Question

求一道线代题

Answer 1

X[E-C^(-1)B]^TC^T = E, 两边转置得
C[E-C^(-1)B]X^T = E
(C-B)X^E = E, 两边转置得
X(C-B)^T = E
X = [(C-B)^T]^(-1)
[(C-B)^T , E] =
[1 0 0 0 1 0 0 0]
[2 1 0 0 0 1 0 0]
[3 2 1 0 0 0 1 0]
[4 3 2 1 0 0 0 1]
初等行变换为
[1 0 0 0 1 0 0 0]
[0 1 0 0 -2 1 0 0]
[0 2 1 0 -3 0 1 0]
[0 3 2 1 -4 0 0 1]
初等行变换为
[1 0 0 0 1 0 0 0]
[0 1 0 0 -2 1 0 0]
[0 0 1 0 -3 -2 1 0]
[0 0 2 1 -4 -3 0 1]
初等行变换为
[1 0 0 0 1 0 0 0]
[0 1 0 0 -2 1 0 0]
[0 0 1 0 -3 -2 1 0]
[0 0 0 1 -4 -3 -2 1]
X = [(C-B)^T]^(-1) =
[ 1 0 0 0]
[-2 1 0 0]
[-3 -2 1 0]
[-4 -3 -2 1]