We will use Gauss-Jordan elimination to determine the solution set of this linear system

We will use Gauss-Jordan elimination to determine the solution set of this linear system.

STEP 1: multiply the first row by -2 and add it to the second row.  Do the same for the third row.

STEP 2: Multiply row 2 by 1/5 to get a 1 in the pivot position

STEP 3: Add the second row to the first to eliminate the entry above the pivot.  Multiply the second row by -4 and add it to the third to eliminate the entry below the pivot

STEP 4: Multiply the third row by -5/14 to get a 1 in the pivot position

STEP 5: Multiply the third row by 4/5 and add it to the second row to eliminate the entry above the pivot.  Multiply the third row by -6/5 and add it to the first row to do the same

We now have a matrix in Reduced Row Echelon Form.

Let us suppose that the columns corresponded to variables x1, x2, x3, and x4.

Then we have one free variable, x4

We set it to an arbitrary parameter, m1

And we have the following equations for one representation of this system's solutions

• x4 = m1 • x1 = 8/7•  x2 = -3/7•  x3 = 3/14 - 1/2m1

• x4 = m1 • x1 = 8/7•  x2 = -3/7•  x3 = 3/14 - 1/2m1

We may now express the system's solutions as a set of vectors.  

Solutions are thus linear combinations of the above two vectors and for any value of our free variable m1.  That is, when we have free variables, there are infinitely many solutions to the system.