1
MATH 221 - Fall, 2015
Beer is proof that God loves us and
wants us to be happy. Benjamin Franklin
(
)
↓
Errata (fancy word for screw ups) for chapter 1 and 2. The notation q, p
↑
(
)
down from the top of the page
th
(correction) means on page q on the p line
there is
up from the bottom of the page
this (correction).
page line down or up
5
10 ↓
11
8 ↓
18
10 ↑
29
29
32
32
34
34
40
37
40
41
41
41
46
48
49
49
51
57
57
57
57
57
253
254
correction
after the third equal replace u1 with u2 (twice)
||~v || in (1.7) should be ||~v ||2
+ ± should be +
2
= −1 (missing minus)
4 ↓
−2
13 ↓
(2.2) should be (2.5)
4, 5 ↑
Replace the word vertexes with vertices. See below.
Figure 2.2 Compare below with your text
6 ↓
parallel. (missing period)
Figure 2.4 see the next page
11 ↑
22 − 7x3 . ← remove period
3 ↑
(2.18) 2 (in the matrix) should be a33
11 ↑
22 − 7x3 . ← remove period
5 ↓
Replace (2.18) with (2.19).
8 ↓
Replace (2.24) with (2.17).
9 ↓
Replace x1 + 3x2 + x3 = 8 + 3(−4) + 1 = −3 6= −2
with x1 + 3x2 + x3 = 8 + 3(−4) + 2 = −2 6= 1.
15 ↓
Replace the word right with left.
3 ↑
(2.36) should be (2.19)
3 ↑
(2.36) should be (2.19)
2 ↑
(2.37) b3 should be − 2




0
0


 0 




1 ↑
 0  should be  0 


0
0
3 ↓
Replace 2 with −2 in ~xp
6 ↓
Replace Let with Letting
5 ↑
Replace −1/2 with 1/2
3 ↑
Replace x1 + x2 = 1 with x1 + (1/2)x2 = 1
1 ↑
Replace the [ 1 1 ] in the last row of A with [ 1 1/2 ]
~
4 ↑
b = [ 1 2 3 ]T ~b = is missing
6 ↓
If x2 = −1 then x1 = 1/2
0
7
and finally ... −1/2 − 1 should be −1/2 − 1
2
2.2 Planes and Normals
The geometry used in Figure 2.1 to illustrate the solutions (or lack of solutions) of
equations in two unknowns can be extended to equations in three unknowns. A portion of
the graph (the triangular region) of the plane defined by the equation

 

1
x1

 
~
x1 + 2x2 + 3x3 = N1 · ~x =  2  ·  x2 
=6
3
x3
x3
[0 0 2]T
(1)
~ 1 = [1 2 3]T
N
←− x1 + 2x2 + 3x3 = 6
x2
T
[0 3 0]
[6 0 0]T
x1
Figure 2.2
is shown in Figure 2.2. Notice for example, if x2 = 0, then the graph of the
equation
reads




0
6




x1 + 3x3 = 6 which is the straight line passing through the two vertices  0  and  0 .
2
0
~2
x3 AKA N
A
x1 + 2x2 + 3x3 = 6
A @
@
R
@
@A
2x1 + 5x2 + 4x3 = 4
@ H
HA
@
~
N1
3
@
@
J
line of
intersection
@
@
@
x2
@
@
@
@
@
@
@
x1
@
@
Figure 2.4
The equation of the lines through the other two legs of the triangle are obtained by setting
3


