There are no solutions.
14. The system reduces to
x
+
2
y
=

2
z
=
2
→
x
=

2

2
y
z
= 2
.
Let
y
=
t
.
x
y
z
=

2

2
t
t
2
15. The system reduces to
x
=
4
y
=
2
z
=
1
.
16. The system reduces to
x
1
+ 2
x
2
+ 3
x
3
+5
x
5
=
6
x
4
+2
x
5
=
7
→
x
1
= 6

2
x
2

3
x
3

5
x
5
x
4
= 7

2
x
5
.
Let
x
2
=
r, x
3
=
s
, and
x
5
=
t
.
17
Chapter 1
ISM:
Linear Algebra
x
1
x
2
x
3
x
4
x
5
=
6

2
r

3
s

5
t
r
s
7

2
t
t
17. The system reduces to
x
1
=

8221
4340
x
2
=
8591
8680
x
3
=
4695
434
x
4
=

459
434
x
5
=
699
434
.
18. a. No, since the third column contains two leading ones.
b. Yes
c. No, since the third row contains a leading one, but the second row does not.
d. Yes
19.
0
0
0
0
and
1
0
0
0
20. Four, namely
0
0
0
0
,
1
k
0
0
,
0
1
0
0
,
1
0
0
1
(
k
is an arbitrary constant.)
21. Four, namely
0
0
0
0
0
0
,
1
k
0
0
0
0
,
0
1
0
0
0
0
,
1
0
0
1
0
0
(
k
is an arbitrary constant.)
22. Seven, namely
0
0
0
0
0
0
,
1
a
b
0
0
0
,
0
1
c
0
0
0
,
0
0
1
0
0
0
,
1
0
d
0
1
e
,
1
f
0
0
0
1
,
0
1
0
0
0
1
.
Here,
a, b, . . . , f
are arbitrary constants.
23. We need to show that the matrix has the three properties listed on page 16.
18
ISM:
Linear Algebra
Section 1.2
Property a holds by Step 2 of the GaussJordan algorithm (page 17).
Property b holds by Step 3 of the GaussJordan algorithm.
Property c holds by Steps 1 and 4 of the algorithm.
24. Yes; each elementary row operation is reversible, that is, it can be “undone.” For example,
the operation of row swapping can be undone by swapping the same rows again.
The
operation of dividing a row by a scalar can be reversed by multiplying the same row by
the same scalar.
25. Yes; if
A
is transformed into
B
by a sequence of elementary row operations, then we can
recover
A
from
B
by applying the inverse operations in the reversed order (compare with
Exercise 24).
26. Yes, by Exercise 25, since rref(
A
) is obtained from
A
by a sequence of elementary row
operations.
27. No; whatever elementary row operations you apply to
1
2
3
4
5
6
7
8
9
, you cannot make the
last column equal to zero.
28. Suppose (
c
1
, c
2
, . . . , c
n
) is a solution of the system
a
11
x
1
+
a
12
x
2
+
· · ·
+
a
1
n
x
n
=
b
1
a
21
x
1
+
a
22
x
2
+
· · ·
+
a
2
n
x
n
=
b
2
. . . . . . . . .
.
To keep the notation simple, suppose we add
k
times the first equation to the second;
then the second equation of the new system will be (
a
21
+
ka
11
)
x
1
+
· · ·
+(
a
2
n
+
ka
1
n
)
x
n
=
b
2
+
kb
1
.
We have to verify that (
c
1
, c
2
, . . . , c
n
) is a solution of this new equation. Indeed, (
a
21
+
ka
11
)
c
1
+
· · ·
+ (
a
2
n
+
ka
1
n
)
c
n
=
a
21
c
1
+
· · ·
+
a
2
n
c
n
+
k
(
a
11
c
1
+
· · ·
+
a
1
n
c
n
) =
b
2
+
kb
1
.
We have shown that any solution of the “old” system is also a solution of the “new.” To
see that, conversely, any solution of the new system is also a solution of the old system,
note that elementary row operations are reversible (compare with Exercise 24); we can
obtain the old system by subtracting
k
times the first equation from the second equation
of the new system.