在Matlab中求解线性方程 [英] Solving Linear Equation in Matlab

问题描述

我想在Matlab中求解#n个线性方程组AX=bi(对于#nb而言)，其中b循环变化，而A是常数.

I wanna solve #n linear equations AX=bi(for #n b's) in Matlab which b changes in a loop and A is constant.

一种快速的方法是在循环之前计算A的逆，并且在循环体内只需从inv(A)*b获得X，但是由于矩阵A是奇异的，所以我感到很糟糕回答！ 当然，数值解A/b提供了一个很好的答案，但要点是，在#n循环中计算#n不同的X的时间花费了很长的时间.

One way which is fast, is to compute the inverse of A before the loop and in the loop body just get X from inv(A)*b, but because the matrix A is singular, I get an awful answer! Of course, the numerical solution A/b gives a good answer, but the point is that it takes a long time to compute #n different X's in #n loops.

我想要的是一种既准确又快速的解决方案.

What I want is a solution which can be both accurate and fast.

推荐答案

我实际上认为这是一个好问题，除了错别字和矩阵奇异性问题.有几种解决方法，Tim Davis在MATLAB Central上的 factorize提交覆盖了所有角度.

I actually think this is a good question, typos and issues of matrix singularity aside. There are a few good ways to handle this, and Tim Davis' factorize submission on MATLAB Central covers all the angles.

但是，仅供参考，让我们在本机MATLAB中自行进行操作，从A是正方形的情况开始.首先，有两种建议的方法(inv和\，mldivide):

However, just for reference, let's do it on our own in native MATLAB, starting with the case where A is square. First, there are the two methods you suggested (inv and \,mldivide):

% inv, slow and inacurate
xinvsol = inv(A)*b;
norm(A*xinvsol - b ,'fro')

% mldivide, faster and accurate
xref = A\b;
norm(A*xref - b ,'fro')

但是，如果像您说的那样A不变，只需分解A并求解新的b即可！假设A是对称正定:

But if like you said A does not change, just factorize A and solve for new b! Say A is symmetric positive definite:

L = chol(A,'lower'); % Cholesky factorization

% mldivide, much faster (not counting the chol factorization) and most accurate
xcholbs= L'\(L\b); %'
norm(A*xcholbs - b ,'fro')

% linsolve, fastest (omits checks for matrix configuration) and most accurate
sol1 = linsolve(L, b, struct('LT',true));
xcholsolv = linsolve(L, sol1, struct('LT',true,'TRANSA',true));
norm(A*xcholsolv - b ,'fro')

如果A不是对称正定的，则​​将LU分解用于方阵，否则将QR用于.同样，您可以自己完成所有操作，也可以只使用 Tim Davis出色的分解函数.

If A is not symmetric positive definite, then you'd use LU decomposition for a square matrix or QR otherwise. Again, you can do it all yourself, or you can just use Tim Davis' awesome factorize functions.

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