Question

我的矩阵跟其逆矩阵交换相乘结果不一样

根据逆矩阵定义，我在R中得到一个矩阵x的逆矩阵ix, 然后交换相乘结果是不一样呢，为什么
ix <- solve(x)
x %*% ix
[,1] [,2] [,3] [,4]
[1,] 1.000000e+00 0.000000e+00 1.110223e-16 -5.551115e-17
[2,] -6.661338e-16 1.000000e+00 8.881784e-16 -1.665335e-16
[3,] 0.000000e+00 0.000000e+00 1.000000e+00 -1.665335e-16
[4,] -4.440892e-16 -2.220446e-16 4.440892e-16 1.000000e+00

ix %*% x
[,1] [,2] [,3] [,4]
[1,] 1.000000e+00 -6.661338e-16 -2.220446e-16 0.000000e+00
[2,] 8.326673e-17 1.000000e+00 1.110223e-16 1.665335e-16
[3,] -2.220446e-16 -2.220446e-16 1.000000e+00 -2.220446e-16
[4,] 4.440892e-16 -5.551115e-17 1.665335e-16 1.000000e+00

Answer 1

原式=xcoslnx-∫xdcoslnx+c
=xcoslnx-∫x(-sinlnx*1/x)dx+c
=xcoslnx+∫sinlnxdx+c
=xcoslnx+xsinlnx-∫xdsinlnx+c
=xcoslnx+xsinlnx-∫coslnxdx+c
故2∫coslnxdx=xcoslnx+xsinlnx
所以∫coslnxdx=1/2（xcoslnx+xsinlnx）+c