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inv(A)*B vs A\B - 为什么在 MatLab 中出现这种奇怪的行为?

[英]inv(A)*B vs A\B - Why this weird behavior in MatLab?

Lets create two random matrices,让我们创建两个随机矩阵,

A = randn(2)
B = randn(2)

both inv(A)*B and A\B give the same result inv(A)*BA\B都给出相同的结果

inv(A)*B
A\B

ans =

    0.6175   -2.1988
   -0.7522    5.0343

ans =

    0.6175   -2.1988
   -0.7522    5.0343

unless I multiply with some factor.除非我乘以某个因素。 Why is this?为什么是这样?

.5*A\B
.5*inv(A)*B

ans =

    1.2349   -4.3977
   -1.5045   10.0685

ans =

    0.3087   -1.0994
   -0.3761    2.5171

This is very annoying since MatLab always nudges me to use A\B instead of inv(A)*B and it took me years to figure out why my code was not working.这很烦人,因为 MatLab 总是让我使用A\B而不是inv(A)*B并且我花了好几年才弄清楚为什么我的代码不起作用。

When A is non-singular matrix , then inv(A) * B = A \ B .A是非奇异矩阵时,则inv(A) * B = A \ B

Your calculation is as follows: .5 * A\B = (0.5 * A) \ B vs .5* inv(A) * B = 0.5 * (A\B) .您的计算如下: .5 * A\B = (0.5 * A) \ B vs .5* inv(A) * B = 0.5 * (A\B) As such, it will give your unequal result.因此,它会给你带来不平等的结果。

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