82. Using the Gaussian method, solve the system of equations

$$
\left\{\begin{array}{l}
3 x+2 y-z=4 \\
2 x-y+3 z=9 \\
x-2 y+2 z=3
\end{array}\right.
$$
Solution. Let's rearrange the third equation to the first position:

$$
\left\{\begin{array}{l}
x-2 y+2 z=3 \\
3 x+2 y+z=4 \\
2 x-y+3 z=9
\end{array}\right.
$$

Write the augmented matrix:

$$
\left(\begin{array}{rrr|r}
1 & -2 & 2 & 3 \\
3 & 2 & -1 & 4 \\
2 & -1 & 3 & 9
\end{array}\right)
$$

To get $a_{21}=a_{31}=0$ in the 1st column, multiply the 1st row by 3 and then by 2, and subtract the results from the 2nd and 3rd rows:

$$
\left(\begin{array}{rrr|r}
1 & -2 & 2 & 3 \\
0 & 8 & -7 & -5 \\
0 & 3 & -1 & 3
\end{array}\right)
$$

Divide the 2nd row by 8, multiply the obtained results by 3, and subtract from the 3rd row:

$$
\left(\begin{array}{rrr|r}
1 & -2 & 2 & 3 \\
0 & 1 & -\frac{7}{8} & -\frac{5}{8} \\
0 & 0 & \frac{13}{8} & \frac{39}{8}
\end{array}\right)
$$

Write the new equivalent system corresponding to the augmented matrix:

$$
\left\{\begin{aligned}
x-2 y+2 z & =3 \\
y-\frac{7}{8} z & =-\frac{5}{8} \\
\frac{13}{8} z & =\frac{39}{8}
\end{aligned}\right.
$$

Performing the backward substitution, we find the unknowns sequentially:

$$
\begin{gathered}
\frac{13}{8} z=\frac{39}{8} ; z=3 \\
y-\frac{7}{8} \cdot 3=-\frac{5}{8} ; y=-\frac{5}{8}+\frac{21}{8}=2 \\
x-2 \cdot 2+2 \cdot 3=3 ; x=3+4-6=1
\end{gathered}
$$

Thus, we get the answer: $(1 ; 2 ; 3)$.