Cramer's rule

Cramer's rule
You are encouraged to solve this task according to the task description, using any language you may know.
In linear algebra, Cramer's rule is an explicit formula for the solution of a system of linear equations with as many equations as unknowns, valid whenever the system has a unique solution. It expresses the solution in terms of the determinants of the (square) coefficient matrix and of matrices obtained from it by replacing one column by the vector of right hand sides of the equations.

Given

${\displaystyle \left\{{\begin{matrix}a_{1}x+b_{1}y+c_{1}z&={\color {red}d_{1}}\\a_{2}x+b_{2}y+c_{2}z&={\color {red}d_{2}}\\a_{3}x+b_{3}y+c_{3}z&={\color {red}d_{3}}\end{matrix}}\right.}$

which in matrix format is

${\displaystyle {\begin{bmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{bmatrix}}{\begin{bmatrix}x\\y\\z\end{bmatrix}}={\begin{bmatrix}{\color {red}d_{1}}\\{\color {red}d_{2}}\\{\color {red}d_{3}}\end{bmatrix}}.}$

Then the values of ${\displaystyle x,y}$ and ${\displaystyle z}$ can be found as follows:

${\displaystyle x={\frac {\begin{vmatrix}{\color {red}d_{1}}&b_{1}&c_{1}\\{\color {red}d_{2}}&b_{2}&c_{2}\\{\color {red}d_{3}}&b_{3}&c_{3}\end{vmatrix}}{\begin{vmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{vmatrix}}},\quad y={\frac {\begin{vmatrix}a_{1}&{\color {red}d_{1}}&c_{1}\\a_{2}&{\color {red}d_{2}}&c_{2}\\a_{3}&{\color {red}d_{3}}&c_{3}\end{vmatrix}}{\begin{vmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{vmatrix}}},{\text{ and }}z={\frac {\begin{vmatrix}a_{1}&b_{1}&{\color {red}d_{1}}\\a_{2}&b_{2}&{\color {red}d_{2}}\\a_{3}&b_{3}&{\color {red}d_{3}}\end{vmatrix}}{\begin{vmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{vmatrix}}}.}$

Given the following system of equations:

${\displaystyle {\begin{cases}2w-x+5y+z=-3\\3w+2x+2y-6z=-32\\w+3x+3y-z=-47\\5w-2x-3y+3z=49\\\end{cases}}}$

solve for ${\displaystyle w}$, ${\displaystyle x}$, ${\displaystyle y}$ and ${\displaystyle z}$, using Cramer's rule.

ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.8.3.win32

Uses the non-standard DET operator available in Algol 68G.

# returns the solution of a.x = b via Cramer's rule                    ##         this is for REAL arrays, could define additional operators   ##         for INT, COMPL, etc.                                         #PRIO CRAMER = 1;OP   CRAMER = ( [,]REAL a, []REAL b )[]REAL:     IF 1 UPB a /= 2 UPB a     OR 1 LWB a /= 2 LWB a     OR 1 UPB a /=   UPB b     THEN        # the array sizes and bounds do not match                       #        print( ( "Invaid parameters to CRAMER", newline ) );        stop     ELIF REAL deta = DET a;          det a = 0     THEN        # a is singular                                                 #        print( ( "Singular matrix for CRAMER", newline ) );        stop     ELSE        # the arrays have matching bounds                               #        [ LWB b : UPB b ]REAL result;        FOR col FROM LWB b TO UPB b DO            # form a matrix from a with the col'th column replaced by b #            [ 1 LWB a : 1 UPB a, 2 LWB a : 2 UPB a ]REAL m := a;            m[ : , col ] := b[ : AT 1 ];            # col'th result elemet as per Cramer's rule                 #            result[ col ] := DET m / det a        OD;        result     FI; # CRAMER # # test CRAMER using the matrix and column vector specified in the task  #[,]REAL a = ( (  2, -1,  5,  1 )            , (  3,  2,  2, -6 )            , (  1,  3,  3, -1 )            , (  5, -2, -3,  3 )            );[]REAL  b = (  -3            , -32            , -47            ,  49            );[]REAL  solution = a CRAMER b;FOR c FROM LWB solution TO UPB solution DO    print( ( " ", fixed( solution[ c ], -8, 4 ) ) )OD;print( ( newline ) )
Output:
   2.0000 -12.0000  -4.0000   1.0000


C

#include <math.h>#include <stdio.h>#include <stdlib.h> typedef struct {    int n;    double **elems;} SquareMatrix; SquareMatrix init_square_matrix(int n, double elems[n][n]) {    SquareMatrix A = {        .n = n,        .elems = malloc(n * sizeof(double *))    };    for(int i = 0; i < n; ++i) {        A.elems[i] = malloc(n * sizeof(double));        for(int j = 0; j < n; ++j)            A.elems[i][j] = elems[i][j];    }     return A;} SquareMatrix copy_square_matrix(SquareMatrix src) {    SquareMatrix dest;    dest.n = src.n;    dest.elems = malloc(dest.n * sizeof(double *));    for(int i = 0; i < dest.n; ++i) {        dest.elems[i] = malloc(dest.n * sizeof(double));        for(int j = 0; j < dest.n; ++j)            dest.elems[i][j] = src.elems[i][j];    }     return dest;} double det(SquareMatrix A) {    double det = 1;     for(int j = 0; j < A.n; ++j) {        int i_max = j;        for(int i = j; i < A.n; ++i)            if(A.elems[i][j] > A.elems[i_max][j])                i_max = i;         if(i_max != j) {            for(int k = 0; k < A.n; ++k) {                double tmp = A.elems[i_max][k];                A.elems[i_max][k] = A.elems[j][k];                A.elems[j][k]     = tmp;            }             det *= -1;        }         if(abs(A.elems[j][j]) < 1e-12) {            puts("Singular matrix!");            return NAN;        }         for(int i = j + 1; i < A.n; ++i) {            double mult = -A.elems[i][j] / A.elems[j][j];            for(int k = 0; k < A.n; ++k)                A.elems[i][k] += mult * A.elems[j][k];        }    }     for(int i = 0; i < A.n; ++i)        det *= A.elems[i][i];     return det;} void deinit_square_matrix(SquareMatrix A) {    for(int i = 0; i < A.n; ++i)        free(A.elems[i]);    free(A.elems);} double cramer_solve(SquareMatrix A, double det_A, double *b, int var) {    SquareMatrix tmp = copy_square_matrix(A);    for(int i = 0; i < tmp.n; ++i)        tmp.elems[i][var] = b[i];     double det_tmp = det(tmp);    deinit_square_matrix(tmp);     return det_tmp / det_A;} int main(int argc, char **argv) {#define N 4    double elems[N][N] = {        { 2, -1,  5,  1},        { 3,  2,  2, -6},        { 1,  3,  3, -1},        { 5, -2, -3,  3}    };    SquareMatrix A = init_square_matrix(N, elems);     SquareMatrix tmp = copy_square_matrix(A);    int det_A = det(tmp);    deinit_square_matrix(tmp);     double b[] = {-3, -32, -47, 49};     for(int i = 0; i < N; ++i)        printf("%7.3lf\n", cramer_solve(A, det_A, b, i));     deinit_square_matrix(A);    return EXIT_SUCCESS;}
Output:
  2.000
-12.000
-4.000
1.000

