Page 215 - Numerical Analysis Using MATLAB and Excel

Chapter 5 Differential Equations, State Variables, and State Equations

6. Expansion of the given matrix yields
· · · ·
x = x 2 x = x 3 x = x 2 x = – x – 2x – 3x – 4x + ut()
3
2
1
3
4
2
4
1
Letting x = y we obtain
dy 4 dy 3 dy 2 dy
-------- + 4 -------- + 3-------- + 2------ + y = ut()
dt 4 dt 3 dt 2 dt
7.
a.
(
)
–
A = 1 2 det A λI = det⎜ ⎛ 1 2 – λ 10 ⎞ ⎟ = det 1 – λ 2 = 0
3 – 1 ⎝ 3 – 1 01 ⎠ 3 – 1 – λ
2
)
( 1 λ – – λ – 6 = 0 , 1– – λ + λ λ + 2 – 6 = 0 , λ = 7 , and thus λ = 7 λ = – 7
)
(
–
1
2
1
b.
–
)
(
B = a0 det B λI = det⎜ ⎛ a0 – λ 10 ⎞ ⎟ = det a λ 0 = 0
–
a – b ⎝ a – b 01 ⎠ a – b – λ
)
)
(
( a λ b – λ = 0 , and thus λ = a λ = b
–
2
1
c.
⎛ ⎞
0 1 0 ⎜ 0 1 0 10 0 ⎟
(
)
C = 0 0 1 det C λI = det⎜ 0 0 1 – λ 01 0 ⎟
–
– 6 – 11 – 6 ⎜ ⎝ – 6 – 11 – 6 00 1 ⎠ ⎟
– λ 1 0
= det 0 – λ 1 = 0
–
– 6 – 11 – 6 λ
2
2
3
)
)
(
λ – 6 λ – – – ( 11 – ) ( λ = λ + 6λ + 11λ + 6 = 0 and it is given that λ = – 1 . Then,
6
–
1
2
3
λ + 6λ + 11λ + 6 2
)
---------------------------------------------- = λ + 5λ + ⇒ ( λ + 1 λ ) ( + 2 λ ) ( + 3 = 0
6
( λ + 1 )
and thus λ = – 1 λ = – 2 λ = – 3
1
1
2
8.
a. Matrix is the same as Matrix C in Exercise 7. Then,
A
λ = – 1 λ = – 2 λ = – 3
1
1
2
5−54 Numerical Analysis Using MATLAB® and Excel®, Third Edition
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