Chiang/Wainwright: Fundamental Methods of Mathematical Economics
¯ ¯ ¯ −r 3 equation of this matrix is |D − rI| = 0 or ¯¯ ¯ −1 −4 − r with (19.16’).
Instructor’s Manual
¯ ¯ ¯ ¯ = r2 + 4r + 3 = 0, which checks ¯ ¯
Exercise 19.3 ⎡
1. Since dt = ⎣
λ1
⎤
⎡
δ − a11
⎤⎡
−a12
β1
⎤
⎡
λ1
⎤
⎦ δ t , we have ⎣ ⎦⎣ ⎦=⎣ ⎦. Thus β 1 = 1 [λ1 (δ − ∆ λ2 −a21 δ − a22 β2 λ2 1 a22 ) + λ2 a12 ] and β 2 = ∆ [λ2 (δ − a11 ) + λ1 a21 ], where ∆ = (δ − a11 )(δ − a22 ) − a12 a21 . It is
clear that the answers in Example 1 are the special case where λ1 = λ2 = 1. ⎤ ⎡ δ 0 ⎦. The rest follows easily. 2. (a) The key to rewriting process is the fact that δI = ⎣ 0 δ (b) Scalar: δ. Vectors: β, u. Matrices: I, A. (c) β = (δI − A)−1 u ⎡ ⎤ ⎡ ⎤ ⎡ ρ 0 1 0 a ⎦+⎣ ⎦ − ⎣ 11 3. (a) ρI + I − A = ⎣ 0 ρ 0 1 a21 rest follows easily.
a12 a22
⎤
⎡
⎦=⎣
ρ + 1 − a11
−a12
−a21
ρ + 1 − a22
⎤
⎦. The
(b) Scalar: ρ. Vectors: β, λ. Matrices: I, A. (c) β = (ρI + I − A)−1 λ. t 4. (a) With trial solution β i δ t = β i ( 10 12 ) , we find from (19.22’) that β 1 =
x1p =
70 12 t 39 ( 10 )
20 12 t 13 ( 10 ) . 3 b − 10
and x2p =¯ ¯ 4 ¯ − 10 (b) From the equation ¯¯ 3 2 ¯ − 10 b − 10 These give us m1 = 4A1 , n1 = 3A1 ; m2
6 t 1 t and x2c = 3A1 ( 10 ) − A2 (− 10 )
¯ ¯ ¯ ¯] = b2 − ¯ ¯
5 10 b
−
6 100
70 39
and β 2 =
= 0, we find b1 =
6 10 ,b2
20 13 .
So
1 = − 10 .
6 t 1 t = A2 , n2 = −A2 . Thus x1c = 4A1 ( 10 ) + A2 (− 10 )
(c) Combining the above results, and utilizing the initial conditions, we find A1 = 1 and A2 = −1. Thus the time paths are x1,t = 4(
6 t 1 70 12 ) − (− )t + ( )t 10 10 39 10
x2,t = 3(
6 t 1 20 12 ) − (− )t + ( )t 10 10 13 10
t
5. (a) With trial solution β i eρt = β i e 10 , we find from (19.25’) that β 1 = x1p =
17 t/10 6 e
and x2p =
19 t/10 . 6 e
131
17 6
and β 2 =
19 6 .
So