Common Lisp

(defun minor (m col)  (loop with dim = (1- (array-dimension m 0))        with result = (make-array (list dim dim))        for i from 1 to dim        for r = (1- i)        do (loop with c = 0                 for j to dim                 when (/= j col)                   do (setf (aref result r c) (aref m i j))                      (incf c))        finally (return result))) (defun det (m)  (assert (= (array-rank m) 2))  (assert (= (array-dimension m 0) (array-dimension m 1)))  (let ((dim (array-dimension m 0)))    (if (= dim 1)        (aref m 0 0)        (loop for col below dim              for sign = 1 then (- sign)              sum (* sign (aref m 0 col) (det (minor m col))))))) (defun replace-column (m col values)  (let* ((dim (array-dimension m 0))         (result (make-array (list dim dim))))    (dotimes (r dim result)      (dotimes (c dim)        (setf (aref result r c)              (if (= c col) (aref values r) (aref m r c))))))) (defun solve (m v)  (loop with dim = (array-dimension m 0)        with det = (det m)        for col below dim        collect (/ (det (replace-column m col v)) det))) (solve #2A((2 -1  5  1)           (3  2  2 -6)           (1  3  3 -1)           (5 -2 -3  3))       #(-3 -32 -47 49))
Output:
(2 -12 -4 1)

EchoLisp

 (lib 'matrix)(string-delimiter "")(define (cramer A B (X)) ;; --> vector X	(define ∆ (determinant A))	(for/vector [(i (matrix-col-num A))]		(set! X (matrix-set-col! (array-copy A) i B))		(// (determinant X) ∆))) (define (task)	(define A (list->array   	'( 2 -1 5 1 3 2 2 -6 1 3 3 -1 5 -2 -3 3) 4 4))	(define B #(-3 -32 -47 49))	(writeln "Solving A * X = B")	(array-print A)	(writeln "B = " B)	(writeln "X = " (cramer A B)))
Output:
(task)
Solving A * X = B
2   -1   5    1
3   2    2    -6
1   3    3    -1
5   -2   -3   3
B =      #( -3 -32 -47 49)
X =      #( 2 -12 -4 1)


Fortran

In Numerical Methods That Work (Usually), in the section What not to compute, F. S. Acton remarks "...perhaps we should be glad he didn't resort to Cramer's rule (still taught as the practical method in some high schools) and solve his equations as the ratios of determinants - a process that requires labor proportional to N! if done in the schoolboy manner. The contrast with N3 can be startling!" And further on, "Having hinted darkly at my computational fundamentalism, it is probably time to commit to a public heresy by denouncing recursive calculations. I have never seen a numerical problem arising from the physical world that was best calculated by a recursive subroutine..."

Since this problem requires use of Cramer's rule, one might as well be hung for a sheep instead of a lamb, so the traditions of Old Fortran and heavy computation will be ignored and the fearsome RECURSIVE specification employed so that the determinants will be calculated recursively, all the way down to N = 1 even though the N = 2 case is easy. This requires F90 and later. Similarly, the MODULE protocol will be employed, even though there is no significant context to share. The alternative method for calculating a determinant involves generating permutations, a tiresome process.

Array passing via the modern arrangements of F90 is a source of novel difficulty to set against the slight convenience of not having to pass an additional parameter, N. Explicitly, at least. There are "secret" additional parameters when an array is being passed in the modern way, which are referred to by the new SIZE function. Anyway, for an order N square matrix, the array must be declared A(N,N), and specifically not something like A(100,100) with usage only of elements up to N = 7, say, because the locations in storage of elements in use would be quite different from those used by an array declared A(7,7). This means that the array must be re-declared for each different size usage, a tiresome and error-inviting task. One-dimensional arrays do not have this problem, but they do have to be "long enough" so B and X might as well be included. This also means that the auxiliary matrices needed within the routines have to be made the right size, and fortunately they can be declared in a way that requests this without the blather of ALLOCATE, this being a protocol introduced by Algol in the 1960s. Unfortunately, there is no scheme such as in pl/i to declare AUX "like" A, so some grotesquery results. And in the case of function DET which needs an array of order N - 1, when its recursion bottoms out with N = 1 it will have declared MINOR(0,0), a rather odd situation that fortunately evokes no complaint, and a test run in which its "value" was written out by WRITE (6,*) MINOR produced a blank line: no complaint there either, presumably because zero elements were being sent forth and so there was no improper access of ... nothing.

With matrices, there is a problem all the way from the start in 1958. Everyone agrees that a matrix should be indexed as A(row,column) and that when written out, rows should run down the page and columns across. This is unexceptional and the F90 function MATMUL follows this usage. However, Fortran has always stored its array elements in what is called "column major" order, which is to say that element A(1,1) is followed by element A(2,1) in storage, not A(1,2). Thus, if an array is written (or read) by something like WRITE (6,*) A, consecutive elements, written along a line, will be A(1,1), A(2,1), A(3,1), ... So, subroutine SHOWMATRIX is employed to write the matrix out in the desired form, and to read the values into the array, an explicit loop is used to place them where expected rather than just READ(INF,*) A

