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NOTES 567 " 1 2" " l 9" " 1 4 " .0 1. . 0 1 . . 0 1 . " 1 7" " l 5" " 1 3 " . 0 1. . 0 1 . . 0 1 . " 1 6" " l l " " 1 8" 0 1 . 0 1. . 0 1 . which has constant sum For this square, A = and AJ = 9 135 0 9 3 15 1 15 and constant product 0 3 0 0 3 30 0 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 In fact, it is easily shown by induction that A" = 3"- 1 5n3fl-» 0 3"-1 1 1 1 1 1 1 1 1 1 forn > 2. The fact that A has distinct entries, but A3 does not, shows that the above theorem does not hold for 3 x 3 magic squares of matrices (with magic squares defined to satisfy (i) to (iv)). References 1. J. M. H. Peters, Inverses and cubes, Math. Gaz. 65 (1982), pp. 253-4. 2. T. J. Fletcher, Linear algebra through its applications, van Nostrand Reinhold (1972). RAY HILL Department of Mathematics and Computer Science, University ofSalford M5 4WT S. M. ELZAIDI Department of Mathematics, University of Nottingham NG7 2RD 80.47 An iterative algorithm for matrix inversion which is always convergent The problem of numerically solving a set of linear equations, or equivalently of inverting a matrix, can be tackled either by a direct method such as Gaussian elimination or by an iterative technique [1]. For a general matrix the former is usually more efficient, but for particular types, such as sparse matrices, an iterative approach can be more efficient. This is because the latter can be advantageous, both in terms of the total amount of computation involved and also in the amount of computer storage required, Downloaded from https:/www.cambridge.org/core. IP address: 88.99.165.207, on 13 Jul 2017 at 01:14:27, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.2307/3618529 THE MATHEMATICAL GAZETTE 568 as discussed for example in [1]. Furthermore, iterative techniques can be more accurate as they are not affected by the cumulative effect of roundingoff errors and the corresponding pivoting problems inherent in direct methods. Now, a well-known iterative procedure is the Gauss-Seidel method, but this is restricted in application to those situations for which it is convergent; this corresponds to the relevant matrix being in some suitable sense diagonally dominant. The purpose of the present note is to point out a simple way by which any matrix inversion problem may be transformed into one for which the Gauss-Seidel method converges. We consider a set of n simultaneous linear equations of the form n ^cipfy = bp (1 < p < ri) q=\ for the n unknowns xp (1 < p < n). If xpr) is the rth iterate, the GaussSeidel method gives X P (r+l) = 1 n 7> - 1 2^apqxq + U=i 2 , <WV - bP > i=p+i and it is known that this yields a convergent algorithm if the n x n matrix A = [apq] is both symmetric and positive-definite (see, for example, [1]). Let us now consider the set of linear equations Cx = d where C is non-singular, in a case where the Gauss-Seidel algorithm applied to the matrix C does not converge. Operating on both sides of this equation with the transposed matrix CT immediately transforms it into Ax = b where A = C C and b = C d, and we can now easily show that A is both symmetric and positive definite. To prove the former, we note that r T AT = (CTC)T = C^C™ = CTC = A while the latter property follows from yTAy = yTCTCy = (Cy)TCy > 0, for y non-zero, since Cy = 0 => y = 0, because C is non-singular. It follows immediately that A = CTC is a positive definite matrix and hence that the Gauss-Seidel method applied to the above equation Ax = b will always converge and will yield the solution of our original equation Cx = d. This technique may easily be extended to calculating C~l, the inverse of C, which satisfies CC~l = I. Since AC~X = CTCC~l = CTI = CT we consider in turn each column of C_1 and the corresponding column of CT, and use the Gauss-Seidel method to solve for the former. Downloaded from https:/www.cambridge.org/core. IP address: 88.99.165.207, on 13 Jul 2017 at 01:14:27, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.2307/3618529 NOTES 569 Finally we give an example of the above technique applied to a 3 x 3 matrix for which the Gauss-Seidel method cannot even be formulated, let alone converge, since the diagonal elements of the matrix are all zero. The equations we take are l5 / 0 6 l \ lx\ 4 0 3 y = 13 \5 2 o / \z, l\ H which, in fact, have the exact solution (x, y, z) = (1, 2, 3). Following our technique we multiply both sides of the above equation by the transpose of the matrix above and hence obtain ' 41 10 12 \ M 10 40 6 12 6 10/ / 97\ 108 54 We begin the iteration by taking x ( ' = 0 , followed by repeated \0/ application of the Gauss-Seidel iteration formula as given above. The first / 2.366 \ iterate is x(1) = 2.109 which clearly differs significantly from the exact \ 1.300/ solution. Further iterations, however, converge quite rapidly leading to /1.002\ x<7) = 2.001 with the error decreasing by a factor of about 3 at each step. \2.998/ We conclude that our modification of the Gauss-Seidel method is successful in yielding an iterative algorithm which always converges. 0 6 1 T1 The calculation of | 4 0 3 by solving 5 2 0 ' 41 10 12 \ X[ X2 X-i 10 40 6 y\ yi yi ,12 6 10 , z l z 2 z 3 0 4 5 6 0 2 1 3 0 is left as an exercise to the reader. Reference 1. A. Ralston and P. Rabinowitz, A first course in numerical analysis, McGraw-Hill (1978). S. SIMONS School of Mathematical Sciences, Queen Mary and Westfield College, Mile End Road, London El 4NS Downloaded from https:/www.cambridge.org/core. IP address: 88.99.165.207, on 13 Jul 2017 at 01:14:27, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.2307/3618529