# Please give me the steps Solve the system of linear equations, using the Gauss-Jordan elimination method. (If there is no solution, enter NO SOLUTION. If there are infinitely many solutions, express your answer in terms of the parameters t and/or s.) x+ 2y + z = -2; -2x -3y - z = 3; 3x + 6y + 3z = -6

The system is consistent and dependent and has the solutions (t,-t-1,t) for all real t. (An alternative is (t,-t,t)+(0,-1,0).) Use an augmented matrix in row-echelon form to solve this problem. We are asked to solve the system given below:

x+2y+z=-2
-2x-3y-z=3
3x+6y+3z=-6

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:

`([1,2,1,|,-2],[-2,-3,-1,|,3],[3,6,3,|,-6])`

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:

`=>([1,2,1,|,-2],[0,1,1,|,-1],[0,0,0,|,0])`

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

`([1,0,-1,|,0],[0,1,1,|,-1],[0,0,0,|,0])`

Let z=t.

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.