Similarly, if instead a DATA statement were used to initialise the array for the example problem, and it looked something like
      DATA A/2, -1,  5,  1     1       3,  2,  2, -6     2       1,  3,  3, -1     3       5, -2, -3,  3/
(ignoring integer/floating-point issues) thus corresponding to the layout of the example problem, there would need to be a statement A = TRANSPOSE(A) to obtain the required order. I have never seen an explanation of why this choice was made for Fortran.
      MODULE BAD IDEA	!Employ Cramer's rule for solving A.x = b...       INTEGER MSG	!Might as well be here.       CONTAINS		!The details.        SUBROUTINE SHOWMATRIX(A)	!With nice vertical bars.         DOUBLE PRECISION A(:,:)	!The matrix.         INTEGER R,N			!Assistants.          N = SIZE(A, DIM = 1)		!Instead of passing an explicit parameter.          DO R = 1,N			!Work down the rows.            WRITE (MSG,1) A(R,:)		!Fling forth a row at a go.    1       FORMAT ("|",<N>F12.3,"|")		!Bounded by bars.          END DO			!On to the next row.        END SUBROUTINE SHOWMATRIX	!Furrytran's default order is the transpose.         RECURSIVE DOUBLE PRECISION FUNCTION DET(A)	!Determine the determinant.         DOUBLE PRECISION A(:,:)	!The square matrix, order N.         DOUBLE PRECISION MINOR(SIZE(A,DIM=1) - 1,SIZE(A,DIM=1) - 1)	!Order N - 1.         DOUBLE PRECISION D		!A waystation.         INTEGER C,N			!Steppers.          N = SIZE(A, DIM = 1)		!Suplied via secret parameters.          IF (N .LE. 0) THEN		!This really ought not happen.            STOP "DET: null array!"		!But I'm endlessly suspicious.          ELSE IF (N .NE. SIZE(A, DIM = 2)) THEN	!And I'd rather have a decent message            STOP "DET: not a square array!"	!In place of a crashed run.          ELSE IF (N .EQ. 1) THEN	!Alright, now get on with it.            DET = A(1,1)		!This is really easy.          ELSE				!But otherwise,            D = 0			!Here we go.            DO C = 1,N			!Step along the columns of the first row.              CALL FILLMINOR(C)			!Produce the auxiliary array for each column.              IF (MOD(C,2) .EQ. 0) THEN		!Odd or even case?                D = D - A(1,C)*DET(MINOR)	!Even: subtract.               ELSE				!Otherwise,                D = D + A(1,C)*DET(MINOR)	!Odd: add.              END IF				!So much for that term.            END DO			!On to the next.            DET = D		!Declare the result.          END IF	!So much for that.         CONTAINS	!An assistant.           SUBROUTINE FILLMINOR(CC)	!Corresponding to A(1,CC).           INTEGER CC	!The column being omitted.           INTEGER R	!A stepper.           DO R = 2,N		!Ignoring the first row,             MINOR(R - 1,1:CC - 1) = A(R,1:CC - 1)	!Copy columns 1 to CC - 1. There may be none.             MINOR(R - 1,CC:) = A(R,CC + 1:)		!And from CC + 1 to N. There may be none.           END DO		!On to the next row.          END SUBROUTINE FILLMINOR	!Divide and conquer.        END FUNCTION DET	!Rather than mess with permutations.         SUBROUTINE CRAMER(A,X,B)	!Solve A.x = b, where A is a matrix...Careful! The array must be A(N,N), and not say A(100,100) of which only up to N = 6 are in use.         DOUBLE PRECISION A(:,:)	!A square matrix. I hope.         DOUBLE PRECISION X(:),B(:)	!Column matrices look rather like 1-D arrays.         DOUBLE PRECISION AUX(SIZE(A,DIM=1),SIZE(A,DIM=1))	!Can't say "LIKE A", as in pl/i, alas.         DOUBLE PRECISION D	!To be calculated once.         INTEGER N		!The order of the square matrix. I hope.         INTEGER C		!A stepper.          N = SIZE(A, DIM = 1)	!Alright, what's the order of battle?          D = DET(A)		!Once only.          IF (D.EQ.0) STOP "Cramer: zero determinant!"	!Surely, this won't happen...          AUX = A		!Prepare the assistant.          DO C = 1,N		!Step across the columns.            IF (C.GT.1) AUX(1:N,C - 1) = A(1:N,C - 1)	!Repair previous damage.            AUX(1:N,C) = B(1:N)		!Place the current damage.            X(C) = DET(AUX)/D		!The result!          END DO		!On to the next column.        END SUBROUTINE CRAMER	!This looks really easy!      END MODULE BAD IDEA	!But actually, it is a bad idea for N > 2.       PROGRAM TEST	!Try it and see.      USE BAD IDEA	!Just so.      DOUBLE PRECISION, ALLOCATABLE ::A(:,:), B(:), X(:)	!Custom work areas.      INTEGER N,R	!Assistants..      INTEGER INF	!An I/O unit.       MSG = 6		!Output.      INF = 10		!For reading test data.      OPEN (INF,FILE="Test.dat",STATUS="OLD",ACTION="READ")	!As in this file.. Chew into the next problem.   10 IF (ALLOCATED(A)) DEALLOCATE(A)	!First,      IF (ALLOCATED(B)) DEALLOCATE(B)	!Get rid of      IF (ALLOCATED(X)) DEALLOCATE(X)	!The hired help.      READ (INF,*,END = 100) N		!Since there is a new order.      IF (N.LE.0) GO TO 100		!Perhaps a final order.      WRITE (MSG,11) N			!Othewise, announce prior to acting.   11 FORMAT ("Order ",I0," matrix A, as follows...")	!In case something goes wrong.      ALLOCATE(A(N,N))			!For instance,      ALLOCATE(B(N))			!Out of memory.      ALLOCATE(X(N))			!But otherwise, a tailored fit.      DO R = 1,N			!Now read in the values for the matrix.        READ(INF,*,END=667,ERR=665) A(R,:),B(R)	!One row of A at a go, followed by B's value.      END DO				!In free format.      CALL SHOWMATRIX(A)		!Show what we have managed to obtain.      WRITE (MSG,12) "In the scheme A.x = b, b = ",B	!In case something goes wrong.   12 FORMAT (A,<N>F12.6)		!How many would be too many?      CALL CRAMER(A,X,B)		!The deed!      WRITE (MSG,12) "    Via Cramer's rule, x = ",X	!The result!      GO TO 10				!And try for another test problem. Complaints.  665 WRITE (MSG,666) "Format error",R	!I know where I came from.  666 FORMAT (A," while reading row ",I0,"!")	!So I can refer to R.      GO TO 100		!So much for that.  667 WRITE (MSG,666) "End-of-file", R		!Some hint as to where. Closedown.  100 WRITE (6,*) "That was interesting."	!Quite.      END	!Open files are closed, allocated memory is released.

Oddly, the Compaq Visual Fortran F90/95 compiler is confused by the usage "BAD IDEA" instead of "BADIDEA" - spaces are not normally relevant in Fortran source. Anyway, file Test.dat has been filled with the numbers of the example, as follows:

4           /The order, for A.x = b.
2  -1   5   1,  -3    /First row of A, b
3   2   2  -6, -32    /Second row...
1   3   3  -1, -47     third row.
5  -2  -3   3,  49    /Last row.



Fortran's free-form allows a comma, a tab, and spaces between numbers, and regards the / as starting a comment, but, because each row is read separately, once the required five (N + 1) values have been read, no further scanning of the line takes place and the next READ statement will start with a new line of input. So the / isn't needed for the third row, as shown. Omitted values lead to confusion as the input process would read additional lines to fill the required count and everything gets out of step. Echoing input very soon after it is obtained is helpful in making sense of such mistakes.

For more practical use it would probably be better to constrain the freedom somewhat, perhaps requiring that all the N + 1 values for a row appear on one input record. In such a case, the record could first be read into a text variable (from which the data would be read) so that if a problem arises the text could be printed as a part of the error message. But, this requires guessing a suitably large length for the text variable so as to accommodate the longest possible input line.

Output:

Order 4 matrix A, as follows...
|       2.000      -1.000       5.000       1.000|
|       3.000       2.000       2.000      -6.000|
|       1.000       3.000       3.000      -1.000|
|       5.000      -2.000      -3.000       3.000|
In the scheme A.x = b, b =    -3.000000  -32.000000  -47.000000   49.000000
Via Cramer's rule, x =     2.000000  -12.000000   -4.000000    1.000000
That was interesting.


And at this point I suddenly noticed that the habits of Old Fortran are not so easily suppressed: all calculations are done with double precision. Curiously enough, for the specific example data, the same results are obtained if all variables are integer.

