ABSTRACT
In the previous five chapters we have discussed a
wide range of topics, from what we called “the essential
essentials” of MATLAB and Octave to the usage of various
different objects that the software provides. We have also
highlighted the flexibility that is provided by the scripting
In this chapter we present a few examples that showcase the
use of both MATLAB and Octave in context. This should
not be taken as a thorough and rigorous discussion of
the topics used, but rather as a small smörgåsbord that
represents the handling of some of the techniques discussed
earlier on in the book. We therefore recommend reading
this chapter with the help of a suitable textbook, which may
help clarify some of the concepts and ideas, as well as a
running instance of MATLAB or Octave. We have provided
some basic references in each example with the intention
of guiding you to find further information on each of the
feel
Given that the most basic object in MATLAB and
Octave is a matrix, it seems natural to use the software for
linear algebra applications first. Consider the following
and solve the corresponding linear system to find the coeffi-
c1(2, 0,−1, 7) + c2(−1, 3, 4, 7) +c3(0, 2, 5, 7) + c4(0, 1, 3, 5) = (5, 4, 12, 33). (6.7)
We can write the system of equations in terms of matrices as
Notice that matrix A is made out of the transpose of the original vectors. We solve this problem in MATLAB and
It follows that the system is consistent with the solution
and hence v is a linear combination of v1 , v2 , v3 and v4 with
A =
; (6.19) we would like to find its eigenvalues and eigenvectors2. 2 Strang, G. (2003). Introduction
to Linear Algebra. WellesleyCambridge PressWe need to find the values λ that satisfy the characteristic
A− λI =
=
, (6.21) and then calculate its determinant. Using MATLAB and
(A− λI) x = 0, (6.22)
The command eig returns a diagonal matrix, lambda, that
contains the eigenvalues, and a matrix V whose columns
correspond to the eigenvectors. This means that the eigen-
λ2 = −19.6154, (6.24) λ3 = −3.7534, (6.25)
−0.456435−0.099460 −0.884180
With the aid of Equation (6.22) we can verify that the results
given by MATLAB and Octave are correct. In this case, we
can recast the equation in terms of the eigenvector v1 to verify that Av1 = λ1v1 :
the dif-
precision used you may see a different result but very close
to zero. Finally, similar relationships can be verified for the
other eigenvectors and eigenvalues, i.e. Av2 = λ2v2 and Av3 = λ3v3 .
