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 solution to the system is x=0, y=-31, z=60. It can be found using a matrix in row-echelon form.

We are asked to solve the following system using Gauss-Jordan elimination:

x+2y+z=-2
-2x-3y-z=33
x+6y+3z=-6

We create the matrix equation:

`([1,2,1],[-2,-3,-1],[1,6,3])([x],[y],[z])=([-2],[33],[-6])`

To use Gaussian elimination, we create an augmented matrix (the coefficient matrix augmented with the solution vector) and then use elementary row/column operations to reduce the matrix to row-echelon form.

`([1,2,1,|,-2],[-2,-3,-1,|,33],[1,6,3,|,-6])`

To begin we want a "1" in the upper left corner, with zeros underneath. We can take R2=2R1+R2 (two times row 1 + row 2) and R3=R3-R1:

`([1,2,1,|,-2],[0,1,1,|,29],[0,4,2,|,-4])`

Now we need the middle element of row 2 to be 1 with zero underneath it. Let R3=-4R2+R3

`([1,2,1,|,-2],[0,1,1,|,29],[0,0,-2,|,-120])`

Now let R3=-1/2(R3) so that the element is 1.

`([1,2,1,|,-2],[0,1,1,|,29],[0,0,1,|,60])`

This is in row-echelon form. We could solve the system from here using back substitution. If we continue and put it in reduced row-echelon form we will be able to read the answer directly. (Note that we have z=60, so y+60=29 => y=-31 and x+2(-31)+60=-2 => x=0.)

To put in reduced row-echelon form each column should have zeros above the ones. Let R2=R2-R3:

`([1,2,1,|,-2],[0,1,0,|,-31],[0,0,1,|,60])`

Let R1=-2R2+R1

`([1,0,1,|,60],[0,1,0,|,-31],[0,0,1,|,60])`

Now let R1=R1-R3

`([1,0,0,|,0],[0,1,0,|,-31],[0,0,1,|,60])`

So `([1,2,1],[-2,-3,-1],[1,6,3])([0],[-31],[60])=([-2],[33],[-6])`

and x=0, y=-31, z=60.