Modern Control Engineering

Example Problems and Solutions 705

A–9–10. Show that by use of the minimal polynomial, the inverse of a nonsingular matrix Acan be ex-
pressed as a polynomial in Awith scalar coefficients as follows:

(9–100)

wherea 1 ,a 2 ,p,amare coefficients of the minimal polynomial

Then obtain the inverse of the following matrix A:

Solution.For a nonsingular matrix A, its minimal polynomial f(A)can be written as

whereamZ0. Hence,

Premultiplying by A–1, we obtain

which is Equation (9–100).

For the given matrix A, adj(lI-A)can be given as

Clearly, there is no common divisor d(l)of all elements of adj(lI-A). Hence,d(l)=1.

Consequently, the minimal polynomial f(l)is given by

Thus, the minimal polynomial f(l)is the same as the characteristic polynomial.

Since the characteristic polynomial is

we obtain

f(l)=l^3 + 3 l^2 - 7 l- 17

∑l I-A∑=l^3 + 3 l^2 - 7 l- 17

f(l)=

∑l I-A∑

d(l)

=∑l I-A∑

adj(I-A)=C

l^2 + 4 l+ 3

3 l+ 7

l+ 1

2 l+ 6

l^2 + 2 l- 3

2

- 4

• 2 l+ 2
  l^2 - 7

S

A-^1 =-

1

am

AAm-^1 +a 1 Am-^2 +p+am- 2 A+am- 1 IB

I=-

1

am

AAm+a 1 Am-^1 +p+am- 2 A^2 +am- 1 AB

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

A= C

1

3

1

2

- 1

0

0

- 2

- 3

S

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

A-^1 =-

1

am

AAm-^1 +a 1 Am-^2 +p+am- 2 A+am- 1 IB