Tutorial 5 mth 3201
-
Upload
drradz-maths -
Category
Documents
-
view
697 -
download
5
Transcript of Tutorial 5 mth 3201
Tutorial MTH 3201 Linear Algebras
Tutorial 5
π΄π₯ = 0
3 4β1 2
π₯1
π₯2=
00
3 4β1 2
00
1 00 1
00
πΈπ π
β΄ π₯1 = π₯2 = 0, ππ’πππ ππππ, π₯ =00
π΄π₯ = 0
1 0 3β1 1 β12 4 0
π₯1
π₯2
π₯3
=000
1 0 3β1 1 β12 4 0
000
πΈπ π 1 0 30 1 20 0 1
000
β΄ π₯1 = π₯3 = π₯2 = 0, ππ’πππ ππππ, π₯ =000
π΄π₯ = 0
β2 61 β3
π₯1
π₯2=
00
β2 61 β3
00
1 β30 0
00
πΈπ π
π₯1 β 3π₯2 = 0, π₯1 = 3π₯2
Let π₯2 = π‘, π₯1 = 3π‘
ππ’πππ ππππ, π₯ =3π‘π‘
= π‘31
π΄π₯ = 0
1 5 10 2 β36 β2 4
π₯1
π₯2
π₯3
=000
1 5 10 2 β36 β2 4
000
πΈπ π 1 5 1
0 1 β3
20 0 1
000
β΄ π₯1 = π₯3 = π₯2 = 0, ππ’πππ ππππ, π₯ =000
π΄π₯ = 0
1 β2 32 1 β11 β3 4
π₯1
π₯2
π₯3
=000
1 β2 32 1 β11 β3 4
000
πΈπ π 1 β2 3
0 1 β7
50 0 1
000
π₯3 = 0,
π₯2 β7π₯3
5= 0 β π₯2 = 0,
π₯1 β 2π₯2 + 3π₯3 β π₯1 = 0
π₯ =000
, So there are no basis.
Dimension=0
1 β3 1β1 β5 13 1 β2
1
β1β2
000
πΈπ π
1 β3 10 1 β1/40 0 1
102
000
π₯3 + 2π₯4 = 0 β π₯3 = β2π₯4 π₯2 β π₯3/4 = 0 β π₯3 = 4π₯2
π₯1 β 3π₯2 + π₯3 + π₯4 = 0 β π₯1 = π₯3/4
Let π₯3 = π‘, π₯1 =π‘
4, π₯2 =
π‘
4, π₯4 = βπ‘/2
π₯ = π‘
1/41/41
β1/2
= π‘
β1/2β1/2β21
β΄ π£ =
β1/2β1/2β21
span solution space, dimension = 1
π΄π₯ = π
β5 23 7
π₯1
π₯2=
8β1
β5 23 7
8
β1
1 β2/50 1
β8/519/41
πΈπ π
π₯2 =19
41, π₯1 = β
58
41
π is in column space of A.
π =81
=β58
41β53
+19
4127
linear combination of column space of A.
