From 7ce331c01ce6eb7b3f5c715a38a24359da9c6ee2 Mon Sep 17 00:00:00 2001
From: Eric Andersen <andersen@codepoet.org>
Date: Thu, 22 Nov 2001 14:04:29 +0000
Subject: Totally rework the math library, this time based on the MacOs X math
 library (which is itself based on the math lib from FreeBSD).  -Erik

---
 libm/double/minv.c | 61 ------------------------------------------------------
 1 file changed, 61 deletions(-)
 delete mode 100644 libm/double/minv.c

(limited to 'libm/double/minv.c')

diff --git a/libm/double/minv.c b/libm/double/minv.c
deleted file mode 100644
index df788fecf..000000000
--- a/libm/double/minv.c
+++ /dev/null
@@ -1,61 +0,0 @@
-/*							minv.c
- *
- *	Matrix inversion
- *
- *
- *
- * SYNOPSIS:
- *
- * int n, errcod;
- * double A[n*n], X[n*n];
- * double B[n];
- * int IPS[n];
- * int minv();
- *
- * errcod = minv( A, X, n, B, IPS );
- *
- *
- *
- * DESCRIPTION:
- *
- * Finds the inverse of the n by n matrix A.  The result goes
- * to X.   B and IPS are scratch pad arrays of length n.
- * The contents of matrix A are destroyed.
- *
- * The routine returns nonzero on error; error messages are printed
- * by subroutine simq().
- *
- */
-
-minv( A, X, n, B, IPS )
-double A[], X[];
-int n;
-double B[];
-int IPS[];
-{
-double *pX;
-int i, j, k;
-
-for( i=1; i<n; i++ )
-	B[i] = 0.0;
-B[0] = 1.0;
-/* Reduce the matrix and solve for first right hand side vector */
-pX = X;
-k = simq( A, B, pX, n, 1, IPS );
-if( k )
-	return(-1);
-/* Solve for the remaining right hand side vectors */
-for( i=1; i<n; i++ )
-	{
-	B[i-1] = 0.0;
-	B[i] = 1.0;
-	pX += n;
-	k = simq( A, B, pX, n, -1, IPS );
-	if( k )
-		return(-1);
-	}
-/* Transpose the array of solution vectors */
-mtransp( n, X, X );
-return(0);
-}
-
-- 
cgit v1.2.3