Question

linear algebra

Use the age transition matrix L and the age distribution vector x1 to find the age distribution vectors X2 and X3. Then find a stable age distribution vector. 0 3 18 0 0 1 450 450 L = X1 = 0 1 9 0 450 X2 X3 Find a stable age distribution vector. X=t

Answer 1

Given age transition matux L= O 3 18 1 0 0 0 1 o 9 age distibution vector 450 450 450 The age distribution matrix X₂ is given by (vector) X = LEXI D 18 4.50 - O 0 450 =10 0 450 0X 450+ 3x450 +18X450 1x 450 + 0x450 +0X450 0x450 + 2X 450 + 0x450 9 X2 9450 450 50

Finding roots of characteristic polynomial -23 +31 +2=0 put Al -(1837 +360+2 = -1+3+2 = 4 Put A=2 -(203+3(2)+2 = -87672=0 x=2 i the root of the equation simple division me thod =2 0 3 2 0 -2 - 4 ادم ہے -1 -2 27-12-21-1=0 n²7214120 all » (x+y)2 = 7 7 =-1,-1 Nawes are x=2,4,- Ergen Take the positive Eigenvalue x=2 and find the Eigenweltor.

Eigenvector for 1:2 L-XI 3 13 З 1g 1 р-) о - 2 о — о 40-) о а - 2 О-гр р, - р. + 1 ві 2 - ) 18 3 12 -] а б 6+12 1+-) ) - + 3 2 о - 2 о - 2 а агт Pz - P3+ Хр, - 3 18 -1 2 о об apply R,= Rit 6xp, Ri=Rila р: 24р., - 2 D 121 о - 36 О - ) 9 2. Г. - 18 о О б боб (L-AI) - Х - б О - 20 — О -18 І:) : - ОО D

- X, - 36X3 =0 X2-18X₂=0 let X3=t => X, 360 20 - x = 36t and X2 - 18t - => X₂ = 1st 9Xz=t The eigenwector rector for for das is 3bt x = 36 18 + 18t - t (The distribution vector) stable age