Systems of Linear Algebraic Equations - Numerical Methods in Engineering with MATLAB - page 83

check =
1.0000 -0.0000 -0.0000

0 1.0000 0.0000

0 -0.0000 1.0000

EXAMPLE 2.14
Invert the matrix

⎡
⎣

⎤
⎦

−

2 1 0000

− 12 − 1 000

0 − 12 − 1 00

A =

00 − 12 − 1 0

000 − 12 − 1

0000 − 15

Solution Since the matrix istridiagonal, we solve AX I using the functions LUdec3

and LUsol3 (LU decomposition for tridiagonal matrices):

=

% Example 2.14 (Matrix inversion)
n=6;
d = ones(n,1)*2;
e = -ones(n-1,1);
c=e;
d(n) = 5;
[c,d,e] = LUdec3(c,d,e);
fori=1:n
b = zeros(n,1);
b(i) = 1;
Ainv(:,i) = LUsol3(c,d,e,b);

end
Ainv

The result is

>> Ainv =
0.8400 0.6800 0.5200 0.3600 0.2000 0.0400

0.6800 1.3600 1.0400 0.7200 0.4000 0.0800

0.5200 1.0400 1.5600 1.0800 0.6000 0.1200

0.3600 0.7200 1.0800 1.4400 0.8000 0.1600

0.2000 0.4000 0.6000 0.8000 1.0000 0.2000

0.0400 0.0800 0.1200 0.1600 0.2000 0.2400

Note that although A istridiagonal, A − 1 isfullypopulated.