Go

Library gonum:

package main import (    "fmt"     "gonum.org/v1/gonum/mat") var m = mat.NewDense(4, 4, []float64{    2, -1, 5, 1,    3, 2, 2, -6,    1, 3, 3, -1,    5, -2, -3, 3,}) var v = []float64{-3, -32, -47, 49} func main() {    x := make([]float64, len(v))    b := make([]float64, len(v))    d := mat.Det(m)    for c := range v {        mat.Col(b, c, m)        m.SetCol(c, v)        x[c] = mat.Det(m) / d        m.SetCol(c, b)    }    fmt.Println(x)}
Output:
[2 -12.000000000000007 -4.000000000000001 1.0000000000000009]


Library go.matrix:

package main import (    "fmt"     "github.com/skelterjohn/go.matrix") var m = matrix.MakeDenseMatrixStacked([][]float64{    {2, -1, 5, 1},    {3, 2, 2, -6},    {1, 3, 3, -1},    {5, -2, -3, 3},}) var v = []float64{-3, -32, -47, 49} func main() {    x := make([]float64, len(v))    b := make([]float64, len(v))    d := m.Det()    for c := range v {        m.BufferCol(c, b)        m.FillCol(c, v)        x[c] = m.Det() / d        m.FillCol(c, b)    }    fmt.Println(x)}
Output:
[2.0000000000000004 -11.999999999999998 -4 0.9999999999999999]