π΄π₯ = π 7 2 12 3 41 1 0
π₯1
π₯2
π₯3
=28
β1
7 2 12 3 41 1 0
28
β1 πΈπ π 1 1 0
0 1 40 0 1
β110
59/21
π₯3 =59
21, π₯2 =
β26
21, π₯1 =
5
21
π =28
β1=
5
21
721
β26
21
231
+59
21
140
π΄π₯ = 0
3 β1 01 7 02 4 0
000
πΈπ π 1 7 0
0 1 00 0 0
000
π₯3 = π‘, π₯2 = 0, π₯1 = 0
π₯ = π‘001
001
is a basis for nullspace
β1 3 5β6 4 1β3 β2 1
000
πΈπ π 1 β3 β5
0 1 29/140 0 1
000
π₯3 = 0, π₯2 = 0, π₯1 = 0
π₯ =000
There is no basis
1 2 β1β3 3 45 1 1
β212
000
πΈπ π
1 2 β10 1 1/90 0 1
β2
β5/91
000
π₯4 = π‘, π₯3 = βπ‘, π₯2 =2π‘
3, π₯1 = βπ‘/3
π₯ =1
3π‘
β12
β33
π£1 =
β12
β33
is the basis for the nullspace
7 1 61 4 52 2 β1
000
πΈπ π 1 0 0
0 1 00 0 1
000
π₯ =000
There is no basis
π΄π₯ = 0
3 β1 01 7 02 4 0
πΈπ π 1 0 0
0 1 00 0 0
β΄ r 1 and r 2 are bases for row space of A
β1 3 5β6 4 1β3 β2 1
πΈπ π 1 0 0
0 1 00 0 1
π 1 = 1 0 0 , π 2 = 0 1 0
π β²1 =100
πππ π β²2 =010
π 1 =312
πππ π 2 =β174
β΄ c 1 and c 2 are bases for column space of A
β΄ r 1, r 2and r 3 are bases for row space of A
π 1 = 1 0 0 , π 2 = 0 1 0 , π 3 = 0 0 1
π 1 =β1β6β3
, π 2 =34
β2πππ π 3 =
511
β΄ c 1 , c 2 and c 3 are bases for column space of A
1 2 β1β3 3 45 1 1
β212
πΈπ π
1 0 00 1 00 0 1
1/3
β2/31
7 1 61 4 52 2 β1
πΈπ π 1 0 0
0 1 00 0 1
β΄ r 1, r 2and r 3 are bases for row space of A
π 1 = 1 0 0 1/3 , π 2 = 0 1 0 β 2/3 , π 3 = 0 0 1 1
π 1 =1
β35
, π 2 =231
πππ π 3 =β141
β΄ c 1 , c 2 and c 3 are bases for column space of A
β΄ r 1, r 2and r 3 are bases for row space of A
π 1 = 1 0 0 , π 2 = 0 1 0 , π 3 = 0 0 1
π 1 =712
, π 2 =142
πππ π 3 =65
β1
β΄ c 1 , c 2 and c 3 are bases for column space of A
π΄π =3 1 2
β1 7 40 0 0
πΈπ π 1 15 10
0 1 β14/220 0 0
π1 β² =
100
, π2 β² =
1510
π1 =3
β10
, π2 =170
c1 and c2 form bases for column space ofAT
basis of C. S of AT= basis of R. S of AT
π1 = 3 β1 0 , π2 = 1 7 0
r1 and r2 form bases of row space consisting entirely of row vector π΄
π΄π =β1 β6 β33 4 β25 1 1
πΈπ π
1 6 30 1 11/40 0 1
π1 β² =
100
, π2 β² =
610
, π3 β² =
311/4
1
π1 =β135
, π2 =β641
, π2 =β3β21
c1 , c2 and c3 form bases for column space of AT
basis of C. S of AT= basis of R. S of AT
π1 = β1 3 5 , π2 = β6 4 1
r1 , r2 and r3 form bases of row space consisting entirely of row vector π΄
π3 = β3 β2 1
π΄π =
1 β3 52 3 1
β1β2
41
12
πΈπ π 1 β3 50 1 β100
00
10
π1 =
12
β1β2
, π2 =
β3341
, π2 =
5112
c1 , c2 and c3 form bases for column space of AT
basis of C. S of AT= basis of R. S of AT
π1 = 1 2 β1 β2 , π2 = β3 3 4 1
r1 , r2 and r3 form bases of row space consisting entirely of row vector π΄
π3 = 5 1 1 2
π΄π =7 1 21 4 26 5 β1
πΈπ π
1 4 20 1 4/90 0 1
π1 =716
, π2 =145
, π2 =22
β1
c1 , c2 and c3 form bases for column space of AT
