是否有逆 np.dot 函数? [英] Is there any inverse np.dot function?

问题描述

如果我有两个矩阵 a 和 b，是否有任何函数可以找到矩阵 x，当点乘以 a 时得到 b?寻找python解决方案，用于numpy数组形式的矩阵.

If I have two matrices a and b, is there any function I can find the matrix x, that when dot multiplied by a makes b? Looking for python solutions, for matrices in the form of numpy arrays.

推荐答案

这个求X如A*X=B的问题等价于搜索A的逆"，即一个矩阵如X = Ainverse * B.

This problem of finding X such as A*X=B is equivalent to search the "inverse of A", i.e. a matrix such as X = Ainverse * B.

有关信息，在数学 Ainverse 中注明 A^(-1)(A 的幂 -1"，但您可以说A 逆"; 代替).

For information, in math Ainverse is noted A^(-1) ("A to the power -1", but you can say "A inverse" instead).

在 numpy 中，这是一个内置函数，用于求矩阵的逆a:

In numpy, this is a builtin function to find the inverse of a matrix a:

import numpy as np

ainv = np.linalg.inv(a)

例如参见本教程了解说明.

您需要注意一些矩阵不是可逆的"，最明显的例子(大致)是:

You need to be aware that some matrices are not "invertible", most obvious examples (roughly) are:

• 非方阵
• 表示投影的矩阵

numpy 在某些情况下仍然可以近似某个值.

numpy can still approximate some value in certain cases.

这篇关于是否有逆 np.dot 函数?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！