import Data.Matrix solveCramer :: (Ord a, Fractional a) => Matrix a -> Matrix a -> Maybe [a]solveCramer a y  | da == 0 = Nothing  | otherwise = Just $map (\i -> d i / da) [1..n] where da = detLU a d i = detLU$ submatrix 1 n 1 n $switchCols i (n+1) ay ay = a <|> y n = ncols a task = solveCramer a y where a = fromLists [[2,-1, 5, 1] ,[3, 2, 2,-6] ,[1, 3, 3,-1] ,[5,-2,-3, 3]] y = fromLists [[-3], [-32], [-47], [49]] Output: λ> task Just [2.0000000000000004,-11.999999999999998,-4.0,0.9999999999999999]  Version 2 We use Rational numbers for having more precision. a % b is the rational a / b.  s_permutations :: [a] -> [([a], Int)]s_permutations = flip zip (cycle [1, -1]) . (foldl aux [[]]) where aux items x = do (f,item) <- zip (cycle [reverse,id]) items f (insertEv x item) insertEv x [] = [[x]] insertEv x l@(y:ys) = (x:l) : map (y:) (insertEv x ys) foldlZipWith::(a -> b -> c) -> (d -> c -> d) -> d -> [a] -> [b] -> dfoldlZipWith _ _ u [] _ = ufoldlZipWith _ _ u _ [] = ufoldlZipWith f g u (x:xs) (y:ys) = foldlZipWith f g (g u (f x y)) xs ys foldl1ZipWith::(a -> b -> c) -> (c -> c -> c) -> [a] -> [b] -> cfoldl1ZipWith _ _ [] _ = error "First list is empty"foldl1ZipWith _ _ _ [] = error "Second list is empty"foldl1ZipWith f g (x:xs) (y:ys) = foldlZipWith f g (f x y) xs ys multAdd::(a -> b -> c) -> (c -> c -> c) -> [[a]] -> [[b]] -> [[c]]multAdd f g xs ys = map (\us -> foldl1ZipWith (\u vs -> map (f u) vs) (zipWith g) us ys) xs mult:: Num a => [[a]] -> [[a]] -> [[a]]mult xs ys = multAdd (*) (+) xs ys elemPos::[[a]] -> Int -> Int -> aelemPos ms i j = (ms !! i) !! j prod:: Num a => ([[a]] -> Int -> Int -> a) -> [[a]] -> [Int] -> aprod f ms = product.zipWith (f ms) [0..] s_determinant:: Num a => ([[a]] -> Int -> Int -> a) -> [[a]] -> [([Int],Int)] -> as_determinant f ms = sum.map (\(is,s) -> fromIntegral s * prod f ms is) elemCramerPos::Int -> Int -> [[a]] -> [[a]] -> Int -> Int -> aelemCramerPos l k ks ms i j = if j /= l then elemPos ms i j else elemPos ks i k solveCramer:: [[Rational]] -> [[Rational]] -> [[Rational]]solveCramer ms ks = xs where xs | d /= 0 = go (reverse [0..pred.length.head$ ks])     | otherwise = []  go (u:us) = foldl glue (col u) us  glue us u = zipWith (\ys (y:_) -> y:ys) us (col u)  col k = map (\l -> [(/d) $s_determinant (elemCramerPos l k ks) ms ps])$ ls  ls = [0..pred.length $ms] ps = s_permutations ls d = s_determinant elemPos ms psmatI::(Num a) => Int -> [[a]]matI n = [ [fromIntegral.fromEnum$ i == j | i <- [1..n]] | j <- [1..n]] task::[[Rational]] -> [[Rational]] -> IO()task a b = do  let x         = solveCramer a b  let u         = map (map fromRational) x  let y         = mult a x  let identity  = matI (length x)  let a1        = solveCramer a identity  let h         = mult a a1  let z         = mult a1 b  putStrLn "a ="  mapM_ print a  putStrLn "b ="  mapM_ print b  putStrLn "solve: a * x = b => x = solveCramer a b ="  mapM_ print x  putStrLn "u = fromRationaltoDouble x ="  mapM_ print u  putStrLn "verification: y = a * x = mult a x ="  mapM_ print y  putStrLn $"test: y == b = " print$ y == b  putStrLn "identity matrix: identity ="  mapM_ print identity  putStrLn "find: a1 = inv(a) => solve: a * a1 = identity => a1 = solveCramer a identity ="  mapM_ print a1  putStrLn "verification: h = a * a1 = mult a a1 ="  mapM_ print h  putStrLn $"test: h == identity = " print$ h == identity  putStrLn "z = a1 * b = mult a1 b ="  mapM_ print z  putStrLn "test: z == x ="  print $z == x main = do let a = [[2,-1, 5, 1] ,[3, 2, 2,-6] ,[1, 3, 3,-1] ,[5,-2,-3, 3]] let b = [[-3], [-32], [-47], [49]] task a b  Output: a = [2 % 1,(-1) % 1,5 % 1,1 % 1] [3 % 1,2 % 1,2 % 1,(-6) % 1] [1 % 1,3 % 1,3 % 1,(-1) % 1] [5 % 1,(-2) % 1,(-3) % 1,3 % 1] b = [(-3) % 1] [(-32) % 1] [(-47) % 1] [49 % 1] solve: a * x = b => x = solveCramer a b = [2 % 1] [(-12) % 1] [(-4) % 1] [1 % 1] u = fromRationaltoDouble x = [2.0] [-12.0] [-4.0] [1.0] verification: y = a * x = mult a x = [(-3) % 1] [(-32) % 1] [(-47) % 1] [49 % 1] test: y == b = True identity matrix: identity = [1 % 1,0 % 1,0 % 1,0 % 1] [0 % 1,1 % 1,0 % 1,0 % 1] [0 % 1,0 % 1,1 % 1,0 % 1] [0 % 1,0 % 1,0 % 1,1 % 1] find: a1 = inv(a) => solve: a * a1 = identity => a1 = solveCramer a identity = [4 % 171,11 % 171,10 % 171,8 % 57] [(-55) % 342,(-23) % 342,119 % 342,2 % 57] [107 % 684,(-5) % 684,11 % 684,(-7) % 114] [7 % 684,(-109) % 684,103 % 684,7 % 114] verification: h = a * a1 = mult a a1 = [1 % 1,0 % 1,0 % 1,0 % 1] [0 % 1,1 % 1,0 % 1,0 % 1] [0 % 1,0 % 1,1 % 1,0 % 1] [0 % 1,0 % 1,0 % 1,1 % 1] test: h == identity = True z = a1 * b = mult a1 b = [2 % 1] [(-12) % 1] [(-4) % 1] [1 % 1] test: z == x = True  Version 3  import Data.List determinant::(Fractional a, Ord a) => [[a]] -> adeterminant ls = if null ls then 0 else pivot 1 (zip ls [(0::Int)..]) where good rs ts = (abs.head.fst$ ts) <= (abs.head.fst $rs) go us (vs,i) = if v == 0 then (ws,i) else (zipWith (\x y -> y - x*v) us ws,i) where (v,ws) = (head$ vs,tail vs)  change i (ys:zs) = map (\xs -> if (==i).snd $xs then ys else xs) zs pivot d [] = d pivot d zs@((_,j):ys) = if 0 == u then 0 else pivot e ws where e = if i == j then u*d else -u*d ((u:us),i) = foldl1 (\rs ts -> if good rs ts then rs else ts) zs ws = map (go (map (/u) us))$ if i == j then ys else change i zs solveCramer::(Fractional a, Ord a) => [[a]] -> [[a]] -> [[a]]solveCramer as bs = if 0 == d then [] else ans bs  where   d = determinant as  ans = transpose.map go.transpose    where    ms = zip [0..] (transpose as)    go us =  [ (/d) $determinant [if i /= j then vs else us | (j,vs) <- ms] | (i,_) <- ms] matI::(Num a) => Int -> [[a]]matI n = [ [fromIntegral.fromEnum$ i == j | i <- [1..n]] | j <- [1..n]] foldlZipWith::(a -> b -> c) -> (d -> c -> d) -> d -> [a] -> [b]  -> dfoldlZipWith _ _ u [] _          = ufoldlZipWith _ _ u _ []          = ufoldlZipWith f g u (x:xs) (y:ys) = foldlZipWith f g (g u (f x y)) xs ys foldl1ZipWith::(a -> b -> c) -> (c -> c -> c) -> [a] -> [b] -> cfoldl1ZipWith _ _ [] _          = error "First list is empty"foldl1ZipWith _ _ _ []          = error "Second list is empty"foldl1ZipWith f g (x:xs) (y:ys) = foldlZipWith f g (f x y) xs ys multAdd::(a -> b -> c) -> (c -> c -> c) -> [[a]] -> [[b]] -> [[c]]multAdd f g xs ys = map (\us -> foldl1ZipWith (\u vs -> map (f u) vs) (zipWith g) us ys) xs mult:: Num a => [[a]] -> [[a]] -> [[a]]mult xs ys = multAdd (*) (+) xs ys task::[[Rational]] -> [[Rational]] -> IO()task a b = do  let x         = solveCramer a b  let u         = map (map fromRational) x  let y         = mult a x  let identity  = matI (length x)  let a1        = solveCramer a identity  let h         = mult a a1  let z         = mult a1 b  putStrLn "a ="  mapM_ print a  putStrLn "b ="  mapM_ print b  putStrLn "solve: a * x = b => x = solveCramer a b ="  mapM_ print x  putStrLn "u = fromRationaltoDouble x ="  mapM_ print u  putStrLn "verification: y = a * x = mult a x ="  mapM_ print y  putStrLn $"test: y == b = " print$ y == b  putStrLn "identity matrix: identity ="  mapM_ print identity  putStrLn "find: a1 = inv(a) => solve: a * a1 = identity => a1 = solveCramer a identity ="  mapM_ print a1  putStrLn "verification: h = a * a1 = mult a a1 ="  mapM_ print h  putStrLn $"test: h == identity = " print$ h == identity  putStrLn "z = a1 * b = mult a1 b ="  mapM_ print z  putStrLn "test: z == x ="  print $z == x main = do let a = [[2,-1, 5, 1] ,[3, 2, 2,-6] ,[1, 3, 3,-1] ,[5,-2,-3, 3]] let b = [[-3], [-32], [-47], [49]] task a b  Output: a = [2 % 1,(-1) % 1,5 % 1,1 % 1] [3 % 1,2 % 1,2 % 1,(-6) % 1] [1 % 1,3 % 1,3 % 1,(-1) % 1] [5 % 1,(-2) % 1,(-3) % 1,3 % 1] b = [(-3) % 1] [(-32) % 1] [(-47) % 1] [49 % 1] solve: a * x = b => x = solveCramer a b = [2 % 1] [(-12) % 1] [(-4) % 1] [1 % 1] u = fromRationaltoDouble x = [2.0] [-12.0] [-4.0] [1.0] verification: y = a * x = mult a x = [(-3) % 1] [(-32) % 1] [(-47) % 1] [49 % 1] test: y == b = True identity matrix: identity = [1 % 1,0 % 1,0 % 1,0 % 1] [0 % 1,1 % 1,0 % 1,0 % 1] [0 % 1,0 % 1,1 % 1,0 % 1] [0 % 1,0 % 1,0 % 1,1 % 1] find: a1 = inv(a) => solve: a * a1 = identity => a1 = solveCramer a identity = [4 % 171,11 % 171,10 % 171,8 % 57] [(-55) % 342,(-23) % 342,119 % 342,2 % 57] [107 % 684,(-5) % 684,11 % 684,(-7) % 114] [7 % 684,(-109) % 684,103 % 684,7 % 114] verification: h = a * a1 = mult a a1 = [1 % 1,0 % 1,0 % 1,0 % 1] [0 % 1,1 % 1,0 % 1,0 % 1] [0 % 1,0 % 1,1 % 1,0 % 1] [0 % 1,0 % 1,0 % 1,1 % 1] test: h == identity = True z = a1 * b = mult a1 b = [2 % 1] [(-12) % 1] [(-4) % 1] [1 % 1] test: z == x = True  J Implementation: cramer=:4 :0 A=. x [ b=. y det=. -/ .