TASK MATHEMATICS for ECONOMIC APPLICATIONS 07/02/2017
I M 1) $ 3 "# & 3 "# œ$ " " & 3 #3 "# œ
œ$ # & " #3 " œ$ #&† #3œ$ # 3 œ #' #$ 3 œ œ %$ cos $ sin $ %cos # sin #
#1 3 #1 œ 1# 5$ 1 3 1# 5$ 1 ,
! Ÿ 5 Ÿ #. The three roots are:
B œ % 3 œ %3
# #
" cos 1 sin 1 ;
B œ % ( 3 ( œ % $ 3" œ # $ 3
' ' # #
# cos 1 sin 1 ;
B œ % "" 3 "" œ % $ 3" œ # $ 3
' ' # #
$ cos 1 sin 1 .
I M 2) The characteristic polynomial of is
: œ œ œ
$ # "
" %
" #
- -ˆ
-
-
-
5
$
œ $ % # " " % œ
# " "
- - -
- -
5 5
$ $ #
œ $ - - #- "# #5 # $ -5 # % -œ
œ -$ %-# # ' 5- % ) 5 œ - #-# # # ) - 5.
If we define ; - œ -# # # ) - 5, - œ # is a multiple eingenvalue for iff
; # œ # ) 5œ ! or 5 œ ); in such case the characteristic polynomial : - can be factorized as : - œ - - ## and the three eingevalues are -"ß# œ # and -$ œ ! . To check if is a diagonalizable matrix we need the geometric multiplicity of - œ # to be equal to , its algebraic multiplicity.#
For this goal we consider a generic element —œ −Xf , must satisfy the— B
B B
"
#
$
-œ#
condition i.e. the linear system:
B #B B œ ! B #B B œ ! B #B )B œ ! *B œ !
B #B &B œ ! 'B œ !
Ê Ê B œ #B
B œ !
" # $ " # $
" # $ $
" # $ $
" #
$ . Every element of
Xf-œ# is a vector - is
#B #
B "
! !
œ B
#
# # ; the geometric multiplicity of œ # only " and the matrix is not diagonalizable.
I M 3) The dim Imm of a linear map is equal to the rank of the matrix associated at the map; so by elementary operations on matrix's lines we get:
" " # # " " # #
" # " " ! " " $
$ # 7 5 ! " 7 ' 5 '
Ä Ä
V V
V $V V V
# "
$ " $ #
Ä
" " # #
! " " $
! ! 7 ( 5 *
, and so we get:
Rank if . By the dimention Theorem:
otherwise
œ #
$ 7 œ ( • 5 œ *
dim Ker dim Imm œdim ‘% œ % and trivially dim Ker is maximum if and only if dim Imm is minimum and so 7 œ ( • 5 œ * with dim Imm œdim Ker œ #. To find a basis for Ker remember that belongs to the kernel of if — 0 —† œ, i.e.:
B B #B #B œ ! B B #B #B œ ! B #B B B œ ! B B $B œ !
$B #B (B *B œ ! B B $B œ !
Ê Ê
" # $ % " # $ %
" # $ % # $ %
" # $ % # $ %
Ê B œ $B &B B œ B $B
"# $ $ % % . Every element of the Kernel is a vector:
— œ $B &B B $B B ß B $ %ß $ %ß $ %œ B$ $ " "ß ! B & $ !ß "ß ß % ß ß and a basis for the Kernel is UO/< 0 œ $ " "ß ! ß & $ !ß "ß ß ß ß .
To find a basis for Imm note that a vector belongs to the image of if ˜ 0 —† œ˜, that in system form is:
B B #B #B œ C B B #B #B œ C B #B B B œ C B B $B œ C C
$B #B (B *B œ C B B $B œ C $C
Ê Ê
" # $ % " " # $ % "
" # $ % # # $ % # "
" # $ % $ # $ % $ "
Ê C C %C œ ! C œ %C C
B B #B #B œ C B B $B œ C C
! œ C C %C
" # $ % "
# $ % # "
$ # "
$ # " $ " #
. From we get
and so every element of Imm is a vector:
˜ œ C C C "ß #ß $ œ C C %C C"ß #ß " #œ C " ! % C ! "" ß ß # ß ß " and a basis for the Image is UM77 0 œ" ! % ß ! "ß ß ß ß ".
