Question

第三题 求 线性代数

Answer 1

（3） （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]
行初等变换为
[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 -2 -2 0 -1 0 0 1]
行初等变换为
[1 1 1 1 1 0 0 0]
[0 -2 -2 0 -1 0 0 1]
[0 0 -2 -2 -1 1 0 0]
[0 -2 0 -2 -1 0 1 0]
行初等变换为

[1 0 0 1 1/2 0 0 1/2]
[0 1 1 0 1/2 0 0 -1/2]
[0 0 1 1 1/2 -1/2 0 0]
[0 0 2 -2 0 0 1 -1]
行初等变换为
[1 0 0 1 1/2 0 0 1/2]
[0 1 0 -1 0 1/2 0 -1/2]
[0 0 1 1 1/2 -1/2 0 0]
[0 0 0 -4 -1 1 1 -1]
行初等变换为
[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]
A^(-1) = (1/4)*
[1 1 1 1]
[1 1 -1 -1]
[1 -1 1 -1]
[1 -1 -1 1]