用初等行变换,将矩阵化为阶梯形及行最简形,并求出矩形的秩

1 -1 3 0
-2 1 -2 1
-1 -1 5 2
r2+2r1,r3+r1
1 -1 3 0
0 -1 4 1
0 -2 8 2
r3-2r2
1 -1 3 0
0 -1 4 1
0 0 0 0 这是梯矩阵,r(A)=2.
r2*(-1),r1+r2
1 0 -1 -1
0 1 -4 -1
0 0 0 0 这是行简化梯矩阵
第二题 用初等行变换求逆矩阵
1 0 0 0 1 0 0 0
1 2 0 0 0 1 0 0
2 1 3 0 0 0 1 0
3 2 1 4 0 0 0 1
r2-r1,r3-2r1,r4-3r1
1 0 0 0 1 0 0 0
0 2 0 0 -1 1 0 0
0 1 3 0 -2 0 1 0
0 2 1 4 -3 0 0 1
r4-r2,r2*(1/2),r3-r2
1 0 0 0 1 0 0 0
0 1 0 0 -1/2 1/2 0 0
0 0 3 0 -3/2 -1/2 1 0
0 0 1 4 -2 -1 0 1
r3*(1/3),r4-r3
1 0 0 0 1 0 0 0
0 1 0 0 -1/2 1/2 0 0
0 0 1 0 -1/2 -1/6 1/3 0
0 0 0 4 -3/2 -5/6 -1/3 1
r4*(1/4)
1 0 0 0 1 0 0 0
0 1 0 0 -1/2 1/2 0 0
0 0 1 0 -1/2 -1/6 1/3 0
0 0 0 1 -3/8 -5/24 -1/12 1/4
A^-1 =
1 0 0 0
-1/2 1/2 0 0
-1/2 -1/6 1/3 0
-3/8 -5/24 -1/12 1/4算这个有什么捷径或者有什么方便的算法吗，我觉得算这个好头痛啊～～