Principles of Mathematics in Operations Research

Solutions 227

A 2 = Al = (A--lf =

00^

00 0

00 0

=> dimM(Ai) = 3 - rank(Ai) = 3-1 = 2.

A\ = 0 => rfim7V(yl?) = 3 =>• m = 3, m(s) = f s - — J.

Choose v 3 e Af(Al) 9» 2 = AiV 3 ^6 ^ Afv 3 = ui-

t; 3 = e^ = (0,0,l)J => v 2 = A 1 v 3 = 0,— ,0

Thus,

A^10 =

1010

5 =

1 10 45

0 1 10

0 0 1

1

00 ool

0^0

0 01

=

• 1
  100
  0
  0

=>

00"

TO 0

0 1_

10'

S~lAS

Vi = A\V 2

Vioo

,0,0

in -^1 -

10

10

10 J-

10 J

10

100

10

1

= Syl^10 ^-^1.

Note that the calculation of A^10 is as hard as that of A^10 since A is not

diagonal. However, because (easy to prove by induction)

A" (i)A"-^1 Q)A"-^2

A" (^)A"-^1

An

[A 1 1

A 1

A

n

=

we have

A^10 =

(^)°io(^)

9

«(^)

8

(TO)^10 IO(^)

(TO)^10.

10 10

1 100 4500

0 1 100

0 0 1

Hence, it is still useful to have Jordan decomposition.

4.4 (a)

—— = -0.03Yi - 0.02F 2 —- = -0.04Yi - 0.0iy 2

at at

^ = -0.05.Yi - 0.02X 2 ~ = -0.03Xi - 0.00X 2

at at

Let WT = [Xi,X2,Yi,Y 2 }. Then, the above equation is rewritten as

dW

dt

= AW,