3 0 -1 设A= 1 4 1 ,求矩阵B,使得AB-2A=2B.1 0 3

题目

3 0 -1 设A= 1 4 1 ,求矩阵B,使得AB-2A=2B.1 0 3

答案

由AB-2A=2B
得 (A-2E)B=2A
(A-2E,A) =
1 0 -1 3 0 -1
1 2 1 1 4 1
1 0 1 1 0 3
r2-r1,r3-r1
1 0 -1 3 0 -1
0 2 2 -2 4 2
0 0 2 -2 0 4
r2*(1/2),r3*(1/2),
1 0 -1 3 0 -1
0 1 1 -1 2 1
0 0 1 -1 0 2
r1+r3,r2-r3
1 0 0 2 0 1
0 1 0 0 2 -1
0 0 1 -1 0 2
2 0 1 4 0 2
所以有 B = 2(A-2E)^-1A= 2* 0 2 -1 = 0 4 -2
-1 0 2 -2 0 4