Section 4-2

(26) 2 R1 ---> R1

(28) 2/3R2 + R1 ---> R1

(30) 4R1 + R2 ---> R2

(38) Solve the system:

2x1
+
x2
=
0
x1
-
2x2
=
-5

[
2
1
|
0
 ]
1
-2
|
-5
R1 <--->R2
[
1
-2
|
-5
 ]
2
1
|
0
-2R1+R2--->R2

[
1
-2
|
-5
 ]
0
5
|
10
1/5R2--->R2
[
1
-2
|
-5
 ]
0
1
|
2
2R2+R1--->R1

[
1
0
|
-1
 ]
0
1
|
2
so the
solution is
x1
=
-1
x2
=
2

(40) Solve the system:

2x1
-
3x2
=
-2
-4x1
+
6x2
=
7

[
2
-3
|
-2
 ]
-4
6
|
7
2R1+R2--->R2
[
2
-3
|
-2
 ]
0
0
|
3

The second row of the final matrix corresponds to the equation 0 = 3. Thus the system has no solutions.

(46) Solve the system:

-6x1
+
2x2
=
4
3x1
-
-1x2
=
-2

First we put this into matrix form, below. We could start with multiplying the first row by -1/6 because that would give a 1 in the upper left. However, we would end up with fractions for the other entries in the frist row. Thus, let's start with getting a 0 under the 6.

[
-6
2
|
4
 ]
3
-1
|
-2
1/2R1+R2--->R2
[
-6
2
|
4
 ]
0
0
|
0
-1/6R1--->R1

[
1
-1/3
|
-2/3
 ]
0
0
|
0
This corresponds
to the system
x1
-
1/3x2
=
-2/3
0x1
+
0x2
=
0

Since any pair solves the second equation, the solutions to the system will be those pairs which solve the first. Thus the solution is

x1
=
1/3t
-
2/3
x2
=
t