I M 4) If two matri es are similar they have the same characteristic polynomial, and so,- calculating the two polynomials we get:
: œ œ " # œ % "
" $
- -ˆ - - -
-
# and
: œ œ œ % " : œ :
"
- -ˆ - - - - -
-
"
%
# , so .
For similarity we need a non singular matrix such that:
† œ † Ê " # † : : œ : : † ! " Ê
" $ : : : : " %
#""" ##"# ""#" ##"#
Ê : #: : #: œ : %: :
: $: : $: : %: :
"""" #"#" "#"# #### ##"# "### ""#" and in system form we get:
: #: œ : : #: œ :
: #: œ %: : : #: œ $: : œ #
: $: œ : : $: œ :
: $: œ %: : : : œ :
Ê Ê
"" #" "# "" #" "#
"# ## "# "" "" ## "# ""
"" #" ## "" #" ##
"# ## ## #" "# #" ##
: : : œ : :
#" "#
## "# #" .
is any matrix of the form œ #: ; ; ; Á „ #:
: : ;
with .
II M 1) It is easy to verify that on point T the given equation is satisfied.
The gradient of the functions is f0 œ #B/C B# # /BCß #C/C B# # /BC from which f0 T œ $ß $ ; since 0Cw T Á ! the equation defines an implicit function
B Ä " œ T œ œ "
C B C 0 T $
0 $
with w Bw . The Hessian matrix
Cw
is:
‡
Bß C œ # #B " / / %BC/ /
%BC/ / # #C " / /
# C B BC C B BC
C B BC # C B BC
# # # #
# # # # , from which:
‡ "ß " œ " $
$ &
.
Since C 0 0 C 0 C we get
0
ww BBww BCww w CCww w
Cw
" œ T # T † " T † "
T
#
Cww # $
" œ " " & œ !
$
† † "
.
II M 2) The Lagrangian function of the problem is
AÐBß Cß ß Ñ œD - B C D# # # Ð- B C D "Ñ and its gradient is:
f œ #A B ß- #C ß- #D ß Ð- B C D " .Ñ
M SÞGÞ Ê
# œ !
# œ ! œ œ œ
# œ ! œ
: , with only one solution, the point
B
C B C D
D
Î#
$ Î#
-
- -
- -
B C D œ "
"
T œ "Î$ß "Î$ß "Î$ß #Î$ .
MM SÞGÞ œ
#
#
#
: ‡ A
!
! !
! !
! !
" " "
"
"
"
, since the problem has three variables
and one constraint, we must consider two principal minors, ‡ A$ and ‡ A% .
‡$ T œ œ œ # # œ % !
#
#
#
#
!
!
!
!
!
" "
"
"
" "
" " ;
‡% T œ œ
#
#
#
!
! !
! !
! !
" " "
"
"
"
œ
! ! ! !
! ! ! ! #
! ! !
# #
#
# # !
œ % % % œ "# !
" " "
" " "
" " "
. On point T, ‡$ T and ‡% T are both negative, so T is a minimum point with
min 0 œ 0 "Î$ "Î$ "Î$ß ß œ"Î$ .
II M 3) The function 0 Bß C œ B C BC # # is a twice differentiable function, so H 0@ T! œ f0 T ! † @ and H@ßA# 0 T! œ@ †f0 T ! † AT.
Since @ œ #ß # , A œ #ß # , 0 Bß C BC C B BC
# # # #
f œ # #ß # # ,
f œ " " Bß C # # # !
# # ! #
0 T C B C T
B C B
! ß ; ‡ œ , ‡ ! œ .
And so:
H 0@ T 0 T † @ † #ß # #
# #
! œ f ! œ " " ß œ ;
H@ßA# 0 T @ † 0 T † A #ß # † †
# #
! !
#
#
#
#
œ f œ # ! œ
! #
T
œ
# #
# ß # † #
# œ # .
II M 4) f0 œ œ
0 0 0
Bw Cw Dw
#C D C # †
† 691 † # † 691 † $ C #
† 691 † $D
$ #CD " $ C# "
#CD C# #
#CD #
B BD D
B B BD BD
B B
$ $
$ $
$
C #$ BD C# "$ †B .
