1
00:00:10,570 --> 00:00:13,000
PROFESSOR: Hi, everyone.

2
00:00:13,000 --> 00:00:17,530
So for the first
part of this course,

3
00:00:17,530 --> 00:00:20,250
we've learned basically
the ins and outs

4
00:00:20,250 --> 00:00:21,990
of solving linear
systems of equations.

5
00:00:24,800 --> 00:00:29,760
Today we're going to do a little
review of the basic concepts.

6
00:00:29,760 --> 00:00:35,520
Hopefully we'll see a few of
them in the following problem.

7
00:00:35,520 --> 00:00:37,660
We're given a square
matrix A-- a three

8
00:00:37,660 --> 00:00:42,040
by three square matrix A-- where
the last entry is a parameter

9
00:00:42,040 --> 00:00:43,570
k.

10
00:00:43,570 --> 00:00:46,200
And this parameter will vary.

11
00:00:46,200 --> 00:00:54,620
And we'll see what happens to
the system of equations A*x

12
00:00:54,620 --> 00:01:01,180
equal to [2, 3, 7], for which
k it has a unique solution,

13
00:01:01,180 --> 00:01:04,260
for which k it has
infinitely many solutions.

14
00:01:04,260 --> 00:01:06,710
Then we'll find the
LU decomposition.

15
00:01:06,710 --> 00:01:09,800
And finally, we'll write
down the complete solution,

16
00:01:09,800 --> 00:01:11,087
the system.

17
00:01:11,087 --> 00:01:12,670
So I'll give you a
few minutes to work

18
00:01:12,670 --> 00:01:14,630
this problem on your own.

19
00:01:14,630 --> 00:01:16,710
And then please come
back and see how I do it.

20
00:01:24,810 --> 00:01:26,980
All right, welcome back.

21
00:01:26,980 --> 00:01:33,510
So let's start with part A.
For which k does this system,

22
00:01:33,510 --> 00:01:37,720
A*x equal [2, 3, 7],
have a unique solution?

23
00:01:37,720 --> 00:01:46,310
So what do we know about square
systems of linear equations?

24
00:01:46,310 --> 00:01:51,630
They have a unique solution
when the matrix A is invertible.

25
00:01:51,630 --> 00:01:55,230
So now, when is A invertible?

26
00:01:55,230 --> 00:01:58,590
It is invertible when
it is of full rank.

27
00:01:58,590 --> 00:02:01,240
And how do we figure this out?

28
00:02:01,240 --> 00:02:04,770
We do it by performing
row operations.

29
00:02:04,770 --> 00:02:07,245
We do it by doing
eliminations on the matrix.

30
00:02:10,490 --> 00:02:14,360
But since we want to
simulate an exam setting,

31
00:02:14,360 --> 00:02:18,910
it always pays off to see
what tasks lie ahead of us.

32
00:02:18,910 --> 00:02:25,000
So in part C, we're asked to
find the LU decomposition.

33
00:02:25,000 --> 00:02:28,540
This means that when
we do row operations,

34
00:02:28,540 --> 00:02:34,020
we'd better keep track what row
operations we're doing exactly.

35
00:02:34,020 --> 00:02:38,660
In particular, we'll write them
down as elementary matrices.

36
00:02:38,660 --> 00:02:45,070
And in part D, we'll be asked to
compute the complete solution.

37
00:02:45,070 --> 00:02:51,520
And therefore it's good
to do row operations

38
00:02:51,520 --> 00:02:57,985
on the augmented matrix
A. So let's do this.

39
00:02:57,985 --> 00:03:00,330
I'm going to write this.

40
00:03:00,330 --> 00:03:12,210
The augmented matrix A is the
following beast, 1, 1, 1; 1 2,

41
00:03:12,210 --> 00:03:21,440
3; 3, 4, k; and then 2, 3, 7.

42
00:03:25,280 --> 00:03:31,430
So first thing, we subtract
a multiple of row one

43
00:03:31,430 --> 00:03:32,400
from row two.

