Gaussian Elimination, Question Check. I'm going through my practice problems, and just want to know if I am doing this right:
$$ 2x_1 - 3x_2 = -2$$
$$ 2x_1 + x_2 = 1$$
$$ 3x_1 + 2x_2 = 1$$
And this is my solution:
$$
\begin{align}
\begin{bmatrix}
2 & 3 & -2\\
2 & 1 & 1\\
3 & 2 & 1
\end{bmatrix}
\end{align}
$$
$$
\begin{align}
\begin{bmatrix}
1 & -3/2 & -1\\
2 & 1 & 1\\
3 & 2 & 1
\end{bmatrix}
\end{align}
$$
$$
\begin{align}
\begin{bmatrix}
1 & -3/2 & -1\\
0 & -1 & -1\\
3 & 2 & 1
\end{bmatrix}
\end{align}
$$
$$
\begin{align}
\begin{bmatrix}
1 & -3/2 & -1\\
0 & -1 & -1\\
0 & -1 & -2
\end{bmatrix}
\end{align}
$$
$$
\begin{align}
\begin{bmatrix}
1 & -3/2 & -1\\
0 & 1 & 1\\
0 & -1 & -2
\end{bmatrix}
\end{align}
$$
$$
\begin{align}
\begin{bmatrix}
1 & -3/2 & -1\\
0 & 1 & 1\\
0 & 0 & -1
\end{bmatrix}
\end{align}
$$
Can anyone tell me if I did this right, or if I did a mistake where? Thanks.
 A: In your second matrix, assuming you meant to divide by 2 in the first row, you incorrectly negated the second entry. If you meant to divide by $-2$, then the first entry should be negated, and the third entry should be positive.
In your third matrix, when you doubled row 1 and added to row 2, your entry in the middle row, middle column should be $-3 + 1 =-2$. I'm afraid then, that the error will "trickle down" in later reductions using that row. 
Likewise, in the fourth matrix, when you tripled row 1 and added to row 3, your middle value in the bottom row should be $-\frac 92 + 2 = -\frac 52$.
$$
\begin{align}
\begin{bmatrix}
2 & 3 & -2\\
2 & 1 & 1\\
3 & 2 & 1
\end{bmatrix}
\end{align}
$$
Subtract row 1 from row 2:
$$
\begin{align}
\begin{bmatrix}
2 & 3 & -2\\
0 & -2 & 3\\
3 & 2 & 1
\end{bmatrix}
\end{align}
$$
1/2 row 1:
$$
\begin{align}
\begin{bmatrix}
1 & 3/2 & -1\\
0 & -2 & 3\\
3 & 2 & 1
\end{bmatrix}
\end{align}
$$
Add $-3 R1$ to row 3
$$
\begin{align}
\begin{bmatrix}
1 & 3/2 & -1\\
0 & -2 &  3\\
0 & -5/2 & 4
\end{bmatrix}
\end{align}
$$
-1/2 Row 2, 2 times Row 3:
$$
\begin{align}
\begin{bmatrix}
1 & 3/2 & -1\\
0 & 1 & -3/2\\
0 & -5 & 8
\end{bmatrix}
\end{align}
$$
5 Row 2 + row 3 = row 3
$$
\begin{align}
\begin{bmatrix}
1 & 3/2 & -1\\
0 & 1 & 1\\
0 & 0 & 1/2
\end{bmatrix}
\end{align}
$$
We can now multiply the last row by 2:
$$
\begin{align}
\begin{bmatrix}
1 & 3/2 & -1\\
0 & 1 & 1\\
0 & 0 & 1
\end{bmatrix}
\end{align}
$$
A: first  consider $2x_1 - 3x_2 = -2$ and $2x_1 + x_2 = 1$ your solution is $x_1=\frac{1}{8} , x_2=\frac{3}{4}$ 
now consider :
$2x_1 + x_2 = 1$ and $3x_1 + 2x_2 = 1$ your solution is $x_1=1,x_2=-1$ 
so your system does not an answer
and 3 th row of  your last matrix \begin{align}
$\begin{bmatrix}
1 & -3/2 & -1\\
0 & 1 & 1\\
0 & 0 & -1
\end{bmatrix}
\end{align}$
say that your matrix  has not answer too  
($ 0x_1+0x_2=1 $ is impossible)
