TASK MATHEMATICS for ECONOMIC APPLICATIONS 11/07/2017

(1)

TASK MATHEMATICS for ECONOMIC APPLICATIONS 11/07/2017

I M 1) D  D  %D  % œ D D  "  % D  " œ D  " D  % œ^$ ^# ^#      ^# 

œD  " D  #3 D  #3  ; so D  D  %D  % œ D  " D  #3 D  #3 œ !^$ ^#     . The three roots are D œ " D œ #3 D œ  #3_" , _# , _$ . Since D œ " œ "_" cos!  3sin!;

D œ #3 œ #  3 D œ  #3 œ #  3

# # # #

$ $

# cos1 sin 1 and $ cos 1 sin 1 we get:

D † D † D œ " † # † # † !    3 !   œ

# # # #

$ $

" # $ cos 1 1 sin 1 1

œ %cos 21 3sin 21œ % /^{# 3}¹ œ /^{691 %# 3}¹ .

I M 2) Since  œ

" " " "

"  " "  "

7 7 " "

 

 , the image of B ß B ß B ß B" # $ % can be written

as J B ß B ß B ß B œ ; B B B B

 " # $ %

"

#

$

%

 †

  

using elementary operations on the rows we get:

 

" " " " " " " "

"  " "  " !  # !  # 7 7 " " 7  " 7  " ! !

Ê

V  V V  V

# "

$ "

 

  from which we easily see

that Dim Imm if while Dim Ker if ; the dimension

otherwise otherwise

 œ#  œ#

$ "

7 œ " 7 œ "

of the Kernel is maximum if 7 œ " .

For a basis of the Kernel remember that B ß B ß B ß B_" _# _$ _% belongs to the Kernel of ifJ

† œ

  

   B B B B

"

#

$

%

; in system form we get:



  

B  B  B  B B  B  B  B B  B  B  B

B  B  B  B B B

B  B  B  B

" # $ %

" # $ % $ "

" # $ %

œ ! œ ! œ !

Ê œ ! Ê œ 

œ ! B œ  B_% _# , every ele-

ment of the Kernel is , so a

   

B B " !

B B ! "

B B " !

B !  "

B B

" "

# #

$ "

%

" #

œ œ 

 

 B_#

   

basis for the

Kernel is U_{O/< J}_{ }œ" !ß ß "ß ! ß ! " !ß  "  ß ß . For a basis of the Image remember

that C ß C ß C  belongs to the Image of if , that in system form is B

B B B

C C C

" # $

"

#

$

%

"

#

$

J † œ

    

   

  



B  B  B  B C B  B  B  B C B  B  B  B C

C C

" # $ % "

" # $ % #

" # $ % $

" $

œ œ œ

Ê œ ; so every element of the Image is a vector

(2)

 

   

C C C_"ß _#ß _$ with C œC œ œC C and a basis for the

C C " !

C C ! "

C C " !

$ " " #

" "

# #

$ "

or

   

Image is U_{M77 J}_{ } œ" ! " ß ! "ß ß   ß ß !. I M 3) The characteristic polynomial of is:

: œ  œ  œ #   œ  %  $

  - -ˆ  -    - - -

 # " -

" # ^# " ^#

while the characteristic polyinomial of is:

: œ  œ  œ +  ,  œ  +  ,  +,

  - -ˆ  -    - - -  -

 + !-

! ,

# . The

two matrices have the same caractheristic polynomial if and only if +  , œ % and +, œ $; so +œ " and , œ $ or +œ $ and ,œ ".

For the dimentions of the eigenspaces associated to the eigenvalues of the matrix note that the two eigenvalues are simple, i.e. they have algebraic multiplicity equal to and"

so also their geometric multiplicity is equal to . So any eigenspace has " dimension equal to ".

I M 4) A modal matrix that diagonalizes , if it exists, is a non singular matrix having columns that are linearly independent eigenvectors associated to the matrix . To find   the first step is to calculate the characteristic polynomial of :

:  œ  œ œ   œ

 

   

-  -ˆ -

"  "  "

!  !

# #  # 

"   "

#  # 

-

- œ  -" - # - #œ -^$-^# œ  -^#- " Þ

From - -^#  " œ ! we get the three eigenvalues -" œ-# œ ! and -$ œ  " . Since 7 œ $ ¹_! Rank !ˆœ $ Rank  œ $  " œ # we get 7 œ 7_!¹ ⁺_! and the matrix is a diagonalizable one.

To find two eigenvectors @ œ@ ß @ ß @_" _# _$ associated to the eigenvalue we have to! solve the system   ! †ˆ† œ@ , and so:





@  @  @ !

! !

@  #@  #@ !

