++++⋅⋅⋅⋅−−−−−−−−⋅⋅⋅⋅ ⇒⇒⇒⇒ 6−−−− (2,0) 
Outro modo 







====++++++++++++
====++++++++++++
====++++++++++++
====++++++++++++
4w2zyx
3wz2yx
2wzy2x
1wzyx2
2222
2222
2222
2222
 ⇒⇒⇒⇒ 












42111
31211
21121
11112
 ⇒⇒⇒⇒ 
144
133
122
LL2L
LL2L
LL2L
73110
51310
31130
11112
−−−−⋅⋅⋅⋅====′′′′
−−−−⋅⋅⋅⋅====′′′′
−−−−⋅⋅⋅⋅====′′′′












 (1,05) ⇒⇒⇒⇒ 
244
233
LL3L
LL3L
188200
122800
31130
11112
−−−−⋅⋅⋅⋅====′′′′′′′′
−−−−⋅⋅⋅⋅====′′′′′′′′












 (1,10) ⇒⇒⇒⇒ 
344 LL4L6030000
122800
31130
11112
−−−−⋅⋅⋅⋅====′′′′′′′′′′′′












 (1,15) ⇒⇒⇒⇒ 







====
====++++
====++++++++
====++++++++++++
60w30
12w2z8
3wzy3
1wzyx2
2
22
222
2222
 (1,25) ⇒⇒⇒⇒ 







====
====++++
====++++++++
====++++++++++++
2w
12w2z8
3wzy3
1wzyx2
2
22
222
2222
 (1,5) ⇒⇒⇒⇒ 







====
====→→→→====++++
====→→→→====++++++++
−−−−====→→→→====++++++++++++
2w
1z12)2(2z8
0y321y
1x1210x2
2
22
22
22
(1,75) ⇒⇒⇒⇒ })2,1,0,1({S2 −−−−==== 
 
Questão 4 (3,0 pontos) 
Seja a matriz formada pelos elementos da matriz 3x3ij )a(A ==== , definida por (((( ))))



====
≠≠≠≠−−−−
====
++++
jise,0
jise,1
a
ji
ij , 
determine sua inversa. 
Seja 










====
333231
232221
131211
aaa
aaa
aaa
A 
1)1(a
1)1(a
0a
31
13
21
12
11
====−−−−====
−−−−====−−−−====
====
++++
++++
 
1)1(a
0a
1)1(a
32
23
22
12
21
−−−−====−−−−========
−−−−====−−−−====
++++
++++
 
0a
1)1(a
1)1(a
23
23
32
13
31
====
−−−−====−−−−====
====−−−−====
++++
++++
 (0,75) 










−−−−
−−−−−−−−
−−−−
====
011
101
110
A (1,0). 
Resolução da P1 de álgebra linear 
Marcelo Torraca 
 
Página 4 de 5 
131312121111 AaAaAaAdet ⋅⋅⋅⋅++++⋅⋅⋅⋅++++⋅⋅⋅⋅==== ⇒⇒⇒⇒ 
(((( )))) (((( )))) (((( ))))
11
01
1
01
11
1)1(
01
10
10Adet 312111
−−−−
−−−−
⋅⋅⋅⋅−−−−++++
−−−−−−−−
⋅⋅⋅⋅−−−−⋅⋅⋅⋅−−−−++++
−−−−
−−−−
⋅⋅⋅⋅−−−−⋅⋅⋅⋅====
++++++++++++
 ⇒⇒⇒⇒ 2110Adet ====++++++++==== 
(1,25) 
Como determinante é diferente de zero, logo a matriz é invertível. 
1
11
01)1(A
1
01
11)1(A
1
01
10)1(A
31
13
21
12
11
11
====
−−−−
−−−−
⋅⋅⋅⋅−−−−====
−−−−====
−−−−−−−−
⋅⋅⋅⋅−−−−====
−−−−====
−−−−
−−−−
⋅⋅⋅⋅−−−−====
++++
++++
++++
 
1
11
10)1(A
1
01
10)1(A
1
01
11)1(A
32
23
22
22
12
21
−−−−====
−−−−
−−−−
⋅⋅⋅⋅−−−−====
−−−−====⋅⋅⋅⋅−−−−====
−−−−====
−−−−
−−−−
⋅⋅⋅⋅−−−−====
++++
++++
++++
 (((( ))))
(((( )))) 1
01
10
1A
1
11
10
1A
1
10
11)1(A
33
33
23
32
13
31
−−−−====
−−−−
−−−−
⋅⋅⋅⋅−−−−====
−−−−====
−−−−−−−−
⋅⋅⋅⋅−−−−====
====
−−−−
−−−−
⋅⋅⋅⋅−−−−====
++++
++++
++++
 (1,75) 










−−−−−−−−
−−−−−−−−−−−−
−−−−−−−−
====
111
111
111
Acof (2,0) ⇒⇒⇒⇒ 










−−−−−−−−
−−−−−−−−−−−−
−−−−−−−−
====
111
111
111
)Acof( t (2,5) 
t1 )Acof(
Adet
1A ====−−−− ⇒⇒⇒⇒ 










−−−−−−−−
−−−−−−−−−−−−
−−−−−−−−
⋅⋅⋅⋅====
−−−−
111
111
111
2
1A 1 (2,75) ⇒⇒⇒⇒ 
















−−−−−−−−
−−−−−−−−−−−−
−−−−−−−−
====
−−−−
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
A 1
 ou 










−−−−−−−−
−−−−−−−−−−−−
−−−−−−−−
====
−−−−
5,05,05,0
5,05,05,0
5,05,05,0
A 1 (3,0) 
 
Questão 5 (2,0 pontos) 
Dada a matriz 










−−−−
−−−−
====
210
432
011
A , obtenha a matriz X tal que t2 AAX −−−−==== . 










−−−−
−−−−
⋅⋅⋅⋅










−−−−
−−−−
====⋅⋅⋅⋅====
210
432
011
210
432
011
AAA2 ⇒⇒⇒⇒ 









 −−−−−−−−−−−−
====
812
4118
441
A2 (1,0) 










−−−−
−−−−====
240
131
021
At (1,5) 
Resolução da P1 de álgebra linear 
Marcelo Torraca 
 
Página 5 de 5 
t2 AAX −−−−==== ⇒⇒⇒⇒ 










−−−−
−−−−−−−−









 −−−−−−−−−−−−
====
240
131
021
812
4118
441
X ⇒⇒⇒⇒ 










−−−−
−−−−−−−−−−−−
====
1032
389
462
X (2,0)