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