Systems of Linear Algebraic Equations - Numerical Methods in Engineering with MATLAB - page 39

fork=1:n-1 %Eliminationphase

for i= k+1:n
if A(i,k) ˜= 0
lambda = A(i,k)/A(k,k);
A(i,k+1:n) = A(i,k+1:n) - lambda*A(k,k+1:n);
b(i)= b(i) - lambda*b(k);

end

end

end
if nargout == 2; det = prod(diag(A)); end
fork=n:-1:1 %Backsubstitutionphase
b(k) = (b(k) - A(k,k+1:n)*b(k+1:n))/A(k,k);

end
x=b;

Multiple Sets of Equations

As mentionedbefore, it isfrequentlynecessary to solve the equations Ax = b for

severalconstant vectors.Let there be m such constant vectors, denotedby

b 1 , b 2 ,..., b m and let the corresponding solutionvectors be x 1 , x 2 ,..., x m .We denote

multiple sets of equations by AX = B , where

x 1

x m

b 1

b m X = x 2 ··· B = b 2 ···

are n m matrices whose columnsconsist of solutionvectors and constant vectors,

respectively.
An economical way to handle such equations during the eliminationphase is
to include all m constant vectors in the augmented coefficient matrix,sothat they
aretransformed simultaneouslywith the coefficient matrix. The solutions are then
obtainedbyback substitutioninthe usual manner,one vectorat a time. It would quite
easy to make the corresponding changes in gauss . However, the LU decomposition
method, describedinthe next article, is more versatile in handling multiple constant
vectors.

×

EXAMPLE 2.3
Use Gauss elimination to solve the equations AX = B , where

⎡
⎣

⎤
⎦

⎡
⎣

⎤
⎦

6 − 4 1 − 14 22

A = − 4 6 − 4 B = 36 − 18

1 − 4 6 6 7