Section 4-3

(26) Solve the system:

3x1
+
5x2
-
x3
=
-7
x1
+
x2
+
x3
=
-1
2x1
+
11x3
=
7

First, we transform the system to an augmented matrix, and then reduce the matrix. In our first step, we switch two rows to get a 1 at the top left corner.

 

[
3
5
-1
|
-7
1
1
1
|
-1
2
0
11
|
7
]
R1 <--->R2
[
1
1
1
|
-1
3
5
-1
|
-7
2
0
11
|
7
]

-3R1+R2--->R2

-2R1+R3--->R3

 

[
1
1
1
|
-1
0
2
-4
|
-4
0
-2
9
|
9
]

1/2R2--->R2
[
1
1
1
|
-1
0
1
-2
|
-2
0
-2
9
|
9
]

-R2+ R1---> R1

2R2+R3--->R3

 

[
1
0
3
|
1
0
1
-2
|
-2
0
0
5
|
5
]

1/5R3--->R3
[
1
0
3
|
1
0
1
-2
|
-2
0
0
1
|
1
]

-3R3+R1---> R1

2R3+R2--->R2

 

[
1
0
0
|
-2
0
1
0
|
0
0
0
1
|
1
]

This reduced matrix corresponds to the following system, which gives the solution.

x1
=
-2
x2
=
0
x3
=
1

 

(30) Solve the system

2x1
+
4x2
-
6x3
=
10
3x1
+
3x2
-
3x3
=
6

 

[
2
4
-6
|
10
 ]
3
3
-3
|
6
1/3R2 --->R2
[
2
4
-6
|
10
 ]
1
1
-1
|
2
R1<--->R2

 

[
1
1
-1
|
2
 ]
2
4
-6
|
10
-2R1+R2--->R2
[
1
1
-1
|
2
 ]
0
2
-4
|
6
1/2R2--->R2

 

[
1
1
-1
|
2
 ]
0
1
-2
|
3
-R2+ R1---> R1
[
1
0
1
|
-1
 ]
0
1
-2
|
3
(reduced)

Since the matrix is reduced we can now write down the solution to the original system. The reduced matrix corresponds to the system

x1
+
x3
=
-1
x2
-
2x3
=
3
or
x1
=
-1
-
x3
x2
=
3
+
2x3
so the solutions are
x1= -1-t
x2= 3+2t
x3= t

 

 

(32) Solve the system

2x1
-x2
=
0
3x1
+2x2
=
7
x1
-x2
=
-2

 

[
2
-1
|
0
 ]
3
2
|
7
1
-1
|
-2
R1 <---->R3
[
1
-1
|
-2
 ]
3
2
|
7
2
-1
|
-0

-3R1+R2--->R2

-2R1+R3--->R3

 

[
1
-1
|
-2
 ]
0
5
|
13
0
1
|
4
-5R3+R2--->R2
[
1
-1
|
-2
 ]
0
0
|
-7
0
1
|
4

 

Even though the last matrix is not yet reduced, we can stop, because the second row corresponds to the equation 0x1+0x2=-7, or 0=-7. Therefore it is impossible for the system to have any solutions. NO SOLUTIONS