使用eig函数，复模态正交的问题
动力方程使用状态空间法求解复模态；(Ar+B)Q=0;A=[ 7.6623e+000 0 -1.1931e+0008.6491e-001 02.0000e+000 0 -2.0000e+0001.0000e+000 007.6623e+0002.7733e+000 08.6491e-001 02.0000e+0003.0000e+000 01.0000e+000
-1.1931e+0002.7733e+0002.2559e+001 -2.2526e+0003.3790e+000 -2.0000e+0003.0000e+0001.4000e+001 -2.0000e+0003.0000e+000
8.6491e-001 0 -2.2526e+0001.2649e+000 01.0000e+000 0 -2.0000e+0001.0000e+000 0
08.6491e-0013.3790e+000 01.2649e+000 01.0000e+0003.0000e+000 01.0000e+000
2.0000e+000 0 -2.0000e+0001.0000e+000 0 0 0 0 0 0
02.0000e+0003.0000e+000 01.0000e+000 0 0 0 0 0
-2.0000e+0003.0000e+0001.4000e+001 -2.0000e+0003.0000e+000 0 0 0 0 0
1.0000e+000 0 -2.0000e+0001.0000e+000 0 0 0 0 0 0
01.0000e+0003.0000e+000 01.0000e+000 0 0 0 0 0];
B=[ 1.5675e+003 0 -9.0588e+0026.8377e+002 0 0 0 0 0 0
01.5675e+0032.2423e+003 06.8377e+002 0 0 0 0 0
-9.0588e+0022.2423e+0031.3609e+004 -1.7809e+0032.6713e+003 0 0 0 0 0
6.8377e+002 0 -1.7809e+0031.0000e+003 0 0 0 0 0 0
06.8377e+0022.6713e+003 01.0000e+003 0 0 0 0 0
0 0 0 0 0 -2.0000e+000 02.0000e+000 -1.0000e+000 0
0 0 0 0 0 0 -2.0000e+000 -3.0000e+000 0 -1.0000e+000
0 0 0 0 02.0000e+000 -3.0000e+000 -1.4000e+0012.0000e+000 -3.0000e+000
0 0 0 0 0 -1.0000e+000 02.0000e+000 -1.0000e+000 0
0 0 0 0 0 0 -1.0000e+000 -3.0000e+000 0 -1.0000e+000
];
=eig(B,-A);
x'*x
模态结果不正交，求解答，恳请大家能给出解释！
求解答。。。。。
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