1

~
x1 = 0 (2x2 + 3x3 = 6) and x3 = 0 (x1 + 2x2 = 6), respectively. The vector N1 =  2 
 is
3
called a normal to the plane. It is orthogonal to every vector in the plane. This will be
shown for arbitrary planes. Here are a few more revelations from Thirsty Bud.
page line down or up correction
69
2 ↓
The resulting matrix is m × p. (It should read
m × n. The second m × p in the line.)
69
6 ↓
In part, (The comma is missing.)
69
7 ↓
The resulting matrix is m × p. (It should read
m × n.)
70
7 ↓
The a32 element in the matrix product A3 A4
should read 1(1) + 3(2) + 0(0) = 6.
71
13
↑
[B T AT ]nm , where the product (This is in part 3
of Theorem 3.7. The word where is missing.)
75
5
↓
77
1
↑
85
7
↓
91
8
↑
93
93
1
6
↓
↑
94
95
95
1
4
5
↑
↓
↑
99
100
106
256
11
5
1
7
↑
↑
↓
↑
256
4
↑
Replace b1 by 3 (twice) in the displayed
equation (3.8)
the matrix "A.. (Delete one
of the two periods).
#
αa11 αa12
write B =
delete the αA =
a21
a22
This is in Example 3.21 in the first displayed line.
In the matrix A change a22 = 3 to a22 = 5
and the last row from 3 4 5 to 1 3 1.
will also give (The word give is missing).
The displayed line should
read
"
#
"
#
1
1
1
1
1
1
p~1 = √
p~2 = √
~q1 =
and ~q2 =
||p~1 ||
||p~2 ||
2 −1
2 1
respectively. (The period is missing).
The (3.8) should be (3.6). In Exercise 3.3 part 3.
Definition(3.1) should be Definition(3.4).
In the first line of Exercise 3.5
is the inverse of U (The of is missing).
Definition(2.10ed) should be Definition(2.10)
it’s should be its
Parts 1,2, and 3 should read Parts 1,2,3 and 4
1
1
1
1
replace by in Part 3., replace by in Part 4.
4
2
2
4
The 3, 1 element of L−1 should be 2 (not a 1
The 1,1 and 2,1 elements of DC should be 5 and 9, respectively.
not 1 and 7. Same change in C T DT .
258
264
11 ↓
4/6 ↓
4
“24 hours in a day, 24 beers in a case. Coincidence, I think not...
Stephen Wright
The work of old Thirsty Bud continues. He doesn’t tell me anything anymore.
page line down or up correction
114
8 ↑
satisfy following (It should read: satisfy the following
115 13 ↑
~v , −~v , is unique. (The comma is missing.)
115 10 ↑
In words, Va (The comma is missing.)
116 10 ↓
Leave out the first sentence of Defnition 4.4.
Replace it with: Let S = {~v1 , ~v2 , . . . , ~vm } denote a set of
m elements in a vector space V. Now continue: The vector ~u = . . . .
119
8 ↑
span{~a∗2 , ~a∗2 } should read span{~a∗2 , ~a∗3 }
4 ↑
span{~a∗1 , ~a∗1 } should read span{~a∗1 , ~a∗2 }
121 16 ↑
Occurs in displayed equation (4.12). Replace in CA the
vector [1 2 3 ]T with [1 2 1 ]T and [2 3 4 ]T with [2 5 3 ]T
and in C the vector [0 1 2 ]T with [0 1 1 ]T .0
142
3
↓
142
7/8 ↑
142
146
151
152
154
156
158
158
164
266
268
270
271
271
272
1
6
7
1
3
8
6
11
4
6
2
4
11
10
4
↑
↑
↑
↓
↓
↓
↓
↓
↓
↑
↑
↓
↑
↑
↑
273
4
↓
275
7
↑
276
6
↑
⊥
⊥
Replace R3 = NA ⊕ NA⊥ = NA ⊕ RA
= RA
⊕ RA should read
3
⊥
⊥
R = N (A) ⊕ N (A) = N (A) ⊕ RA = RA ⊕ RA
⊥
⊕ RA
or more simply R3 = N (A) ⊕ N (A)⊥ = RA
⊥
⊥
Replace R4 = NA ⊕ NA⊥ = NA ⊕ RA
= RA
⊕ RA should read
4
⊥
⊥
R = N (A) ⊕ N (A) = N (A) ⊕ RA = RA ⊕ RA
⊥
⊕ RA
or more simply R4 = N (A) ⊕ N (A)⊥ = RA
~v ∈ V ⊥ should read ~v ⊥ ∈ V ⊥
NA should read N (A). Two times in this line.
Need a period at the end of the displayed equation.
parameter in A (the in is missing).
NA should be N (A)
There should be a comma between w
~ 1 and w
~ 2.
Replace parts 1, 2, and 3. by parts 1 and 2.
Replace part 3 by part 2.
A = should read A~x =
a33 = 2 should read a33 = 1.
impliesδ should read implies δ
Need a period at the end of the displayed equation.
Part 2 should read In Part 2,
N (B T b) should read N (B T B)
In this case it is also the solution which ... should read
In this case ~xls is also the traditional solution which ...
normal
equations# reads should read normal equations
"
1 3 5
A=
.The period is missing
2 6 10
−2(2x3 + b2 − 2b1 ) − 3x3 + b1 = −7x3 + 5b1 − 2b2 .The − should be =.
5
“A woman drove me to drink, and I hadn’t even the courtesy to thank her.
W.C. Fields
page line down or up correction
175
4 ↑
A~pj = λj p~j , j = 1, 2 The comma is missing.
194 10 ↑
Displayed equation (5.28): with inverse Q−1 ...
instead of: has the inverse
205
8 ↑
to closely should be too closely.
211
3 ↓
it’s should be its too closely.
215
3 ↑
Theorem(5.11) should be Theorem(5.10)
216 11 ↓
and a22 be real... the word be is missing.
216 13 ↓
delete the second A in equation(5.45)
216 16 ↓
a22 > 0 not a12 > 0
217 12 ↓
The
first#a33 should" be a#23
"
b
b
232
5 ↓
P
= (~u~uT )
, replace a with − a.
−a
−a
232 10 ↑
Need a period at the end of part 1.
288
7 ↓
p~∗1 p~2 = 0 missing = sign
p~1
p~2
8 ↓
the second
should be
||~p1 ||
||~p2 ||
Thirsty Bud’s work is now done.
Math made simple
An amazing formula attributed to Euler is
eπi + 1 = 0
which connects everybodies favorite constants: e is the Naperian base for logarithms, π is
the √
area of a circle of radius 1, 0 is the additive identity, 1 is the multiplicative identity and
i = −1 is the complex unit. Actually, and this really torques me off, the number 2 didn’t
make the formula and, to this day, I’m still pissed off at Euler. After all, the number 2 is
the only even prime. It makes one think that maybe Euler was prejudiced. Prejudice is not
just a 21st (there’s that number 2 again) century phenomena. I mean Euler could have
written
eπi + 1
= 0 or 2 eπi + 1 = 0
2
st
which, and this is a 21 century phenomena, is more inclusive. It’s all just so depressing.