Systems of Linear Algebraic Equations - Numerical Methods in Engineering with MATLAB - page 38

the lower triangular portion of the coefficient matrix anyway, its contents are
irrelevant.

Back substitution phase After Gauss elimination the augmented coefficient ma-
trix has the form

⎡
⎣

⎤
⎦

A 11 A 12 A 13 ··· A 1 n b 1

0 A 22 A 23 ··· A 2 n b 2

A

b

00 A 33 ··· A 3 n b 3

=

. . . . .

000 ··· A nn b n

The last equation, A nn x n = b n , is solvedfirst, yielding

x n = b n / A nn (2.9)

x k + 1 have been
already been computed (in thatorder), and we are abouttodetermine x k from the k th
equation

Considernow the stageofback substitutionwhere x n , x n − 1 ,...,

A kk x k + A k , k + 1 x k + 1 +···+ A kn x n = b k

The solutionis

b k −

A kj x j 1

n

x k =

A kk , k

= n − 1 , n − 2 ,..., 1 (2.10)

j = k + 1

The corresponding algorithm forback substitutionis:

fork=n:-1:1
b(k) = (b(k) - A(k,k+1:n)*b(k+1:n))/A(k,k);

end

gauss

The function gauss combines the elimination and the back substitutionphases.Dur-
ing back substitution b isoverwrittenbythe solutionvector x ,sothat b contains the
solutionupon exit.

function [x,det] = gauss(A,b)
%SolvesA*x=bbyGausseliminationandcomputesdet(A).
% USAGE: [x,det] = gauss(A,b)

if size(b,2) > 1; b = b'; end % b must be column vector

n = length(b);