Question 442935
We will start by making an augmented matrix.

{{{matrix(3,4,3,-1,2,3,2,3,-1,-3,1,-2,3,2)}}}

Let's switch some rows so that we can get leading ones.

{{{matrix(3,4,1,-2,3,2,3,-1,2,3,2,3,-1,-3)}}}

Multiply row 2 and row 3 by -1.

{{{matrix(3,4,1,-2,3,2,-3,1,-2,-3,-2,-3,1,3)}}}

Notice we have leading ones now.

{{{matrix(3,4,highlight(1),-2,3,2,-3,highlight(1),-2,-3,-2,-3,highlight(1),3)}}}


Multiply row 1 by 3 and add it to row 2

{{{matrix(3,4,1,-2,3,2,0,-5,7,3,-2,-3,1,3)}}}

Multiply row 1 by 2 and add it to row 3

{{{matrix(3,4,1,-2,3,2,0,-5,7,3,0,-7,7,7)}}}

Multiply row 3 by -1 and add it to row 2

{{{matrix(3,4,1,-2,3,2,0,2,0,-4,0,-7,7,7)}}}

Divide row 2 by 2, and row 3 by 7.

{{{matrix(3,4,1,-2,3,2,0,1,0,-2,0,-1,1,1)}}}

Multiply row 2 by 2 add it to row 1 ... also add row 2 to row 3.

{{{matrix(3,4,1,0,3,-2,0,1,0,-2,0,0,1,-1)}}}

Multiply row 3 by -3 and add it to row 1.

{{{matrix(3,4,1,0,0,1,0,1,0,-2,0,0,1,-1)}}}

x = 1, y = -2, z = -1.

Let's check:

3x + -y + 2z   = 3(1) +2 -2 = 3
2x + 3y -z = 2(1) +3(-2) -(-1) = 2 -6 +1 = -3
x - 2y + 3z = 1 +4 -3 = 2

So yes, the solution works.