非齐次线性方程组 ) 2(x)_(1)-(x)_(2)-(x)_(3)+(x)_(4)=2 (x)_(1)+(x)_(2)-2(x)_(3)+(x)_(4)=4 4(x)_(1)-6(x)_(2)+2(x)_(3)-2 .的通解可表示为 ) 2(x)_(1)-(x)_(2)-(x)_(3)+(x)_(4)=2 (x)_(1)+(x)_(2)-2(x)_(3)+(x)_(4)=4 4(x)_(1)-6(x)_(2)+2(x)_(3)-2 . ) 2(x)_(1)-(x)_(2)-(x)_(3)+(x)_(4)=2 (x)_(1)+(x)_(2)-2(x)_(3)+(x)_(4)=4 4(x)_(1)-6(x)_(2)+2(x)_(3)-2 . ) 2(x)_(1)-(x)_(2)-(x)_(3)+(x)_(4)=2 (x)_(1)+(x)_(2)-2(x)_(3)+(x)_(4)=4 4(x)_(1)-6(x)_(2)+2(x)_(3)-2 .

步骤 1：写出增广矩阵
将非齐次线性方程组写成增广矩阵的形式，即：
$$
\left[\begin{array}{cccc|c}
2 & -1 & -1 & 1 & 2 \\
1 & 1 & -2 & 1 & 4 \\
4 & -6 & 2 & -2 & 0
\end{array}\right]
$$
步骤 2：化简增广矩阵
对增广矩阵进行行变换，化简为阶梯形矩阵。首先，将第一行乘以$\frac{1}{2}$，得到：
$$
\left[\begin{array}{cccc|c}
1 & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & 1 \\
1 & 1 & -2 & 1 & 4 \\
4 & -6 & 2 & -2 & 0
\end{array}\right]
$$
然后，将第二行减去第一行，第三行减去4倍的第一行，得到：
$$
\left[\begin{array}{cccc|c}
1 & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & 1 \\
0 & \frac{3}{2} & -\frac{3}{2} & \frac{1}{2} & 3 \\
0 & -4 & 4 & -4 & -4
\end{array}\right]
$$
接着，将第二行乘以$\frac{2}{3}$，第三行加上$\frac{8}{3}$倍的第二行，得到：
$$
\left[\begin{array}{cccc|c}
1 & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & 1 \\
0 & 1 & -1 & \frac{1}{3} & 2 \\
0 & 0 & 0 & 0 & 0
\end{array}\right]
$$
步骤 3：求解方程组
从化简后的增广矩阵中，可以得到方程组的解。令${x}_{3}=t$，则有：
$$
\begin{cases}
{x}_{1} - \frac{1}{2}{x}_{2} - \frac{1}{2}{x}_{3} + \frac{1}{2}{x}_{4} = 1 \\
{x}_{2} - {x}_{3} + \frac{1}{3}{x}_{4} = 2
\end{cases}
$$
将${x}_{3}=t$代入，得到：
$$
\begin{cases}
{x}_{1} - \frac{1}{2}{x}_{2} - \frac{1}{2}t + \frac{1}{2}{x}_{4} = 1 \\
{x}_{2} - t + \frac{1}{3}{x}_{4} = 2
\end{cases}
$$
解得：
$$
\begin{cases}
{x}_{1} = 4 + t \\
{x}_{2} = 3 + t \\
{x}_{4} = -3
\end{cases}
$$