线性代数矩阵的一道求解题设A=101020101矩阵X满足AX+E=A^2+X求X

Question

【线性代数】矩阵的一道求解题
设A= 1 0 1
0 2 0
1 0 1
矩阵X满足AX+E=A^2+X
求X

Answer 1

∵A^2=A*A=
0 4 0
2 0 2
2 0 2
又∵A^2与X必须是同型矩阵才能相加
∴设:X=
a b c
d e f
g h i
亦知E=
1 0 0
0 1 0
0 0 1
∴AX+E=
a+g+1 ,b+h ,c+f
2d ,2*e+1 ,2*f
a+g ,b+h ,c+f+1
A^2+X=
a+2 b c+2
d e+4 f
g+2 h i+2
∵AX+E=A^2+X
∴ a+g+1=a+2 => g=1
b+h=b => h=0
c+i=c+2 => i=2
2*d=d => d=0
2*e+1=e+4 => e=3
2*f=f => f=0
a+g=g+2 => a=2
b+h=h => b=0
c+i+1=i+2 => c=1
∴X=
2 0 1
0 3 0
1 0 2