@ @  @ !

" # $

$ " #

œ œ

œ

Ê œ

#

, so every eigenvector associated to is a vec-

tor @ œ œ œ 

   

@ @ " !

@ @ ! "

@ @  @ " "

@ @

" "

# #

$ " #

" #

   

   .

To find one eigenvector associated to the eigenvalue  " we solve the system:

 



 

   "

#

  † † œ Ê #

œ œ

œ

œ ˆ @ , or œ

@  @  @ !

@ !

@  #@  @ !

@ @

@ !

" # $

#

" # $

$ "

# , so every eigenvector associated to  " @ œ œ œ

#

is a vector .

   

@ @ "

@ ! !

@ @ #

@

" "

#

$ "

"

  

So a modal matrix to diagonalize may be  T œ .

" ! "

! " !

" " #

 

(3)

II M 1) 0 Bß C  ia a differentiable function with f0 œ $B ß  $C ^# ^#, so:

H 0 T œ_@   f0 † @ œ $B ß  $C ^# ^#† "ß ! œ $B  ^#, H 0 T œ_A   f0 † A œ $B ß  $C ^# ^#† !ß " œ  $C  ^#.

From H 0 T œ H 0 T@   A   it follows $B œ  $C# # ÊB  C œ !# # whose unique so- lution is T œ !ß ! .

II M 2) The Lagrangian function of the problem is:

ABß Cß Dß-œB  #C  %D-B  C  D  #"

%

# # # .

M SG f œ: A " # Bß - # # Cß- % #-D  B  C  D ß  #".

%

# # #

  

   

   

   

  





"  œ !

#  œ ! œ 

%  D œ ! B  C  D œ

Ê Ê

œ D œ

œ

œ œ 

œ  " œ "

D œ # D œ  #

œ " œ  "

# B

 # C C

#

B B B

C C

- - -

- -

# # #

"

#

#" #"

" "

# #

#"

%

#

% %

- - - -^#

or . The problem

has two constrained critical points: T_" œ "ß  ß" # T_# œ  "ß ß " #

# #

  and  .

MM SG œ

D



D 

: ‡ A -

-

 

 

!

! !

#B #C #

#B #

#C #

# #

.

Since the problem has three variables and one constraint, we must consider two leading principal minors: ‡ A_$  and ‡ A_% .

‡ A  -

- -

-

$  œ  œ    #  œ

 



 

!

#B #C

#B #

#C #

#B #B C #B #

#C # #C

œ ) B  ) C œ )- ^# - ^# -B  C^# ^# ;

‡ A  -

-

%  œ œ 

D



D 



D 

 

   

 

!

! !

!



#B #C #

#B #

#C #

# #

#B

#C #

# #

 # D œ

 

D  D

! 

C #

#B # #B #

#C #C #

# # #

   

- -

-

! !

! ! !



œ  "' B  "' C  "' D œ  "'-^{# #} -^{# #} -^{# #} -^#B  C  D^# ^# ^# . Since:

‡$ T"  ‡% T"  T" 0 T " œ #"

 !à  ! Ê is the maximum point with #

‡_$ T_#  ‡_% T_#  T_# 0 T _# œ  #"

 !à  ! Ê is the minimum point with # . II M 3) is a differentiable function with 0 0 !ß !ß ! œ !  .

f0 œ/  D B/  /^C ß ^C ^Dß  C/  B^D , f !ß !ß !0 œ"  " !ß ß ; ‡ 0 œ ‡0 Ð!ß !ß !Ñ œ

! / " ! " "

/ B/  / " !  "

"  /  C/ "  " !

   

C

C C D

D D

, .

The request polinomial is À

(4)

T_#B C D œ 0 !ß !ß !  f !ß !ß !ß ß  0 B C Dß ß  "B C D† 0 Ð!ß !ß !Ñ † œ

  #

  

  † ‡  B

C D œ B  C  BC  BD  CD .

II M 4) The Jacobian matrix ‰ œ ` 0 ß 1

` Bß Cß D

 

  is :

‰ œ sinC  D$ sinB  D/^{D BC}^ ^ B cosC  D/^{D BC}^ ^ cosB  B  C /  ^{D BC}^ ^

B C  CD^# B  D  BD^$  C  BC with

‰  T œ "  "  "

 "

! ! .

So ` Cß D

` B

 

  œ‰^" T _CßD †‰ T _B œ œ

"

 _{ }  _{ }   "  "  "

 " † !

! œ ! "  " œ ! 

"  † ! 

"   " .

Or .C ; .D .

.B œ  œ ! .B œ  œ  "

"  "  " "

! !  " !

 "  "  "  "

 " !  " !

   