大一线性代数第二题，用高斯消元法解该线性方程组，麻烦大家把过程写一下，我刚开始学还不太会，谢谢了。

A =
1 -2 3 -4 4
0 1 -1 1 -3
1 3 0 -3 1
0 -7 3 1 -3

A =
1 -2 3 -4 4
0 1 -1 1 -3
0 5 -3 1 -3
0 -7 3 1 -3

A =
1 -2 3 -4 4
0 1 -1 1 -3
0 0 2 -4 12
0 0 -4 8 -24

A =
1 -2 3 -4 4
0 1 -1 1 -3
0 0 2 -4 12
0 0 0 0 0

A =
1 -2 3 -4 4
0 1 -1 1 -3
0 0 1 -2 6
0 0 0 0 0

A =
1 -2 0 2 -14
0 1 0 -1 3
0 0 1 -2 6
0 0 0 0 0

A =
1 0 0 0 -8
0 1 0 -1 3
0 0 1 -2 6
0 0 0 0 0

所以：
x1 = -8
x2 = 3 + x4
x3 = 6 + 2x4
(x4为自由变量)