matrix-inverse

问题 I am new to world of matrix, sorry for this basic question I could not figure out: I have four matrix (one unknown). Matrix X x <- c(44.412, 0.238, -0.027, 93.128, 0.238, 0.427, -0.193, 0.673, 0.027, -0.193, 0.094, -0.428, 93.128, 0.673, -0.428, 224.099) X <- matrix(x, ncol = 4 ) Matrix B : need to be solved , 1 X 4 (column x nrows), with b1, b2, b3, b4 values Matrix G g <- c(33.575, 0.080, -0.006, 68.123, 0.080, 0.238, -0.033, 0.468, -0.006, -0.033, 0.084, -0.764, 68.123, 0.468, -0.764, 205

问题 I am developing an algorithm which solves Ax = b , where A and b are known. There are two ways to do this x = A -1 b or using Cholesky. I know the matrix will always be square and positive definite although the det( A ) maybe zero. In those rare cases I can just ignore it. But from a computation point of view and efficiency, is creating an inverse matrix too inefficient? 回答1: In general, you always want to use a solver; the actual solver should run about as fast as multiplying by an inverse.

问题 In a numerical solver I am working on in C, I need to invert a 2x2 matrix and it then gets multiplied on the right side by another matrix: C = B . inv(A) I have been using the following definition of an inverted 2x2 matrix: a = A[0][0]; b = A[0][1]; c = A[1][0]; d = A[1][1]; invA[0][0] = d/(a*d-b*c); invA[0][1] = -b/(a*d-b*c); invA[1][0] = -c/(a*d-b*c); invA[1][1] = a/(a*d-b*c); In the first few iterations of my solver this seems to give the correct answers, however, after a few steps things

问题 Normally I would invert an array of 3x3 matrices in a for loop like in the example below. Unfortunately for loops are slow. Is there a faster, more efficient way to do this? import numpy as np A = np.random.rand(3,3,100) Ainv = np.zeros_like(A) for i in range(100): Ainv[:,:,i] = np.linalg.inv(A[:,:,i]) 回答1: It turns out that you're getting burned two levels down in the numpy.linalg code. If you look at numpy.linalg.inv, you can see it's just a call to numpy.linalg.solve(A, inv(A.shape[0]).

问题 Normally I would invert an array of 3x3 matrices in a for loop like in the example below. Unfortunately for loops are slow. Is there a faster, more efficient way to do this? import numpy as np A = np.random.rand(3,3,100) Ainv = np.zeros_like(A) for i in range(100): Ainv[:,:,i] = np.linalg.inv(A[:,:,i]) 回答1: It turns out that you're getting burned two levels down in the numpy.linalg code. If you look at numpy.linalg.inv, you can see it's just a call to numpy.linalg.solve(A, inv(A.shape[0]).

问题 updated on this question: I have closed this question and I will post a new question focus on R package Rmpfr. To conclude this question and to help others, I will post my codes of the inverse of a Vandermonde Matrix from its explicit inverse formula. The generation terms are the x's in [here]1. I am not a skilled programmer. Therefore I don't expect my codes to be the most efficient one. I post the codes here because it is better than nothing. library(gtools) #input is the generation vector

问题 What is the fastest algorithm (a link to C or C++ example would be cool) to check if a small square matrix (<16*16 elements) is singular (non-invertible, det = 0) ? 回答1: Best way is to compute the condition number via SVD and check if it is greater than 1 / epsilon, where epsilon is the machine precision. If you allow false negatives (ie. a matrix is defective, but your algorithm may not detect it), you can use the max(a_ii) / min(a_ii) formula from the Wikipedia article as a proxy for the

问题 I'd like to take the modular inverse of a matrix like [[1,2],[3,4]] mod 7 in Python. I've looked at numpy (which does matrix inversion but not modular matrix inversion) and I saw a few number theory packages online, but nothing that seems to do this relatively common procedure (at least, it seems relatively common to me). By the way, the inverse of the above matrix is [[5,1],[5,3]] (mod 7). I'd like Python to do it for me though. 回答1: A hackish trick which works when rounding errors aren't an

问题 I have a badly conditioned matrix, whose rcond() is close to zero, and therefore, the inverse of that matrix does not come out to be correct. I have tried using pinv() but that does not solve the problem. This is how I am taking the inverse: X = (A)\(b); I looked up for a solution to this problem and found this link (last solution) for improving the matrix. The solution there suggests to use this: A_new = A_old + c*eye(size(A_old)); Where c > 0 . So far employing this technique works in

问题 What's the easiest way to compute a 3x3 matrix inverse? I'm just looking for a short code snippet that'll do the trick for non-singular matrices, possibly using Cramer's rule. It doesn't need to be highly optimized. I'd prefer simplicity over speed. I'd rather not link in additional libraries. 回答1: Why don't you try to code it yourself? Take it as a challenge. :) For a 3×3 matrix (source: wolfram.com) the matrix inverse is (source: wolfram.com) I'm assuming you know what the determinant of a