Page 121 - Algorithm Collections for Digital Signal Processing Applications using MATLAB

3. Numerical Linear Algebra 109

T
E2= [-0.4484 -0.5592 -0.6973]

T
E3= [0.2600 -0.4686 0.8443]

Thus V= U’’
U’ is obtained as follows
U

(-0.4484) + c3 e
U’’= c1 e -2.2470 t (-0.8990) + c2 e 0.8019 t -0.5550 t (0.2600)

(-0.5592) + c3 e
U’= c1 e -2.2470 t (0.4001) + c2 e 0.8019 t -0.5550 t (-0.4686)

(-0.6973) + c3 e
U= c1 e -2.2470 t (-0.1781) + c2 e 0.8019 t -0.5550 t (0.8443)

The values of the constants can be solved using the initial conditions
using the set of equations given below.

U’’(0) =3 =(-0.8990) c1 + (-0.4484) c2 + (0.2600) c3

U’(0) =1 = (0.4001) c1 + (-0.5592) c2 + (-0.4686) c3

U(0) = -1 = (-0.1781)c1 + (-0.6973)c2 + (0.8443) c3

Thus c1= -3.4815, c2= -1.5790, c3= -3.2227

10. COMPUTATION OF PSEUDO INVERSE
OF THE MATRIX

Let us consider the matrix A = 2 3 4
3 4 5

Decompose the matrix ‘A’ using Singular Value Decomposition.