$$\begin{cases}
2x_1+5x_2-8x_3=8\\
4x_1+3x_2-9x_3=9\\
2x_1+3x_2-5x_3=7\\
x_1+8x_2-7x_3=12
\end{cases}$$
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?
Best Answer
Hint: Try inputing the solution $(3,2,1)$ into every step. That will allow you to identify the step where you went wrong.