We are asked to solve the system given below:
We can see by inspection that if the system is consistent, there will be infinite solutions; the third equation is a multiple of the first equation. We proceed using Gauss-Jordan elimination:
We create the augmented matrix: the 3x3 coefficient matrix augmented with the solution vector:
We want to put this in reduced row-echelon form using elementary row operations. To begin we need a 1 in the upper left position with zeros underneath it.
Let R1=R1; R2=2R1+R2; R3=-3R1+R3 where R2 is row 2:
The third row indicates the system is consistent (there are solutions) and dependent (the number of solutions is infinite.)
The augmented matrix is in row-echelon form. To be in reduced row-echelon form we need zeros in the columns above the leading 1.
Let R3=R3; R2=R2; R1=-2R2+R1
Then y+z=-1 => y+t=-1 or y=-t-1
x-z=0 => x-t=0 or x=t.
Thus the general solution is (t,-t-1,t) for all real values of t.
An alternative expression is (t,-t,t)+(0,-1,0). This is the parametric equation of the line where the "three" planes intersect.
So, for example, (5,-6, 5) is a solution: 5+2(-6) +5=-2 and-2(5)-3(-6)-5=3.