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