44
00:03:32,400 --> 00:03:36,540
And it's exactly
negative 1 times

45
00:03:36,540 --> 00:03:40,360
the first row plus the second.

46
00:03:40,360 --> 00:03:45,990
Let me write down the
corresponding elementary matrix

47
00:03:45,990 --> 00:03:47,340
that does this.

48
00:03:47,340 --> 00:03:51,300
It's E_(2,1).

49
00:03:51,300 --> 00:03:55,730
And it's lower diagonal
with 1's on the diagonal.

50
00:03:55,730 --> 00:03:59,670
And it's going to
be exactly minus 1

51
00:03:59,670 --> 00:04:02,300
in the first entry
of the second row.

52
00:04:04,870 --> 00:04:12,850
So we get 1, 1,
1, 2; 0, 1, 2, 1.

53
00:04:15,460 --> 00:04:18,620
And we copy down the third row.

54
00:04:21,810 --> 00:04:25,660
Now we subtract a
multiple of the first row

55
00:04:25,660 --> 00:04:27,550
from the third one.

56
00:04:27,550 --> 00:04:30,790
And let me write this here.

57
00:04:34,890 --> 00:04:37,640
Yeah, we'll multiply the
first row by negative 3,

58
00:04:37,640 --> 00:04:40,800
and add it to the third one.

59
00:04:40,800 --> 00:04:45,740
This is accomplished by the
elementary matrix E_(3,1) which

60
00:04:45,740 --> 00:04:51,530
is 1, 1, 1, negative
three, and then 0, 0.

61
00:04:54,320 --> 00:05:01,030
OK, 1, 1, 1, 2; 0, 1, 2, 1.

62
00:05:01,030 --> 00:05:03,250
We copy the first two rows.

63
00:05:03,250 --> 00:05:10,630
And then the third one
will be 0, 1, k minus 3,

64
00:05:10,630 --> 00:05:14,750
and 7 minus 3 times
2, 7 minus 6, 1.

65
00:05:17,430 --> 00:05:22,413
We have essentially one last
row operation to perform.

66
00:05:22,413 --> 00:05:23,560
Let me do it here.

67
00:05:26,300 --> 00:05:31,970
So we'll subtract the second
row from the third one.

68
00:05:31,970 --> 00:05:48,450
And we'll get 1, 1, 1, 2;
0, 1, 2, 1; 0, 0, k minus 5,

69
00:05:48,450 --> 00:05:51,300
and then 0.

70
00:05:51,300 --> 00:05:57,150
And this was achieved by the
elementary matrix E_(3,2),

71
00:05:57,150 --> 00:06:05,780
which was 1, 1, 1,
and then negative 1.

72
00:06:05,780 --> 00:06:10,040
Because we multiplied the
second row by negative 1

73
00:06:10,040 --> 00:06:13,410
and added it to the third one.

74
00:06:13,410 --> 00:06:24,410
So we got to a matrix,
which is upper triangular.

75
00:06:24,410 --> 00:06:30,920
And we want to figure out: for
which value of the parameter k

76
00:06:30,920 --> 00:06:35,730
is this matrix of full rank.

77
00:06:38,310 --> 00:06:39,320
This is a pivot.

78
00:06:39,320 --> 00:06:40,440
This is a pivot.

79
00:06:40,440 --> 00:06:43,200
And we want this one
to be a pivot as well.

80
00:06:43,200 --> 00:06:51,410
And that happens when
k minus 5 is not 0.

81
00:06:51,410 --> 00:06:58,000
So when k is different from 5,
the matrix A is of full rank.

82
00:06:58,000 --> 00:07:02,330
And therefore the system
A*x equals to [2, 3,

83
00:07:02,330 --> 00:07:05,220
7] has a unique solution.

84
00:07:05,220 --> 00:07:14,960
Now part B. For which k do we
have infinitely many solutions?

85
00:07:14,960 --> 00:07:21,160
So when are we in
such a situation?

86
00:07:21,160 --> 00:07:25,750
We are in such a situation when
the null space of the matrix A

87
00:07:25,750 --> 00:07:28,990
is nontrivial.

