SOLUTION: Help! I don't know if I'm doing this right. Use the matrix method. x+y-z = -9 2x-y-2z=-3 x+2y+3z=14 I'm confused I seen one way adding R1+R3 to start and another saying interc

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Question 73188: Help! I don't know if I'm doing this right.
Use the matrix method.
x+y-z = -9
2x-y-2z=-3
x+2y+3z=14
I'm confused I seen one way adding R1+R3 to start and another saying interchange row 1 and 3????? This is what I have done:
R1 1 1 -1 | -9
R2 2 -1 -2 | -3
R3 1 2 3 | 14
R1 1 2 3 | 14
R2 2 -1 -2 | -3
R3 1 1 -1 | -9
R1 1 2 3 | 14
R2 0 -5 -8 |-31
R3 1 1 -1 | -9
R1 1 2 3 | 14
R2 0 -5 -8 |-31
R3 0 -1 -4 |-23
R1 1 2 3 | 14
R2 0 1 8/5 | 31/5
R3 0 -1 -4 |-23
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Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Use the matrix method.
x+y-z = -9
2x-y-2z=-3
x+2y+3z=14
I'm confused I seen one way adding R1+R3 to start and another saying interchange row 1 and 3????? This is what I have done:
R1 1 1 -1 | -9
R2 2 -1 -2 | -3
R3 1 2 3 | 14
OK
-------------------------
R1 1 2 3 | 14
R2 2 -1 -2 | -3
R3 1 1 -1 | -9
ok
---------------------
R1 1 2 3 | 14
R2 0 -5 -8 |-31
R3 1 1 -1 | -9
OK
------------------
R1 1 2 3 | 14
R2 0 -5 -8 |-31
R3 0 -1 -4 |-23
OK
-------------------
Multiply thru R3 by -1 and interchange R2 and R3
R1 1 2 3 | 14
R2 0 1 4 | 23
R3 0 -5 -8 |-31
--------------------
Multiply R2 by 5 and add it to R3 to get
R1 1 2 3 | 14
R2 0 1 4 | 23
R3 0 0 12 | 84
OK
-----------------------
Divide R3 by 12 to solve for "z"
z=7
--------
Substitute z=7 into R2 to solve for "y"
y +4*7 =23
y=-5
-------
Substitute y=-5 and z=7 into R1 to solve for "x"
x + 2*-5 +3*7 = 14
x -10+21 = 14
x = 3
---------
Solution:
x=3 ; y=-5 ; z = 7
==============
Cheers,
Stan H.
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