Solve the system by hand using matrices;
x+y−z=2
x +z = 2
y+2z=4
i got:
1 0 1 2
0 1 2 4
0 0 1 1
but my calculator says something different. Please help. Thank you!

Question

Solve the system by hand using matrices;
x+y−z=2 
x +z = 2 
y+2z=4
i got:
1  0  1  2
0  1  2  4 
0  0  1  1 
but my calculator says something different. Please help. Thank you!

anonymous · Answer

Your columns are incorrect.

anonymous · Answer

I think you've done it correctly actually

anonymous · Answer

i did it again and i got 
1  0  1  2 
0  1 -2 0
0  0  1  1
and still not the same answer as the calculator

anonymous · Answer

Rewrite your equations in this format
$$
\begin{array}{cccc}
1x+&1y+&-1z&=2\
1x+&0y+&1z&=2\
0x+&0y+&2z&=4
\end{array}
$$
Now form your determinant.

anonymous · Answer

Above you got the solution x=1, y=2, z=1 right? That is the solution. I don't care what your calculator says :p

anonymous · Answer

@Thomas9, where are your values coming from. . ? @jkd13 hasn't even written the determinant yet to find the values. . .

Also, you should read the Code of Conduct. We are not supposed to post full answers.

anonymous · Answer

jkd13 has already reduced the matrix to upper triangle form. In that form you can just read the solution.

anonymous · Answer

@Thomas9, I understand what we're talking about now. @jkd13 is using Gaussian elimination.

anonymous · Answer

thomas, i don't think that is the answer i am looking for but thank you. 
limitless heres what i did…
I started out with 
1  1  -1  2
1  0  1  2 
0  1  2  4 
i switched row 1&2 
then multiplied -1 to the new row 1+ row 2 
which got me to:
1  0  1  2  
0  1  -2  0
0 1  2  4

anonymous · Answer

then multiplied -1 tho row 2 + row 3 got:
1  0  1  2
0  1  -2  0
0   0  4  4 
divided row 3 by 4

anonymous · Answer

@jkd13, your original 
1  0  1  2 
0  1 -2 0
0  0  1  1
is correct. You can work out the solutions easily now. Start with the bottom row of $z=1$ and work your way up.

anonymous · Answer

1  0  1  2 
0  1 -2 0
0  0  1  1
thats my final answer but i still don't think it is correct...

anonymous · Answer

I had misunderstood what you were doing, @jkd13. That is why I made that comment. I now see what you are doing. You are correct. I checked the implied solutions.

anonymous · Answer

ok thanks for your help!!

anonymous · Answer

You're welcome. Thanks to @Thomas9, as well.

anonymous · Answer

cheers

anonymous · Answer

one more thing if i needed to solve for the values they would be (1,2,1) right?

anonymous · Answer

Yup.

anonymous · Answer

$(x,y,z)=(1,2,1)$

anonymous · Answer

thank you!!