* A %~&det (i.#A) b"_[]}&.|:"0 2 A) Task data: A=: _&".;._2]t=: 0 :0 2 -1 5 1 3 2 2 -6 1 3 3 -1 5 -2 -3 3) b=: _3 _32 _47 49 Task example:  A cramer b2 _12 _4 1 JavaScript  var matrix = [ [2, -1, 5, 1], [3, 2, 2, -6], [1, 3, 3, -1], [5, -2, -3, 3]];var freeTerms = [-3, -32, -47, 49]; var result = cramersRule(matrix,freeTerms);console.log(result); /** * Compute Cramer's Rule * @param {array} matrix x,y,z, etc. terms * @param {array} freeTerms * @return {array} solution for x,y,z, etc. */function cramersRule(matrix,freeTerms) { var det = detr(matrix), returnArray = [], i, tmpMatrix; for(i=0; i < matrix[0].length; i++) { var tmpMatrix = insertInTerms(matrix, freeTerms,i) returnArray.push(detr(tmpMatrix)/det) } return returnArray;} /** * Inserts single dimensional array into * @param {array} matrix multidimensional array to have ins inserted into * @param {array} ins single dimensional array to be inserted vertically into matrix * @param {array} at zero based offset for ins to be inserted into matrix * @return {array} New multidimensional array with ins replacing the at column in matrix */function insertInTerms(matrix, ins, at) { var tmpMatrix = clone(matrix), i; for(i=0; i < matrix.length; i++) { tmpMatrix[i][at] = ins[i]; } return tmpMatrix;}/** * Compute the determinate of a matrix. No protection, assumes square matrix * function borrowed, and adapted from MIT Licensed numericjs library (www.numericjs.com) * @param {array} m Input Matrix (multidimensional array) * @return {number} result rounded to 2 decimal */function detr(m) { var ret = 1, k, A=clone(m), n=m[0].length, alpha; for(var j =0; j < n-1; j++) { k=j; for(i=j+1;i<n;i++) { if(Math.abs(A[i][j]) > Math.abs(A[k][j])) { k = i; } } if(k !== j) { temp = A[k]; A[k] = A[j]; A[j] = temp; ret *= -1; } Aj = A[j]; for(i=j+1;i<n;i++) { Ai = A[i]; alpha = Ai[j]/Aj[j]; for(k=j+1;k<n-1;k+=2) { k1 = k+1; Ai[k] -= Aj[k]*alpha; Ai[k1] -= Aj[k1]*alpha; } if(k!==n) { Ai[k] -= Aj[k]*alpha; } } if(Aj[j] === 0) { return 0; } ret *= Aj[j]; } return Math.round(ret*A[j][j]*100)/100;} /** * Clone two dimensional Array using ECMAScript 5 map function and EcmaScript 3 slice * @param {array} m Input matrix (multidimensional array) to clone * @return {array} New matrix copy */function clone(m) { return m.map(function(a){return a.slice();});}  Output: [ 2, -12, -4, 1 ] Julia Works with: Julia version 0.6 function cramersolve(A::Matrix, b::Vector) return collect(begin B = copy(A); B[:, i] = b; det(B) end for i in eachindex(b)) ./ det(A)end A = [2 -1 5 1 3 2 2 -6 1 3 3 -1 5 -2 -3 3] b = [-3, -32, -47, 49] @show cramersolve(A, b) Output: cramersolve(A, b) = [2.0, -12.0, -4.0, 1.0] Note that it is entirely impractical to use Cramer's rule in this situation. It would be much better to use the built-in operator for solving linear systems. Assuming that the coefficient matrix and constant vector are defined as before, the solution vector is given by: @show A \ b Kotlin As in the case of the Matrix arithmetic task, I've used the Johnson-Trotter permutations generator to assist with the calculation of the determinants for the various matrices: // version 1.1.3 typealias Vector = DoubleArraytypealias Matrix = Array<Vector> fun johnsonTrotter(n: Int): Pair<List<IntArray>, List<Int>> { val p = IntArray(n) { it } // permutation val q = IntArray(n) { it } // inverse permutation val d = IntArray(n) { -1 } // direction = 1 or -1 var sign = 1 val perms = mutableListOf<IntArray>() val signs = mutableListOf<Int>() fun permute(k: Int) { if (k >= n) { perms.add(p.copyOf()) signs.add(sign) sign *= -1 return } permute(k + 1) for (i in 0 until k) { val z = p[q[k] + d[k]] p[q[k]] = z p[q[k] + d[k]] = k q[z] = q[k] q[k] += d[k] permute(k + 1) } d[k] *= -1 } permute(0) return perms to signs} fun determinant(m: Matrix): Double { val (sigmas, signs) = johnsonTrotter(m.size) var sum = 0.0 for ((i, sigma) in sigmas.withIndex()) { var prod = 1.0 for ((j, s) in sigma.withIndex()) prod *= m[j][s] sum += signs[i] * prod } return sum} fun cramer(m: Matrix, d: Vector): Vector { val divisor = determinant(m) val numerators = Array(m.size) { Matrix(m.size) { m[it].copyOf() } } val v = Vector(m.size) for (i in 0 until m.size) { for (j in 0 until m.size) numerators[i][j][i] = d[j] } for (i in 0 until m.size) v[i] = determinant(numerators[i]) / divisor return v} fun main(args: Array<String>) { val m = arrayOf( doubleArrayOf(2.0, -1.0, 5.0, 1.0), doubleArrayOf(3.0, 2.0, 2.0, -6.0), doubleArrayOf(1.0, 3.0, 3.0, -1.0), doubleArrayOf(5.0, -2.0, -3.0, 3.0) ) val d = doubleArrayOf(-3.0, -32.0, -47.0, 49.0) val (w, x, y, z) = cramer(m, d) println("w =$w, x = $x, y =$y, z = $z")} Output: w = 2.0, x = -12.0, y = -4.0, z = 1.0  Lua local matrix = require "matrix" -- https://github.com/davidm/lua-matrix local function cramer(mat, vec) -- Check if matrix is quadratic assert(#mat == #mat[1], "Matrix is not square!") -- Check if vector has the same size of the matrix assert(#mat == #vec, "Vector has not the same size of the matrix!") local size = #mat local main_det = matrix.det(mat) local aux_mats = {} local dets = {} local result = {} for i = 1, size do -- Construct the auxiliary matrixes aux_mats[i] = matrix.copy(mat) for j = 1, size do aux_mats[i][j][i] = vec[j] end -- Calculate the auxiliary determinants dets[i] = matrix.det(aux_mats[i]) -- Calculate results result[i] = dets[i]/main_det end return resultend ----------------------------------------------- local A = {{ 2, -1, 5, 1}, { 3, 2, 2, -6}, { 1, 3, 3, -1}, { 5, -2, -3, 3}}local b = {-3, -32, -47, 49} local result = cramer(A, b)print("Result: " .. table.concat(result, ", "))  Output: Result: 2, -12, -4, 1 Maple A := Matrix([[2,-1,5,1],[3,2,2,-6],[1,3,3,-1],[5,-2,-3,3]]):w := LinearAlgebra:-Determinant(Matrix([[-3,-1,5,1],[-32,2,2,-6],[-47,3,3,-1],[49,-2,-3,3]]))/ LinearAlgebra:-Determinant(A);x := LinearAlgebra:-Determinant(Matrix([[2,-3,5,1],[3,-32,2,-6],[1,-47,3,-1],[5,49,-3,3]]))/LinearAlgebra:-Determinant(A);y := LinearAlgebra:-Determinant(Matrix([[2,-1,-3,1],[3,2,-32,-6],[1,3,-47,-1],[5,-2,49,3]]))/LinearAlgebra:-Determinant(A); z := LinearAlgebra:-Determinant(Matrix([[2,-1,5,-3],[3,2,2,-32],[1,3,3,-47],[5,-2,-3,49]]))/LinearAlgebra:-Determinant(A); Output: w := 2 x := -12 y := -4 z := 1 Mathematica crule[m_, b_] := Module[{d = Det[m], a}, Table[a = m; a[[All, k]] = b; Det[a]/d, {k, Length[m]}]] crule[{ {2, -1, 5, 1}, {3, 2, 2, -6}, {1, 3, 3, -1}, {5, -2, -3, 3} } , {-3, -32, -47, 49}] Output: {2,-12,-4,1} PARI/GP  M = [2,-1,5,1;3,2,2,-6;1,3,3,-1;5,-2,-3,3];V = Col([-3,-32,-47,49]); matadjoint(M) * V / matdet(M)  Output: [2, -12, -4, 1]~ Perl use Math::Matrix; sub cramers_rule { my ($A, $terms) = @_; my @solutions; my$det = $A->determinant; foreach my$i (0 .. $#{$A}) {        my $Ai =$A->clone;        foreach my $j (0 ..$#{$terms}) {$Ai->[$j][$i] = $terms->[$j];        }        push @solutions, $Ai->determinant /$det;    }    @solutions;} my $matrix = Math::Matrix->new( [2, -1, 5, 1], [3, 2, 2, -6], [1, 3, 3, -1], [5, -2, -3, 3],); my$free_terms = [-3, -32, -47, 49];my ($w,$x, $y,$z) = cramers_rule($matrix,$free_terms); print "w = $w\n";print "x =$x\n";print "y = $y\n";print "z =$z\n";
Output:
w = 2
x = -12
y = -4
z = 1


