3.1
6) The following manipulations give the solution set.
1
2
E
1
!
E
1
,
3
E
1
+
E
2
!
E
2
This show that the system is inconsistent and there is no solution.
14) The following manipulations give the solution set.
E
1
,
E
2
!
E
1
(
putsa
1
asthef ir stel ement
)
,
3
E
1
+
E
2
!
E
2
,
2
E
1
+
E
3
!
E
3
9
E
2
,
23
E
3
!
E
3
Toget
z
=
,
4 and back substitution gives
y
= 3 and
x
=5.
22) The following manipulations give the solution set.
E
1
$
E
2
(
putsa
1
asthef ir stel ement
)
,
4
E
1
+
E
2
!
E
2
,
2
E
1
+
E
3
!
E
3
The second rowis a multiple of the third rowsothat wepick
z
to be arbitrary
z
=
t
. Back substituting gives
y
=
,
5
t
and
x
=
,
4
t
for some
t
2
R
.
26) Plugging in the two initial conditions yields the system
A
+
B
= 44
11
A
,
11
B
= 22
The following manipulations give the solution set.
1
=
11
E
2
!
E
2
,
E
1
+
E
2
!
E
2
,
1
=
2
E
2
!
E
2
This shows that
B
=21and
A
=23. The particular solution is
y
(
x
)=23
e
11
x
+21
e
,
11
x
.
3.2
14) Werow reduce the following augmented matrix
2
4
3
,
6
,
21
2
,
4 1 17
1
,
2
,
2
,
9
3
5
using the following elementary row operations.
R
1
$
R
3
,
2
R
1
+
R
2
!
R
2
,
3
R
1
+
R
3
!
R
3
We see that
x
3
= 7 and that
R
3
is a multiple of
R
2
,sowe pick the free variable
x
2
=
t
. Back substituting we get that
x
1
=2
t
+5.
20) Werow reduce the following augmented matrix
2
4
2 4
,
1
,
2 2 6
1 3 2
,
7 3 9
5 8
,
7 6 1 4
3
5
1
