Understanding Engineering Mathematics

Exercise on 13.6

Where possible, solve the systems of Exercise 13.5 using the inverse matrix.

Answer

See answer to Exercise 13.5.

13.7 Eigenvalues and eigenvectors

Consider the system of equations

3 x+ 2 y= 0

6 x+ 4 y= 0

Note that the determinant of coefficients of the left-hand side is
∣∣
∣
∣

32

64

∣∣

∣

∣=^0

If either of the right-hand sides was non-zero this would mean the system has no solution.
However, when the right-hand sides are all zero, we say the system ishomogeneous,and
in this case thereisthe possibility of a solution. In fact, cancelling 2 from the second
equation reveals that the equations are the same – we really only have one equation:

3 x+ 2 y= 0

Thishas an infinite number of solutions – choose any value ofy,y=s,thentakexas

−

2

3

s.
In fact, in the homogeneous case it is essential for a non-trivial solution that the
determinant of the coefficients is zero. We can generalise this. Returning to the original
motivation for the inverse matrix, remember that we wanted to solve

Au=b

by multiplying byA−^1 :

A−^1 (Au)=(A−^1 A)u=Iu=u

=A−^1 b.

So we have to insist in this case on|A|=0, ifb= 0. On the other hand, ifb= 0 ,and
we have thehomogeneous system

Au= 0

then ifAis non-singular we only get thetrivial solutionu= 0. It is called trivial because
if it were, say, to correspond to a physical dynamical system then it would represent
the state of zero motion – no activity. In the case of homogeneous systems, weonlyget