What am I doing wrong solving this system of equations? $$\begin{cases}
2x_1+5x_2-8x_3=8\\
4x_1+3x_2-9x_3=9\\
2x_1+3x_2-5x_3=7\\
x_1+8x_2-7x_3=12
\end{cases}$$
From my elementary row operations, I get that it has no solution. (Row operations are to be read from top to bottom.)
$$\left[\begin{array}{ccc|c}
2 & 5 & -8 & 8  \\
4 & 3 & -9 & 9  \\
2 & 3 & -5 & 7  \\
1 & 8 & -7 & 12
\end{array}\right]
\overset{\overset{\large{R_1\to R_1-R_3}}{{R_2\to R_2-2R_3}}}{\overset{R_3\to R_3-2R_4}{\large\longrightarrow}}
\left[\begin{array}{ccc|c}
0 &  2  & -3 &  1  \\
0 & -3  &  1 & -5  \\
0 & -13 &  9 & -17 \\
1 &  8  & -7 &  12 
\end{array}\right]
\overset{\overset{\large{R_3\,\leftrightarrow\, R_4}}{R_2\,\leftrightarrow\, R_3}}{\overset{R_1\,\leftrightarrow\,R_2}{\large\longrightarrow}}
\left[\begin{array}{ccc|c}
1 &  8  & -7 &  12 \\
0 &  2  & -3 &  1  \\
0 & -3  &  1 & -5  \\
0 & -13 &  9 & -17
\end{array}\right]$$
$$\overset{R_4\to R_4-R_3}{\large\longrightarrow}
\left[\begin{array}{ccc|c}
1 &  8  & -7 &  12 \\
0 &  2  & -3 &  1  \\
0 & -3  &  1 & -5  \\
0 &  10 &  8 & -12
\end{array}\right]
\overset{\overset{\large{R_3\to R_3+R_2}}{R_4\to R_4-5R_2}}{\large\longrightarrow}
\left[\begin{array}{ccc|c}
1 &  8 & -7  &  12 \\
0 &  2 & -3  &  1  \\
0 & -1 & -2  & -4  \\
0 &  0 &  23 & -17
\end{array}\right]
\overset{\overset{\large{R_2\to R_2+2R_3}}{R_3\to-R_3}}{\large\longrightarrow}$$
$$\left[\begin{array}{ccc|c}
1 & 8 & -7  &  12 \\
0 & 0 & -7  & -7  \\
0 & 1 &  2  &  4  \\
0 & 0 &  23 & -17 \\
\end{array}\right]
\overset{R_2\,\leftrightarrow\,R_3}{\large\longrightarrow}
\left[\begin{array}{ccc|c}
1 & 8 & -7  &  12 \\
0 & 1 &  2  &  4  \\
0 & 0 & -7  & -7  \\
0 & 0 &  23 & -17 \\
\end{array}\right]$$
However, the answer in the book $(3, 2, 1)$ fits the system.
Was there an arithmetical mistake, or do I misunderstand something fundamentally?
 A: Hint: Try inputing the solution $(3,2,1)$ into every step. That will allow you to identify the step where you went wrong.
A: You do (in the third matrix): $$L3-L4=(0, -3, 1 \mid -5)-(0, -13, 9 \mid -19)=(0, 10, -8 \mid 12)$$ but you have $(0, 10, 8 \mid -12)$ instead. 
A: Note 1: Referring to 5xum's answer, in case you don't know the final answer, any solution will do, e.g.:
$$(x_1,x_2,x_3)=(0,0,-1);(0,0,-1);(1,0,-1);(12,0,0) \ \ \text{(respectively)}$$
Note 2: In step $3$, you can reduce column $3$ instead of column $2$:
$$\left[
\begin{array}{ccc|c}
  1&8&-7&12\\
  0&2&-3&1\\
  0&-3&1&-5\\
  0&-13&9&-17\\
\end{array}
\right] \Rightarrow \left[
\begin{array}{ccc|c}
  1&8&-7&12\\
  0&-7&0&-14\\
  0&-3&1&-5\\
  0&14&0&28\\
\end{array}
\right] \stackrel{\frac{-R_2}{7};\\ \frac{-3R_2}{7}+R_3}{=}\Rightarrow \left[
\begin{array}{ccc|c}
  1&8&-7&12\\
  0&1&0&2\\
  0&0&1&1\\
  0&14&0&28\\
\end{array}
\right]$$
The second and fourth equations are dependent, therefore they produce the same solution $x_2=2$. You can finish the problem now.
Note 3. When you can not find your mistake (sometimes the brain gets blocked/accustommed and can not see obvious mistakes), leave it for some time (1 hour, 1 day) and return with fresh mind. You will be surprised to easily spot the error.