88
00:07:28,990 --> 00:07:32,450
So the null space
will be nontrivial

89
00:07:32,450 --> 00:07:38,230
when this k minus 5
number here, which

90
00:07:38,230 --> 00:07:42,760
is what's the pivot in
the first case, is 0.

91
00:07:42,760 --> 00:07:50,030
So k minus 5 equals to 0.

92
00:07:50,030 --> 00:07:54,260
You see there's a
little caveat here.

93
00:07:54,260 --> 00:07:57,990
When k is equal to 5,
we get the third row

94
00:07:57,990 --> 00:08:02,760
of the augmented matrix,
0, 0, 0, equal to 0.

95
00:08:02,760 --> 00:08:05,250
This means that the matrix
is actually consistent,

96
00:08:05,250 --> 00:08:10,090
and we indeed have a solution
But if this entry were nonzero,

97
00:08:10,090 --> 00:08:12,000
then we would get no solutions.

98
00:08:15,240 --> 00:08:24,850
Now off to part C. We want to
compute the LU decomposition.

99
00:08:24,850 --> 00:08:30,040
Well, we already got
what the matrix U

100
00:08:30,040 --> 00:08:34,470
is through performing
row operations.

101
00:08:34,470 --> 00:08:35,870
It's this guy here.

102
00:08:35,870 --> 00:08:47,284
Let me write it down-- 1, 1,
1; 0 1, 2; 0, 0, k minus 5.

103
00:08:47,284 --> 00:08:52,370
And when k is equal to 4,
this entry's negative 1.

104
00:08:56,050 --> 00:08:59,330
And now what about the matrix L?

105
00:08:59,330 --> 00:09:02,710
Well, how did we get to this U?

106
00:09:02,710 --> 00:09:07,460
We had the matrix A.
We got the matrix U.

107
00:09:07,460 --> 00:09:11,670
And what we did was
first we applied E_(2,1).

108
00:09:11,670 --> 00:09:16,840
Then we applied E_(3,1)
and then E_(3,2).

109
00:09:24,500 --> 00:09:28,170
We get A by inverting
this equation.

110
00:09:28,170 --> 00:09:38,860
So it's going to be E_(2,1)
inverse E_(3,1) inverse E_(3,2)

111
00:09:38,860 --> 00:09:48,120
inverse times U. And
this is our matrix L.

112
00:09:48,120 --> 00:09:50,710
And we know it's fairly easy
to invert these elementary

113
00:09:50,710 --> 00:09:56,755
matrices E. We flip the signs
of the off-diagonal entries.

114
00:10:01,280 --> 00:10:04,400
I'm not going to
write down again

115
00:10:04,400 --> 00:10:06,570
the inverses of these guys.

116
00:10:09,500 --> 00:10:11,780
I'm just going to write the
product of the inverses.

117
00:10:11,780 --> 00:10:15,480
And that's also very
easy to compute.

118
00:10:15,480 --> 00:10:18,380
Because one, we
invert the signs.

119
00:10:18,380 --> 00:10:28,060
We just send the numbers in
their corresponding entry of L.

120
00:10:28,060 --> 00:10:30,280
I mean the following thing.

121
00:10:30,280 --> 00:10:33,390
So minus 1 becomes a 1.

122
00:10:33,390 --> 00:10:38,945
And it comes here, in
its respective entry.

123
00:10:42,820 --> 00:10:50,720
For E_(3,1), we flip
the sign, first 3,

124
00:10:50,720 --> 00:10:54,350
and we plug it in here.

125
00:10:54,350 --> 00:10:59,570
And for E_(3,2), we flip the
sign of this guy, becomes 1,

126
00:10:59,570 --> 00:11:01,030
and we plug it in here.

127
00:11:03,700 --> 00:11:07,120
So give me a few moments
to erase the board,

128
00:11:07,120 --> 00:11:09,500
and then I'll do part four.

129
00:11:13,520 --> 00:11:18,650
We're back, and we're
going to do part D now.

