1.
由 AX+I=A^2+X 得 (A-I)X = A^2-I = (A-I)(A+I)
因为 A-I =
0 0 -1
1 2 0
0 2 0
可逆 (行列式 = -2)
所以 X = A+I =
2 0 -1
1 4 0
0 2 2
2.
1 -1 0
0 1 -1
0 0 1
r2+r3, r1+r2 即化为
1 0 0
0 1 0
0 0 1
3.
2 2 3
0 -2 -3/2
0 0 1/4
r3*4
2 2 3
0 -2 -3/2
0 0 1
r1-3r3, r2+3/2r3
2 2 0
0 -2 0
0 0 1
r1*(1/2), r2*(-1/2)
1 1 0
0 1 0
0 0 1
r1-r2
1 0 0
0 1 0
0 0 1