6.4 Linear Systems 179
det{A}=∑
p(− 1 )pa 1 p 1 ·a 2 p 2 ···an pn,where the sum runs over all permutationspof the indices 1 , 2 ,...,n, altogethern!terms. To
calculate the inverse ofAis a formidable task. Here we have to calculatethe complementary
cofactorai jof each elementai jwhich is the(n− 1 )determinant obtained by striking out the
rowiand column jin which the elementai jappears. The inverse ofAis then constructed
as the transpose of a matrix with the elements(−)i+jai j. This involves a calculation ofn^2
determinants using the formula above. A simplified method ishighly needed.
With the LU decomposed matrixAin Eq. (6.18) it is rather easy to find the determinant
det{A}=det{L}×det{U}=det{U},since the diagonal elements ofLequal 1. Thus the determinant can be written
det{A}=N
∏
k= 1ukk.The inverse is slightly more difficult. However, with an LU decomposed matrix this reduces
to solving a set of linear equations. To see this, we recall that if the inverse exists then
A−^1 A=I,the identity matrix. With an LU decomposed matrix we can rewrite the last equation as
LUA−^1 =I.If we assume that the first column (that is column 1) of the inverse matrix can be written as
a vector with unknown entries
A− 11 =
a− 111
a− 211
...
a−n 11
,
then we have a linear set of equations
LU
a− 111
a− 211
...
a−n 11
=
1
0
...
0
.
In a similar way we can compute the unknow entries of the second column,
LU
a− 121
a− 221
...
a−n 21
=
0
1
...
0
,
and continue till we have solved allnsets of linear equations.
A calculation of the inverse of a matrix could then be implemented in the following way:
- Set up the matrix to be inverted.
- Call the LU decomposition function.