130
00:11:18,650 --> 00:11:21,640
So we need to find
the complete solution

131
00:11:21,640 --> 00:11:24,130
of the system for all k.

132
00:11:24,130 --> 00:11:26,950
And we saw that
for k equal to 5,

133
00:11:26,950 --> 00:11:30,650
the system had many,
many solutions.

134
00:11:30,650 --> 00:11:34,280
And for k not equal
to 5, it had only one.

135
00:11:34,280 --> 00:11:38,410
So first let's look
at the case k not

136
00:11:38,410 --> 00:11:42,225
equal to 5, when the
matrix A was invertible.

137
00:11:46,840 --> 00:11:51,830
It's not hard to see what the
solution of the system then is.

138
00:11:51,830 --> 00:11:54,960
It's going to be-- let
me just write it down.

139
00:12:00,420 --> 00:12:05,690
So when k is not equal
to 5, this was nonzero.

140
00:12:05,690 --> 00:12:11,520
Therefore, x_3 needs to be 0.

141
00:12:11,520 --> 00:12:16,370
When x_3 is 0, we have
x_2 plus 2 times x_3.

142
00:12:16,370 --> 00:12:20,250
So x_2 plus 0 equals 1.

143
00:12:20,250 --> 00:12:24,050
Therefore x_2 is 1.

144
00:12:24,050 --> 00:12:27,380
And then we go back
to the first row.

145
00:12:27,380 --> 00:12:32,810
We have x_1 plus x_2
plus x_3 equals 2.

146
00:12:32,810 --> 00:12:38,880
So x_1 plus 1 plus 0 equals 2.

147
00:12:38,880 --> 00:12:40,810
So x_1 plus 1 equals 2.

148
00:12:40,810 --> 00:12:43,860
And therefore x_1 is 1.

149
00:12:43,860 --> 00:12:45,600
Good.

150
00:12:45,600 --> 00:12:51,580
Now what about k equal to 5?

151
00:12:51,580 --> 00:12:58,480
Then we see that x_3
is a free variable.

152
00:12:58,480 --> 00:13:04,125
So the solution will
be [x 1,  x 2, x 3].

153
00:13:07,650 --> 00:13:13,800
x_3 can be any number c.

154
00:13:13,800 --> 00:13:16,420
From the second row, we'll
get the value of x_2.

155
00:13:16,420 --> 00:13:19,190
It's 1 minus 2 times x_3.

156
00:13:19,190 --> 00:13:22,430
So 1 minus 2 times c.

157
00:13:22,430 --> 00:13:37,384
And x_1 is 2 minus
x_2 minus x_3.

158
00:13:37,384 --> 00:13:39,480
So let me rewrite this.

159
00:13:39,480 --> 00:13:45,670
It's 2 minus x_2
is 1 minus-- aah,

160
00:13:45,670 --> 00:13:53,051
chalk-- 2c, minus
c, 1 minus 2c, c.

161
00:13:56,760 --> 00:14:00,400
So we'll decompose
this vector here

162
00:14:00,400 --> 00:14:02,615
into a component which
is independent of c

163
00:14:02,615 --> 00:14:06,440
and a component which
is a multiple of c.

164
00:14:06,440 --> 00:14:10,840
So this is 2 minus 1.

165
00:14:10,840 --> 00:14:20,790
It should be [1, 1, 0]
plus c times--

166
00:14:20,790 --> 00:14:25,830
we'll have two c minus
c, c, so c times 1.

167
00:14:25,830 --> 00:14:29,692
Here we'll have minus
2c so negative 2.

168
00:14:29,692 --> 00:14:33,290
And here we'll have 1.

169
00:14:33,290 --> 00:14:38,470
And thus we get the particular
solution for the system

170
00:14:38,470 --> 00:14:43,160
and the special
solution for the system.

171
00:14:43,160 --> 00:14:45,420
We're kind of done here.

172
00:14:45,420 --> 00:14:48,770
If you're at an exam,
you should immediately

173
00:14:48,770 --> 00:14:50,100
start the next problem.

174
00:14:50,100 --> 00:14:52,530
Good luck, and
Ill see you later.