PHP逆矩陣

Question

我看到了this question，並彈出這個想法。

有沒有一種有效的方式來做到這一點在PHP？

編輯

最佳效果演示？

Answer 1

您可以使用梨包Math_Matrix爲此。

Answer 2

這個package claims能夠做你正在尋找的東西。

Answer 3

有這個開源的PHP Library是能夠反轉矩陣。

所有你需要做的是

<?php 
include_once ("Matrix.class.php"); 
$matrixA = new Matrix(array(array(0, 1), array(2, 6))); 
echo $matrixA->getInverse()->getMathMl(); 
?> 

Answer 4

/** 
* matrix_inverse 
* 
* Matrix Inverse 
* Guass-Jordan Elimination Method 
* Reduced Row Eshelon Form (RREF) 
* 
* In linear algebra an n-by-n (square) matrix A is called invertible (some 
* authors use nonsingular or nondegenerate) if there exists an n-by-n matrix B 
* such that AB = BA = In where In denotes the n-by-n identity matrix and the 
* multiplication used is ordinary matrix multiplication. If this is the case, 
* then the matrix B is uniquely determined by A and is called the inverse of A, 
* denoted by A-1. It follows from the theory of matrices that if for finite 
* square matrices A and B, then also non-square matrices (m-by-n matrices for 
* which m ? n) do not have an inverse. However, in some cases such a matrix may 
* have a left inverse or right inverse. If A is m-by-n and the rank of A is 
* equal to n, then A has a left inverse: an n-by-m matrix B such that BA = I. 
* If A has rank m, then it has a right inverse: an n-by-m matrix B such that 
* AB = I. 
* 
* A square matrix that is not invertible is called singular or degenerate. A 
* square matrix is singular if and only if its determinant is 0. Singular 
* matrices are rare in the sense that if you pick a random square matrix over 
* a continuous uniform distribution on its entries, it will almost surely not 
* be singular. 
* 
* While the most common case is that of matrices over the real or complex 
* numbers, all these definitions can be given for matrices over any commutative 
* ring. However, in this case the condition for a square matrix to be 
* invertible is that its determinant is invertible in the ring, which in 
* general is a much stricter requirement than being nonzero. The conditions for 
* existence of left-inverse resp. right-inverse are more complicated since a 
* notion of rank does not exist over rings. 
*/ 
public function matrix_inverse($m1) 
{ 
    $rows = $this->rows($m1); 
    $cols = $this->columns($m1); 
    if ($rows != $cols) 
    { 
     die("Matrim1 is not square. Can not be inverted."); 
    } 

    $m2 = $this->eye($rows); 

    for ($j = 0; $j < $cols; $j++) 
    { 
     $factor = $m1[$j][$j]; 
     if ($this->debug) 
     { 
      fms_writeln('Divide Row [' . $j . '] by ' . $m1[$j][$j] . ' (to 
                give us a "1" in the desired position):'); 
     } 
     $m1 = $this->rref_div($m1, $j, $factor); 
     $m2 = $this->rref_div($m2, $j, $factor); 
     if ($this->debug) 
     { 
      $this->disp2($m1, $m2); 
     } 
     for ($i = 0; $i < $rows; $i++) 
     { 
      if ($i != $j) 
      { 
       $factor = $m1[$i][$j]; 
       if ($this->debug) 
       { 
        $this->writeln('Row[' . $i . '] - ' . number_format($factor, 4) . ' × 
               Row[' . $j . '] (to give us 0 in the desired position):'); 
       } 
       $m1 = $this->rref_sub($m1, $i, $factor, $j); 
       $m2 = $this->rref_sub($m2, $i, $factor, $j); 
       if ($this->debug) 
       { 
        $this->disp2($m1, $m2); 
       } 
      } 
     } 
    } 
    return $m2; 
} 

Answer 5

這裏測試的代碼https://gist.github.com/unix1/7510208 只有identity_matrix（）和反轉（）函數足夠