Question

已知矩阵A=+1+2+1+0+2+2+0+0+4+,求矩阵方程+AX=4E+的解X。

Answer 1

AX = 4E, X = 4A^(-1)
(A, E) =
[1 2 1 1 0 0]
[0 2 2 0 1 0]
[0 0 4 0 0 1]
初等行变换为
[1 2 0 1 0 -1/4]
[0 2 0 0 1 -1/2]
[0 0 1 0 0 1/4]
初等行变换为
[1 0 0 1 -1 1/4]
[0 2 0 0 1 -1/2]
[0 0 1 0 0 1/4]
初等行变换为
[1 0 0 1 -1 1/4]
[0 1 0 0 1/2 -1/4]
[0 0 1 0 0 1/4]
A^(-1）=
[1 -1 1/4]
[0 1/2 -1/4]
[0 0 1/4]
X = 4A^(-1) =
[4 -4 1]
[0 2 -1]
[0 0 1]