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Lista de Exercícios 02 - modulo 4-convertido

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Lista de Exercícios 
1 – Construa uma matriz A quadrada de ordem 2, de modo que aij = 2i – j. Calcule A2. 
 
2 – Sejam as matrizes A = 










−−
−
−−
132
310
423
, B = 










−
−−
320
251
012
 e C = 










−−
−
02
31
52
calcule, possível: 
a) 2A + 3B b) – A – 2B c) At + B 
d) A.B e) B.A f) B.C + A.C 
 
3 – Considere a matriz A = 





22
3 2
x
x
. Determine x, sabendo que At = A. 
 
4 – Determine os valores de a, b, c e d sabendo-se que 





dc
ba
. 





43
32
= 





10
01
. 
 
5 – Considere as matrizes A = 





62
71
, B = 





34
12
 , C = 





02
20
. Determine a matriz X tal que 
32
CXBAX −
=
−−
 
 
GABARITO 
1 – A = 





23
01
, A2 = 





23
01
. 





23
01
= 





++
++
2.20.33.21.3
2.00.13.01.1
= 





49
01
 
2 a) 2A + 3B = 










−−
−
−−
264
620
846
+ 










−
−−
960
6153
036
= 










−
−
−−
7124
0173
8712
 
2b) – A – 2B = 










−
−
−
132
310
423
– 










−
−−
640
4102
024
= 










−−
−−
−
572
1112
447
 
2c) At + 5B = 










−−
−
−−
134
312
203
+ 










−
−−
15100
10255
0510
= 










−
−−−
1474
13267
2513
 
2d) A.B = 










−−
−
−−
132
310
423
. 










−
−−
320
251
012
= 










−++−−++−−−+−+−−
−++−++−−+−+−
+−+−+−+−−+−−+−−
3).1(2.30).2(2).1(5.3)1).(2(0).1()1.(3)2).(2(
3).3(2.10.02).3(5.1)1.(00).3()1.(1)2.(0
3.42).2(0).3(2.45).2()1).(3(0.4)1).(2()2).(3(
=










−−−
3151
711
818
 
2e) B.A =










−
−−
320
251
012
. 










−−
−
−−
132
310
423
= 










−+−+++−−++−
−+−+−++−−−++−−
−+−−+−+−+−−−+−+−−
)1.(3)3.(24.03.31.2)2.(0)2.(30.2)3.(0
)1.(2)3.(54).1(3.21.5)2).(1()2.(20.5)3).(1(
)1.(0)3).(1(4).2(3.01).1()2).(2()2.(00).1()3).(2(
= 










−−
−−
−
9116
21131
536
 
2f) B.C + A.C = 










−
−−
−
253
236
1621
 
3 – Sendo A = 





22
3 2
x
x
, logo At = 





2
23
2x
x
, Fazendo a igualdade de At = A, temos 





2
23
2x
x
= 





22
3 2
x
x
. 
Logo, 2x = x2 ou x2 = 2x . Resolvendo a equação x2 – 2x = 0, temos x = 0 ou x = 2. 
4 – 





dc
ba
. 





43
32
= 





10
01
 






++
++
dcdc
baba
4332
4332
= 





10
01
 



=+
=+
043
132
ba
ba
 



=+
=+
143
032
dc
dc
 . Resolvendo os sistemas encontramos: a = – 4, b = 3, c = 3, d = –2 
5 – Sendo 
32
CXBAX −
=
−−
, temos 
3(X – A – B) = 2(X – C) 
3X – 3A – 3B = 2X – 2C 
3X – 2X = 3A + 3B – 2C 
X = 





186
213
+ 





912
36
– 





04
40
 
X = 





2714
209

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