= Solution
Use <Gaussian elimination> on the <augmented matrix>. First take $R_2\leftarrow R_2-R_1$ and $R_3\leftarrow R_3-2R_1$, then $R_3\leftarrow4R_3-3R_2$:
$$
\left(\begin{array}{ccc|c}1&2&3&14\\1&-2&1&0\\2&1&-1&1\end{array}\right)
\longrightarrow
\left(\begin{array}{ccc|c}1&2&3&14\\0&-4&-2&-14\\0&-3&-7&-27\end{array}\right)
\longrightarrow
\left(\begin{array}{ccc|c}1&2&3&14\\0&-4&-2&-14\\0&0&-22&-66\end{array}\right).
$$
By <back substitution>, the last row gives $z=3$, the second gives $-4y-6=-14$ and hence $y=2$, and the first gives $x+4+9=14$. Thus
$$
\boxed{(x,y,z)=(1,2,3).}
$$
The right-hand side has been transformed in every row operation, preserving the solution set.
Back to article page
