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Chapter 1
Linear Equations and Matrices
Section 1.1, p. 8
2.
x
=1,
y
=2,
z
=
−
2.
4. No solution.
6.
x
=13+10
z
,
y
=
−
8
−
8
z
,
z
= any real number
.
8. No solution.
10.
x
y
=
−
1.
12. No solution.
14.
x
=
−
1,
y
z
=
−
2.
16. (c) Yes. (d) Yes.
18.
x
y
z
=0.
20. There is no such value of
r
.
22. Zero, inﬁnitely many, zero.
24. 1.5 tons of regular and 2.5 tons of special plastic.
26. 20 tons of 2minute developer and a total of 40 tons of 6minute and 9minute developer.
28. $7000, $14,000, $3000.
T.1. The same numbers
s
j
satisfy the system when the
p
th equation is written in place of the
q
th equation
and vice versa.
T.2. If
s
1
,s
2
,...,s
n
is a solution to (2), then the
i
th equation of (2) is satisﬁed:
a
i
1
s
1
+
a
i
2
s
2
+
···
+
a
in
s
n
=
b
i
. Then for any
r
±
=0,
ra
i
1
s
1
+
i
2
s
2
+
+
in
s
n
=
rb
i
. Hence
s
1
2
n
is a solution to the
new system. Conversely, for any solution
s
±
1
±
2
±
n
to the new system,
i
1
s
±
1
+
+
in
s
±
n
=
i
,
and dividing both sides by nonzero
r
we see that
s
±
1
±
n
must be a solution to the original linear
system.
T.3. If
s
1
2
n
is a solution to (2), then the
p
th and
q
th equations are satisﬁed:
a
p
1
s
1
+
a
pn
=
b
p
a
q
1
s
1
+
a
qn
=
b
q
.
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Chapter 1
Thus, for any real number
r
,
(
a
p
1
+
ra
q
1
)
s
1
+
···
+(
a
pn
+
qn
)
s
n
=
b
p
+
rb
q
and so
s
1
,...,s
n
is a solution to the new system. Conversely, any solution to the new system is also
a solution to the original system (2).
T.4. Yes;
x
=0,
y
= 0 is a solution for any values of
a
,
b
,
c
, and
d
.
Section 1.2, p. 19
2.
a
=3,
b
=1,
c
=8,
d
=
−
2.
4. (a)
C
+
E
=
E
+
C
=
⎡
⎣
5
−
58
429
534
⎤
⎦
.
(b) Impossible.
(c)
±
7
−
7
01
²
.
(d)
⎡
⎣
−
93
−
9
−
12
−
3
−
15
−
6
−
3
−
9
⎤
⎦
.
(e)
⎡
⎣
0
−
9
8
−
1
−
2
−
5
−
43
⎤
⎦
.
(f) Impossible.
6. (a)
A
T
=
⎡
⎣
12
21
34
⎤
⎦
,(
A
T
)
T
=
±
123
214
²
.
(b)
⎡
⎣
545
−
523
894
⎤
⎦
.
(c)
±
−
61
0
11
17
²
.
(d)
±
0
−
4
40
²
.
(e)
⎡
⎣
63
91
0
⎤
⎦
.(
f
)
±
17
2
−
16
6
²
.
8. Yes: 2
±
10
²
+1
±
00
²
=
±
30
02
²
.
10.
⎡
⎣
λ
−
1
−
2
−
3
−
6
λ
+2
−
3
−
5
−
2
λ
−
4
⎤
⎦
.
12. (a)
±
11
²
.
(b)
±
²
.
(c)
±
²
.
(d)
±
²
.
(e)
±
²
.
14.
v
=
³
0000
´
.
T.1. Let
A
and
B
each be diagonal
n
×
n
matrices. Let
C
=
A
+
B
,
c
ij
=
a
ij
+
b
ij
.F
o
r
i
±
=
j
,
a
ij
and
b
ij
are each 0, so
c
ij
=0
.Thus
C
is diagonal. If
D
=
A
−
B
,
d
ij
=
a
ij
−
b
ij
, then
d
ij
= 0. Therefore
D
is diagonal.
T.2. Following the notation in the solution of T.1 above, let
A
and
B
be scalar matrices, so that
a
ij
and
b
ij
= 0 for
i
±
=
j
, and
a
ii
=
a
,
b
ii
=
b
.I
f
C
=
A
+
B
and
D
=
A
−
B
, then by Exercise T.1,
C
and
D
are diagonal matrices. Moreover,
c
ii
=
a
ii
+
b
ii
=
a
+
b
and
d
ii
=
a
ii
−
b
ii
=
a
−
b
,so
C
and
D
are scalar matrices.
T.3. (a)
⎡
⎣
0
b
−
cc
−
e
c
−
b
e
−
c
⎤
⎦
.
(b)
⎡
⎣
2
ac
+
be
+
c
b
+
c
2
d
2
e
c
+
e
2
e
2
f
⎤
⎦
.
(c) Same as (b).
T.4. Let
A
=
³
a
ij
´
and
C
=
³
c
ij
´
c
ij
=
ka
ij
f
ij
= 0, then either
k
=0or
a
ij
= 0 for all
i
,
j
.
T.5. (a) Let
A
=
³
a
ij
´
and
B
=
³
b
ij
´
be upper triangular matrices, and let
C
=
A
+
B
. Then for
i>j
,
c
ij
=
a
ij
+
b
ij
= 0 + 0 = 0, and thus
C
is upper triangular. Similarly, if
D
=
A
−
B
, then for
,
d
ij
=
a
ij
−
b
ij
−
0 = 0, so
D
is upper triangular.
Section 1.2
3
(b) Proof is similar to that for (a).
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This note was uploaded on 06/25/2008 for the course MATH 22a taught by Professor Chuchel during the Spring '08 term at UC Davis.
 Spring '08
 chuchel
 Linear Equations, Equations, Matrices

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