Department of Economics
Econ 2750A
Solution to Assignment #2
1. By multiplying out AB and BA, and comparing each element from the resulting
matrices, we get
12  4k 
 1
AB  

 30  20  k 
Thus, (a)
 1
BA  
15  5k
 24 
 20  k 
1=1
(b)
12  4k  24  4k  36
 k 9
(c)
15  5k  30  5k  45
 k 9
(d)
 20  k  20  k
The solution is k  9 will make AB  BA .
2. In class we have the following formula:
x t  P t x0
for t  1, 2,  ,10
Thus, to answer all three parts of the question, we need to find P t for t  2, 5,10 .
(a) The transition matrix is:
0.9 0.7
P

 0.1 0.3
0.88 0.84
(b) P 2  PP  
;
0.12 0.16
0.876 0.868
P3  P2P  

0.124 0.132
0.87504 0.87472
 0.87500001 0.87499991
P5  P3P2  
; P 10  P 5 P 5  


0.12694 0.12528
0.12499999 0.12500009
Hence,
1008
x 2  P 2 x0  
;
 192 
1049.664
1049.9999
; x 10  P 10 x 0  
x 5  P 5 x0  


 150.336 
 150.0001 
So the unemployment people after 2 periods are 192, after 5 periods are
approximately 150 and after 10 periods are 150.
(c) It is obvious that the steady-state unemployment is 150.
3. Given that X is any n  K matrix, then the dimension of M  I  P can be
calculated as follows
(a) dim( P )  (n  K ) (( K  n) (n  K ))( K  n)  n  n which implies dim( I )  n  n
So dim( M )  n  n .
(b) To show M is idempotent we need to show that MM = M. We have already
shown in class that P is idempotent. Thus
MM  ( I  P )( I  P )  I  IP  PI  PP  I  2 P  P  I  P  M
(c) trace (M )  trace ( I  P )  trace (I )  trace (P )
Now trace (P )  trace ( X ( X T X ) 1 X T )  trace (( X T X 1 ) X T X )  trace ( I K ) = K
Thus, trace (M )  n  K
4. The products are given below
 1 15 4 
(a) A B  12 0 2 
 25 25 10 
T
 1 12 25 
B A  15 0
25 
 4  2  10
T
Note that B T A  ( AT B) T so that all you have to multiply is AT B and then take
the transpose of this product to get B T A .
(b) The determinant for two matrices are given below
A  (2)( 1) 3 2
1 2
 (2)[(1)(5)  (3)( 2)]  22
3 5
(by expanding the 2nd
2 3
 (5)[( 2)( 1)  (3)(6)]  100
6 1
(by expanding the 2nd
column)
B  (5)( 1) 2 2
column)