| 12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061 | /*							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);}
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