I need help understanding special cases in matrices

Question

I need to solve this system of equations using a matrix: x+2y-z=2 , -2x+y-3z=6 , -x+3y-4z=8

 

I wound up with my matrix looking like this before I converted it back into equations again:

 

       1  2  -1   |   2

       0  5  -5   |   10

       0  0   0   |   0

 

 

I was a little confused why the answer in the back of the book said that y is equal to all real numbers. Should it not be z equal to all real numbers?

Answer 1

 

 It really does not matter. You may choose any one of the three to be the free variable.

 

 Suppose we choose y to be the free variable as you said.

 

then per the first equation x + 2y - z = 2

 

                                       x - z = 2 - 2y <---- moves the free var to right side; label this equation ALPHA

 

Per the second equation:   5y - 5z = 10

                                      y - z  = 2 <--- divides everything by 5

 

                                      -z = 2 - y <--- subtracts free var y from both sides

 

                                      z = y - 2 <--- multiplies everything by -1

 

Substituting into equation ALPHA labeled above:

 

  x - z = 2 - 2y

 

  x - (y-2) = 2 - 2y

  x = 2 - 2y + y - 2

  x = -y

 

So in terms of the free variable y:

  x= -y and z = y - 2

 

 

The solution can be checked:

 

1st equation: x + 2y - z = -y + 2y -(y-2) = -y + 2y - y + 2 = -2y + 2y + 2 = 2 <--- 1st equation checks

 

2nd equation: -2x + y - 3z = -2(-y) + y - 3(y-2) = 2y + y - 3y + 6 = 6 <--- 2nd equation checks

 

3rd equation:  -x + 3y - 4z = -(-y) + 3y - 4(y-2) = y + 3y - 4y + 8 = 4y - 4y + 8 = 8 <---- 3rd equation checks out

 

So the solutions in bold above can be used to generate infinitely many solutions to the system