Perl 6

sub det(@matrix) {    my @a = @matrix.map: { [|$_] }; my$sign = +1;    my $pivot = 1; for ^@a ->$k {      my @r = ($k+1 .. @a.end); my$previous-pivot = $pivot; if 0 == ($pivot = @a[$k][$k]) {        (my $s = @r.first: { @a[$_][$k] != 0 }) // return 0; (@a[$s],@a[$k]) = (@a[$k], @a[$s]); my$pivot = @a[$k][$k];        $sign = -$sign;      }      for @r X @r -> ($i,$j) {        ((@a[$i][$j] *= $pivot) -= @a[$i][$k]*@a[$k][$j]) /=$previous-pivot;      }    }    $sign *$pivot} sub cramers_rule(@A, @terms) {    gather for ^@A -> $i { my @Ai = @A.map: { [|$_] };        for ^@terms -> $j { @Ai[$j][$i] = @terms[$j];        }        take det(@Ai);    } »/» det(@A);} my @matrix = (    [2, -1,  5,  1],    [3,  2,  2, -6],    [1,  3,  3, -1],    [5, -2, -3,  3],); my @free_terms = (-3, -32, -47, 49);my ($w,$x, $y,$z) = |cramers_rule(@matrix, @free_terms); say "w = $w";say "x =$x";say "y = $y";say "z =$z";
Output:
w = 2
x = -12
y = -4
z = 1


Phix

Translation of: C

The copy-on-write semantics of Phix really shine here; because there is no explicit return/re-assign, you can treat parameters as a private workspace, confident in the knowledge that the updated version will be quietly discarded; all the copying and freeing of the C version is automatic/unnecessary here.

constant inf = 1e300*1e300,         nan = -(inf/inf) function det(sequence a)atom res = 1     for j=1 to length(a) do        integer i_max = j        for i=j+1 to length(a) do            if a[i][j] > a[i_max][j] then                i_max = i            end if        end for        if i_max != j then            {a[i_max],a[j]} = {a[j],a[i_max]}            res *= -1        end if         if abs(a[j][j]) < 1e-12 then            puts(1,"Singular matrix!")            return nan        end if         for i=j+1 to length(a) do            atom mult = -a[i][j] / a[j][j]            for k=1 to length(a) do                a[i][k] += mult * a[j][k]            end for        end for    end for     for i=1 to length(a) do        res *= a[i][i]    end for    return resend function function cramer_solve(sequence a, atom det_a, sequence b, integer var)    for i=1 to length(a) do      a[i][var] = b[i]    end for    return det(a)/det_aend function sequence a = {{2,-1, 5, 1},              {3, 2, 2,-6},              {1, 3, 3,-1},              {5,-2,-3, 3}},         b = {-3,-32,-47,49}integer det_a = det(a)for i=1 to length(a) do    printf(1, "%7.3f\n", cramer_solve(a, det_a, b, i))end for
Output:
  2.000
-12.000
-4.000
1.000


Prolog

Works with: GNU Prolog
removeElement([_|Tail], 0, Tail).removeElement([Head|Tail], J, [Head|X]) :-    J_2 is J - 1,    removeElement(Tail, J_2, X). removeColumn([], _, []).removeColumn([Matrix_head|Matrix_tail], J, [X|Y]) :-    removeElement(Matrix_head, J, X),    removeColumn(Matrix_tail, J, Y). removeRow([_|Matrix_tail], 0, Matrix_tail).removeRow([Matrix_head|Matrix_tail], I, [Matrix_head|X]) :-    I_2 is I - 1,    removeRow(Matrix_tail, I_2, X). cofactor(Matrix, I, J, X) :-    removeRow(Matrix, I, Matrix_2),    removeColumn(Matrix_2, J, Matrix_3),    det(Matrix_3, Y),    X is (-1) ** (I + J) * Y. det_summand(_, _, [], 0).det_summand(Matrix, J, B, X) :-    B = [B_head|B_tail],    cofactor(Matrix, 0, J, Z),    J_2 is J + 1,    det_summand(Matrix, J_2, B_tail, Y),    X is B_head * Z + Y. det([[X]], X).det(Matrix, X) :-    Matrix = [Matrix_head|_],    det_summand(Matrix, 0, Matrix_head, X). replaceElement([_|Tail], 0, New, [New|Tail]).replaceElement([Head|Tail], J, New, [Head|Y]) :-    J_2 is J - 1,    replaceElement(Tail, J_2, New, Y). replaceColumn([], _, _, []).replaceColumn([Matrix_head|Matrix_tail], J, [Column_head|Column_tail], [X|Y]) :-    replaceElement(Matrix_head, J, Column_head, X),    replaceColumn(Matrix_tail, J, Column_tail, Y). cramerElements(_, B, L, []) :- length(B, L).cramerElements(A, B, J, [X_J|Others]) :-    replaceColumn(A, J, B, A_J),    det(A_J, Det_A_J),    det(A, Det_A),    X_J is Det_A_J / Det_A,    J_2 is J + 1,    cramerElements(A, B, J_2, Others). cramer(A, B, X) :- cramerElements(A, B, 0, X). results(X) :-    A = [            [2, -1,  5,  1],            [3,  2,  2, -6],            [1,  3,  3, -1],            [5, -2, -3,  3]        ],    B = [-3, -32, -47, 49],    cramer(A, B, X).
Output:
| ?- results(X).
X = [2.0,-12.0,-4.0,1.0] ?
yes


