Show that the system has an infinite number of solutions.
And we get: . . . . . . . .
This is true, regardless of the values of and .
Therefore, there are an infinite number of solutions.
Perform some elementary row operations on the augmented matrix to get to reduced row echelon form.
Divide R1 by 3
Multiply R1 by -6 and add to R2
If there is a row in the augmented matrix containing all zeros (to the left and right of the line), then it is as if that row could be completely ignored, as it represents 0 = 0, which by itself is inconclusive. Whether or not there are solutions depends on the remaining row. The remaining row consists of an equation in 2 non-zero variables. We have more variables than we have equations, thus indicating an infinite number of solutions.