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INTERMEDIATE TESTMATHEMATICS for ECONOMIC APPLICATIONS 04/12/2017

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INTERMEDIATE TEST

MATHEMATICS for ECONOMIC APPLICATIONS 04/12/2017

I M 1) Using trigonometric form: "  3 œ#cos 1 sen 1 œ# / and

%  3 % 31%

"  $ 3 œ # œ # / "  3 œ # / œ #/

"  $ 3 # / #

 

 

cos 1 sen 1  ; thus

$  3 $ 3 3

3 3

1 1

1

$ 1 "#

%

$

and     

"  3 # " "

"  $ 3 œ / œ / œ œ

# % %

% %

3"#1 31$ cos  3 sen 

$ $

1 1

œ " $ œ $

% "  "

#  # 3 )  ) 3 .

I M 2) The characteristic polynomial of is  -

-

-

-

: œ œ

#  # "

! 5  !

" # # 

 

 

 

 

 

 

 

œ 5  #  " œ 5  #   " œ 5   %  $ œ

" # 

 - -   -  -   - - - 

-

# #

œ 5  - -  $- "; the matrix  has multiple eigenvalue if 5 œ " or 5 œ $. For 5 œ " the multiple eigenvalue is - œ " with algebraic multiplicity equal to .#

The matrix   equal to

 

 

 

 

 

 

 

 

 

 

 

 

 " †ˆ œ "

" # "

! ! !

" # "

has rank , so the geometric multipli- city of - œ " is equal to and so for # 5 œ " the matrix is diagonalizable.

For 5 œ $ the multiple eigenvalue is - œ $ has algebraic multiplicity equal to but# the matrix  

 

 

 

 

 

 

 

 

 

 

 

 

 $ †ˆ œ #

 " # "

! ! !

" #  "

has rank equal to , so its geometric mul- tiplicity is equal to and so for " 5 œ $ the matrix is not diagonalizable.

I M 3) The dimension of the Image of a linear map is equal to the rank of the matrix as- sociated to the map; the matrix , using the elementary operations:

" Ç‘%# Ç‘$% Ñ% 7 †‘"

" ! ! 5

! " ! !

! ! " !

! ! ! "  75

, , , can be reduced to

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and easily we see that: Rank  œ $

% if

if 57 œ ". 57 Á "

So if 57 œ " DimImm 0 œ $ and Dim Ker  0 œ " while DimImm 0 œ % and Dim Ker  0 œ ! if 57 Á ".

I M 4) An element — belongs to the Kernel of the linear map if 0 — œX †—œ or

   

     

     

     

     

     

     

   

 

 

 

 

 

 

 

 

" " # !

# " " !

$ # $ !

" !  " !

† œ

B B B

"

#

$

that in system form is:

(2)

 

 

 

 

 

 

 

B B  B B  B

B B  B B  $B B  B

B B  B B  B B B

B  B B B

" # $ # $

" # $ # $ # $

" # $ # $ " $

" $ " $

 # Ê # Ê

# œ ! $ œ !

# œ ! œ ! œ $

$ $ œ ! ' œ ! œ

œ ! œ

.

Every element of the ernel is K and so a basis for

  

  

  

B B "

B  B 

B B "

œ œ B †

" $

# $

$ $

$

   

   

   

   

   

 $   $

the Kernel is UO/<0 œ"ß  $ß ".

I M 5) Given the matrix  œ we must verify that it has two eigenvalues

" ! "

! " !

" ! "

 

 

 

 

 

 

 

 

 

 

 

 

equal to those of the matrix # œ † .

# œ † œ † œ

" ! " " ! " # ! #

! " ! ! " ! ! " !

" ! " " ! " # ! #

     

     

     

     

     

     

     

     

     

     

     

      .

The characteristic polynomial of is:

: œ  œ œ "  œ

"  ! "

! "  !

" ! " 

"  "

" " 

     

 

 

 

 

 

   

-  -ˆ -

-

-

-

-

- œ "  -  " -# " œ "   - - # #- œ-" - -  # œ ! and so the three eigenvalues of are  -" œ !, -# œ ", -$ œ #.

The characteristic polynomial of # is:

: œ  œ œ "  œ

#  ! #

! "  !

# ! # 

#  #

# # 

#     

 

 

 

 

 

   

-  -ˆ -

-

-

-

-

-

#

œ "  -  # -# % œ "   - - # %- œ-" - -  % œ ! and so the three eigenvalues of # are -" œ !, -# œ ", -$ œ %.

The Matrices and  # have two common eigenvalues: œ !- and - œ ".

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