Example Problems and Solutions 701

Since

we have

Sincex(0)= 0 andk=1,referring to Equation (9–96), we have

Hence, the output y(t)can be given by

A–9–8. The Cayley–Hamilton theorem states that every nnmatrixAsatisfies its own characteristic
equation. The characteristic equation is not, however, necessarily the scalar equation of least
degree that Asatisfies. The least-degree polynomial having Aas a root is called the minimal
polynomial. That is, the minimal polynomial of an nnmatrixAis defined as the polynomial
f(l)of least degree,

such that or

The minimal polynomial plays an important role in the computation of polynomials in an n*n

matrix.

Let us suppose that d(l), a polynomial in l, is the greatest common divisor of all the elements

of adj(lI-A). Show that, if the coefficient of the highest-degree term in lofd(l)is chosen as

1, then the minimal polynomial f(l)is given by

Solution.By assumption, the greatest common divisor of the matrix adj(lI-A)isd(l). Therefore,

adj(lI-A)=d(l)B(l)

f(l)=^2

l I-A

d(l)

2

f(A)=Am+a 1 Am-^1 +p+am- 1 A+am I= 0

f(A)= 0 ,

f(l)=lm+a 1 lm-^1 +p+am- 1 l+am , mn

y(t)=[1 0]B

x 1

x 2

R=x 1 =e-0.5tsin0.5t

=B

e-0.5tsin 0.5t

• e-0.5t(cos 0.5t+sin 0.5t)+ 1

R

=B

0

- 2

1

- 2

RB

0.5e-0.5t(cos 0.5t-sin 0.5t)-0.5

e-0.5tsin 0.5t

R

=A-^1 AeAt-IB B

x(t)=eAt x( 0 )+A-^1 AeAt-IB Bk

= B

e-0.5t(cos0.5t-sin0.5t)

2e-0.5tsin0.5t

• e-0.5tsin0.5t
  e-0.5t(cos0.5t+sin0.5t)

R

(t)=eAt=l-^1 C(s I-A)-^1 D

= D

s+0.5-0.5

(s+0.5)^2 +0.5^2

1

(s+0.5)^2 +0.5^2

- 0.5

(s+0.5)^2 +0.5^2

s+0.5+0.5

(s+0.5)^2 +0.5^2

T

(s I-A)-^1 = B

s+ 1

• 1

0.5

s

R

• 
  

  1

  
  

1

s^2 +s+0.5

B

s

1

- 0.5

s+ 1

R