Graphics Programs Reference
In-Depth Information
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
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