Solve the system by finding the reduced row-echelon form of the augmented matrix.

Question

Solve the system by finding the reduced row-echelon form of the augmented matrix.
x+2y-4z=-4

-x-y+3z=2

x+3y-5z=-6

How many solutions are there to this system?

Answer 1

1    2   -4      -4                     1   2   -4     -4                                  1   2   -4      -4

1   -1    3       2                      0  -3   7       6     -r1 + r2                 0   1   -1      -2interchange r2 and r3

1     3   -5      -6                     0   1   -1     -2    -r1 + r2                 0   -3   7       6

 

 

1   0   -2     0    -2r2 +r1          1   0   -2       0                   1   0   0       0     2r3+ r1

0   1   -1    -2                            0   1   -1      -2                  0   1   0       -2      r2 +r3

0   0    4     0      3r2 +r3           0   0   1       0    r3/4        0   0   1       0

 

 

So x=0, y=-2 and z=0 is the unique solution.

 

Okay, now for the right problem.

 

1   2   -4     -4                     1   2   -4    -4                        1   0   -2     0    -2r2 +r1 

-1  -1  3       2                     0   1   -1    -2   r1+r2            0   1   -1     -2

1   3   -5      -6                    0   1   -1    -2   r3-r1             0   0    0      0    r3-r2 

 

We've got a problem. This has infinite solutions.

x = 2z and y = z -2 for any value of z. Note that my answer to the first problem works in this. I'm glad I caught my mistake.

Here are two other answers that will work. Pick z =10. This makes x=20 and y=8. 

Pick z = 3. This makes x=6 and y=1. Check it, these work too.