basis of C. S of AT= basis of R. S of AT
π1 = 7 1 6 , π2 = 1 4 5
r1 , r2 and r3 form bases of row space consisting entirely of row vector π΄
π3 = 2 2 β1
Row space based on reduced matrix form
Column space based on original matrix
3 0 33 β4 β15 β4 1
β2β11
πΈπ π 1 0 1
0 1 10 0 0
β2/3β1/4
1
Bases for row space A
π€1 = 1 0 1 β2/3 π€2 = 0 1 1 β1/4
π€3 = 0 0 0 1
β2 1 33 β1 3
β1 β5 2
42
β3 πΈπ π 1 5 β2
0 1 9/160 0 1
3
7/161
Bases for row space A
π€1 = 1 5 β2 3 π€2 = 0 1 9/16 7/16
π€3 = 0 0 1 1
0 1 01 1 0
β1 1 0
101
πΈπ π
1 1 00 1 00 0 0
011
Bases for row space A
π€1 = 1 1 0 0 π€2 = 0 1 0 1 π€3 = 0 0 0 1
Let π΄ =
β3 5 1β3 3 β130
β11
31
β5β791
, π΄π =
β3 β3 35 3 β11
β5β1β7
39
0111
πΈπ π
1 0 10 1 β200
00
00
1/2β1/2
00
π€1 =
1000
, π€2 =
0100
w1, and w2 form bases for column space of R
π£1 =
β351
β5
, π£2 =
β33
β1β7
v1, and v2 form bases for column space ofπ΄π
Vectors in the basis
For vectors not in the basis:
π€3 =
1β200
=
1000
+ (β2)
1100
π€3 = π€1 β 2π€2
π€4 =
0β1/2
00
= 1/2
1000
+ (β1/2)
1100
π€4 = (1/2)π€1 β (1
2)π€2
β΄ π£3 = π£1 β 2π£2 , π£4 = (1/2)π£1 β (1/2)π£2
Let π΄ =
1 1 0β1 3 2β13
50
35
16
β2β1
, π΄π =
1 β1 β11 3 501
26
3β2
305
β1
πΈπ π
1 0 00 1 000
00
10
0001
β΄ v1, v2, v3, v4 is a subset of S that forms a basis for the space spanned by the vector in S
= ππππ π΄ = 2 = ππππ π΄ = 2
= π β ππππ π΄ = 4 β 2 = 2
= π β ππππ π΄ = 4 β 2 = 2
= ππππ π΄ = 3 = ππππ π΄ = 3
= π β ππππ π΄ = 5 β 3 = 2
= π β ππππ π΄ = 7 β 3 = 4
= ππππ π΄ = 0 = ππππ π΄ = 0
= π β ππππ π΄ = 3 β 0 = 2
= π β ππππ π΄ = 3 β 0 = 4
π Γ π
ππ’ππππ‘π¦ + ππππ = π
ππ’ππππ‘π¦ + ππππ = π
π΄ =
π11
π21
π31
π12
π22
π32
β¦ β¦ . . π16
β¦ β¦ . . π26
β¦ β¦ . . π36π41 π42 β¦ β¦ . . π46
π51 π52 β¦ β¦ . . π56
ππππ’ππ π£πππ‘ππ π΄:
π1 =
π11
π21π31
π41
π51
, π2 =
π12
π22π32
π42
π52
,β¦β¦., π6 =
π16
π26π36
π46
π56
Consider the equation πΎ1π 1 + πΎ2π 2+ πΎ3π 3+β¦.+ πΎ6π 6 = 0
Then, express both sides of these equation in term of components:
πΎ1π11 + πΎ2π12 + πΎ3π13 + β― + πΎ6π16 = 0
πΎ1π21 + πΎ2π22 + πΎ3π23 + β― + πΎ6π26 = 0
πΎ1π31 + πΎ2π32 + πΎ3π33 + β― + πΎ6π36 = 0
πΎ1π41 + πΎ2π42 + πΎ3π43 + β― + πΎ6π46 = 0
πΎ1π51 + πΎ2π52 + πΎ3π53 + β― + πΎ6π56 = 0
This is a homogenous system with 5 equations and 6 unknowns.
Then, there exists nontrivial solution. Therefore c1 , c2 , β¦ , c6 are linearly dependent.
ππππ 7π΄ = ππππππ πππ πππ€ π ππππ 7π΄ = ππππππ πππ πππ€ π ππππ π΄ = ππππ π΄
πΏππ‘ π΄ = π Γ π πππ‘πππ₯ ππ’ππππ‘π¦ π΄ + ππππ π΄ = ππ. ππ ππππ’πππ = π
ππ’ππππ‘π¦ π΄ = π β ππππ π΄ = π β ππππππ πππ πππ€ π ππππ π΄
= π β ππππππ πππ πππ€ π ππππ π΄π
= π β ππππ π΄π
= ππ’ππππ‘π¦ π΄π
-dr Radz