Python

 # a simple implementation using numpyfrom numpy import linalg A=[[2,-1,5,1],[3,2,2,-6],[1,3,3,-1],[5,-2,-3,3]]B=[-3,-32,-47,49]C=[[2,-1,5,1],[3,2,2,-6],[1,3,3,-1],[5,-2,-3,3]]X=[]for i in range(0,len(B)):    for j in range(0,len(B)):        C[j][i]=B[j]        if i>0:            C[j][i-1]=A[j][i-1]    X.append(round(linalg.det(C)/linalg.det(A),1)) print('w=%s'%X[0],'x=%s'%X[1],'y=%s'%X[2],'z=%s'%X[3])
Output:
w=2.0 x=-12.0 y=-4.0 z=1.0


Racket

#lang typed/racket(require math/matrix) (define A (matrix [[2 -1  5  1]                   [3  2  2 -6]                   [1  3  3 -1]                   [5 -2 -3  3]])) (define B (col-matrix [ -3                       -32                       -47                        49])) (matrix->vector (matrix-solve A B))
Output:
'#(2 -12 -4 1)

REXX

/* REXX Use Cramer's rule to compute solutions of given linear equations  */Numeric Digits 20names='w x y z'M='  2  -1   5   1',  '  3   2   2  -6',  '  1   3   3  -1',  '  5  -2  -3   3'v=' -3',  '-32',  '-47',  ' 49'Call mk_mat(m)                      /* M -> a.i.j                    */Do j=1 To dim                       /* Show the input                */  ol=''  Do i=1 To dim    ol=ol format(a.i.j,6)    End  ol=ol format(word(v,j),6)  Say ol  EndSay copies('-',35) d=det(m)                            /* denominator determinant       */ Do k=1 To dim                       /* construct nominator matrix    */  Do j=1 To dim    Do i=1 To dim      If i=k Then        b.i.j=word(v,j)      Else        b.i.j=a.i.j      End    End  Call show_b  d.k=det(mk_str())                 /* numerator determinant         */  Say word(names,k) '=' d.k/d       /* compute value of variable k   */  EndExit mk_mat: Procedure Expose a. dim     /* Turn list into matrix a.i.j */  Parse Arg list  dim=sqrt(words(list))  k=0  Do j=1 To dim    Do i=1 To dim      k=k+1      a.i.j=word(list,k)      End    End  Return mk_str: Procedure Expose b. dim     /* Turn matrix b.i.j into list   */  str=''Do j=1 To dim  Do i=1 To dim    str=str b.i.j    End  EndReturn str show_b: Procedure Expose b. dim     /* show numerator matrix         */  do j=1 To dim    ol=''    Do i=1 To dim      ol=ol format(b.i.j,6)      end    Call dbg ol    end  Return det: Procedure                      /* compute determinant           */Parse Arg listn=words(list)call dbg 'det:' listdo dim=1 To 10  If dim**2=n Then Leave  Endcall dbg 'dim='dimIf dim=2 Then Do  det=word(list,1)*word(list,4)-word(list,2)*word(list,3)  call dbg 'det=>'det  Return det  Endk=0Do j=1 To dim  Do i=1 To dim    k=k+1    a.i.j=word(list,k)    End  EndDo j=1 To dim  ol=j  Do i=1 To dim    ol=ol format(a.i.j,6)    End  call dbg ol  Enddet=0Do i=1 To dim  ol=''  Do j=2 To dim    Do ii=1 To dim      If ii<>i Then        ol=ol a.ii.j      End    End  call dbg 'i='i 'ol='ol  If i//2 Then    det=det+a.i.1*det(ol)  Else    det=det-a.i.1*det(ol)  EndCall dbg 'det=>>>'detReturn detsqrt: Procedure/* REXX **************************************************************** EXEC to calculate the square root of a = 2 with high precision**********************************************************************/  Parse Arg x,prec  If prec<9 Then prec=9  prec1=2*prec  eps=10**(-prec1)  k = 1  Numeric Digits 3  r0= x  r = 1  Do i=1 By 1 Until r=r0 | (abs(r*r-x)<eps)    r0 = r    r  = (r + x/r) / 2    k  = min(prec1,2*k)    Numeric Digits (k + 5)    End  Numeric Digits prec  r=r+0  Return r  dbg: Return
Output:
      2     -1      5      1     -3
3      2      2     -6    -32
1      3      3     -1    -47
5     -2     -3      3     49
-----------------------------------
w = 2
x = -12
y = -4
z = 1

Ruby

require 'matrix' def cramers_rule(a, terms)  raise ArgumentError, " Matrix not square"  unless a.square?  cols = a.to_a.transpose  cols.each_index.map do |i|    c = cols.dup    c[i] = terms    Matrix.columns(c).det / a.det  endend matrix = Matrix[    [2, -1,  5,  1],    [3,  2,  2, -6],    [1,  3,  3, -1],    [5, -2, -3,  3],] vector = [-3, -32, -47, 49]puts cramers_rule(matrix, vector)
Output:
2
-12
-4
1


Sidef

func cramers_rule(A, terms) {    gather {        for i in ^A {            var Ai = A.map{.map{_}}            for j in ^terms {                Ai[j][i] = terms[j]            }            take(Ai.det)        }    } »/» A.det} var matrix = [    [2, -1,  5,  1],    [3,  2,  2, -6],    [1,  3,  3, -1],    [5, -2, -3,  3],] var free_terms = [-3, -32, -47, 49]var (w, x, y, z) = cramers_rule(matrix, free_terms)... say "w = #{w}"say "x = #{x}"say "y = #{y}"say "z = #{z}"
Output:
w = 2
x = -12
y = -4
z = 1


Tcl

 package require math::linearalgebranamespace path ::math::linearalgebra # Setting matrix to variable A and size to nset A [list { 2 -1  5  1} { 3  2  2 -6} { 1  3  3 -1} { 5 -2 -3  3}]set n [llength $A]# Setting right side of equationset right {-3 -32 -47 49} # Calculating determinant of Aset detA [det$A] # Apply Cramer's rulefor {set i 0} {$i <$n} {incr i} {  set tmp $A ;# copy A to tmp setcol tmp$i $right ;# replace column i with right side vector set detTmp [det$tmp]         ;# calculate determinant of tmp  set v [expr $detTmp /$detA]  ;# divide two determinants  puts [format "%0.4f" \$v]      ;# format and display result}
Output:
2.0000
-12.0000
-4.0000
1.0000


zkl

Using the GNU Scientific Library, we define the values:

var [const] GSL=Import("zklGSL");	// libGSL (GNU Scientific Library)A:=GSL.Matrix(4,4).set(2,-1, 5, 1,		       3, 2, 2,-6,		       1, 3, 3,-1,		       5,-2,-3, 3);b:=GSL.Vector(4).set(-3,-32,-47,49);

First, just let GSL solve:

A.AxEQb(b).format().println();

Actually implement Cramer's rule:

Translation of: Julia
fcn cramersRule(A,b){   b.len().pump(GSL.Vector(b.len()),'wrap(i){ // put calculations into new Vector      A.copy().setColumn(i,b).det();   }).close()/A.det();}cramersRule(A,b).format().println();
Output:
2.00,-12.00,-4.00,1.00