Hence
q=
After some easy transformation we receive:
(I-βP)= |
1 - 0, 5β -0, 5β -0, 4β 1 - 0, 6β |
T 1 - 0, 5β -0, 4β
= -0, 5β 1 - 0, 6β
and
(I - βP)-1
1 1 - 0,6β 0,5β
det (I - βP)[ 0, 4β 1 - 0, 5β
11 β2
det(I-βP) = (1 -0,5β) (1 -0,6β) = 1
-10 β + β0 = (i — β )(1 — 0, ιβ ).
Finally
(I - βP)-1
1-0,6β
(1-β)(1-0,1β)
0,4β
(1-β)(1-0,1β)
0,5β
(1-β)(1-0,1β)
1-0,5β
(1-β)(1-0,1β)
Now as an example, we decompose into partial fractions the first element of ma-
trix (1 - βP)-1.
We receive:
1 - 0,6β = D111 + D21 = 4 + 5
(1 - β)(1 - 0, 1β) (1 - β) (1 - 0, 1β) (1 - e) (1 - 0, 1β)
because
1 - 0, 6β = D111 (1 - 0, 1β) + (1 - β) D121.
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