Question

线性代数求逆矩阵

1题
[ 1 -1 1]
[-2 2 1]
[ 2 0 1]
2题
[ 1 1 1 1]
[ 1 1 -1 -1]
[ 1 -1 1 -1]
[ 1 -1 -1 1]
乘以标准矩阵
经过初等变换求逆矩阵
要步骤

Answer 1

(1) 解: (A,E) =
1 -1 1 1 0 0
-2 2 1 0 1 0
2 0 1 0 0 1

r2+2r1,r3-2r1
1 -1 1 1 0 0
0 0 3 2 1 0
0 2 -1 -2 0 1

r2*(1/3),r3*(1/2)
1 -1 1 1 0 0
0 0 1 2/3 1/3 0
0 1 -1/2 -1 0 1/2

r1-r2,r3+(1/2)r2
1 -1 0 1/3 -1/3 0
0 0 1 2/3 1/3 0
0 1 0 -2/3 1/6 1/2

r1+r3
1 0 0 -1/3 -1/6 1/2
0 0 1 2/3 1/3 0
0 1 0 -2/3 1/6 1/2

r2<->r3
1 0 0 -1/3 -1/6 1/2
0 1 0 -2/3 1/6 1/2
0 0 1 2/3 1/3 0

所以 A^-1 =
-1/3 -1/6 1/2
-2/3 1/6 1/2
2/3 1/3 0

(2) 解: (A,E) =
1 1 1 1 1 0 0 0
1 1 -1 -1 0 1 0 0
1 -1 1 -1 0 0 1 0
1 -1 -1 1 0 0 0 1

r4-r3,r3-r2,r2-r1
1 1 1 1 1 0 0 0
0 0 -2 -2 -1 1 0 0
0 -2 0 -2 -1 0 1 0
0 0 -2 2 0 0 -1 1

r4-r2
1 1 1 1 1 0 0 0
0 0 -2 -2 -1 1 0 0
0 -2 0 -2 -1 0 1 0
0 0 0 4 1 -1 -1 1

r2*(-1/2),r3*(-1/2),r4*(1/4)
1 1 1 1 1 0 0 0
0 0 1 1 1/2 -1/2 0 0
0 1 0 1 1/2 0 -1/2 0
0 0 0 1 1/4 -1/4 -1/4 1/4

ri-r4,i=2,3,4
1 1 1 0 3/4 1/4 1/4 -1/4
0 0 1 0 1/4 -1/4 1/4 -1/4
0 1 0 0 1/4 1/4 -1/4 -1/4
0 0 0 1 1/4 -1/4 -1/4 1/4

r1-r2-r3
1 0 0 0 1/4 1/4 1/4 1/4
0 0 1 0 1/4 -1/4 1/4 -1/4
0 1 0 0 1/4 1/4 -1/4 -1/4
0 0 0 1 1/4 -1/4 -1/4 1/4

r2<->r3
1 0 0 0 1/4 1/4 1/4 1/4
0 1 0 0 1/4 1/4 -1/4 -1/4
0 0 1 0 1/4 -1/4 1/4 -1/4
0 0 0 1 1/4 -1/4 -1/4 1/4

所以 A^-1 =
1/4 1/4 1/4 1/4
1/4 1/4 -1/4 -1/4
1/4 -1/4 1/4 -1/4
1/4 -1/4 -1/4 1/4