ejercicios fase 1
DESCRIPTION
Ejercicios fase 1TRANSCRIPT
1. Para las operaciones indicadas, los vectores dados conviene tenerlos en notación rectangular.
a¿ u=(|u|cosθ ,|u|senθ )=(−34 ,−3√34 )b¿ v=(|v|cosθ ,|v|senθ )=( 32 ,−3√32 )1.1 .u−v=(−34 ,−3√34 )−( 32 ,−3√32 )=(−94 , 3√34 )1.2 .u−2 v=(−34 ,−3√34 )−2( 32 ,−3√32 )=(−154 ,
9√34 )
1.3 . v+u=( 32 ,−3 √32 )+(−34 ,−3√34 )=( 34 ,−9√34 )
1.4 . v−2u=( 32 ,−3√32 )−2(−34 ,−3√34 )=(3 ,0 )
1.5 .4 u−3 v=4 (−34 ,−3√34 )−3 ( 32 ,−3√32 )=(−152 ,3√32 )
2. Para el ángulo entre dos vectores,2.1 .u ∙ v=(−8 i−4 j ) ∙ (−6 i−4 j )=64 ,‖u‖=√80 ,‖v‖=√52
→cosθ= u ∙ v‖u‖‖v‖
= 64
√80√52= 8
√65→θ=cos−1( 8√65 )≈7.1°
2.2 . w ∙ z=(−i+3 j )∙ (−i−5 j )=−14 ,‖w‖=√10 ,‖z‖=√26
→cosθ= w ∙ z‖w‖‖z‖
= −14√10 √26
= −7√65
→θ=cos−1( −7√65 )≈150.3 °
2.3 . s ∙ t=(−i+3 j+2 k ) ∙ (−i−5 j−k )=−16 ,‖s‖=√14 ,‖t‖=√27
→cosθ= s ∙ t‖s‖‖t‖
= −16√14√27
= −163√42
→θ=cos−1( −7√65 )≈145.4 °
3. Inversa de la matriz C,
(−1 5 107 −3 −10 4 −3|
1 0 00 1 00 0 1) f 1→−f 1
(1 −5 −107 −3 −10 4 −3 |−1 0 0
0 1 00 0 1) f 2−7 f 1
(1 −5 −100 32 690 4 −3 |−1 0 0
7 1 00 0 1) f 2− 1
32f 2
(1 −5 −10
0 16932
0 4 −3|−1 0 0732
132
0
0 0 1) f 1+5 f 2f 3−4 f 2
(1 02532
0 16932
0 0−938
|332
532
0
732
132
0
−78
−18
1) f 3→− 893f 3
(1 02532
0 16932
0 0 1|332
532
0
732
132
0
793
193
−893
) f 1−2532 f 3f 2−6932f 3
(1 0 00 1 00 0 1|
13372
55372
−25372
7124
1124
−23124
793
193
−893
)→C−1=(
13372
55372
−25372
7124
1124
−23124
793
193
−893
)4. Determinante de la matriz A,
|A|=|0 0 0 0 −10 0 −1 −2 −10 2 1 5 74 1 −2 6 −21 0 2 3 4
|=(−1 )|0 0 −1 −20 2 1 54 1 −2 61 0 2 3
|f 3−4 f 4|A|=−|0 0 −1 −2
0 2 1 50 1 −10 −61 0 2 3
|=−(−1|0 −1 −22 1 51 −10 −6|) f 2−2 f 3
|A|=|0 −1 −20 21 171 −10 −6|=1|−1 −2
21 17|=−17−(−42 )=25
5. Inversa de la matriz C,
|C|=|−5 −2 −13 0 5
−8 1 −5|=−3|−2 −11 −5|−5|−5 −2
−8 1 |=−3 (10+1 )−5 (−5−16 )=72
C11=|0 51 −5|=−5 ,C12=| 3 5
−8 −5|=25 ,C13=| 3 0−8 1|=3
C21=|−2 −11 −5|=11 ,C22=|−5 −1
−8 −5|=17 ,C23=|−5 −2−8 1 |=−21
C31=|−2 −10 5 |=−10 ,C32=|−5 −1
3 5 |=−22 ,C33=|−5 −23 0 |=6
→C−1= 1|C|adj (C )= 1
72 (C11 C21 C31C12 C22 C32C31 C23 C33
)=(−572
1172
−536
2572
1772
−1136
124
−724
112
)