TASK 8/1/2020
QUANTITATIVE METHODS for ECONOMIC APPLICATIONS MATHEMATICS for ECONOMIC APPLICATIONS
I M 1) If D œ # # $ / 1% 3 # " $ / 1$ 3, calcolate D. Since / œ cos sin " " and
# #
1% 3 1 1
% 3 % œ 3
/ œ " $
# #
1$ 3
cos1 sin1 it is:
$ 3 $ œ 3
D œ # # $ " " # " $ " $ œ
# # # #
3 3
œ # $ " 3 " $" 3$ œ
œ # $ " $ 3 # $ $ $ œ " 3 .
From " 3 œ # " " œ # † cos sin we get:
# #
3 ( 3 (
% %
1 1
D œ" 3 œ % # † cos( 5 # sin( 5 # Ÿ Ÿ à )1 #1 3 )1 #1 ß ! 5 "
And so - œ" % # †cos( sin ( and - œ# % # †cos "& sin"& .
)1 3 )1 )1 3 )1
I M 2) Given the matrix œ determine the two values of the parameter for5
" " "
" ! "
" 5 "
which the matrix admits multiple eigenvalues. For these values of , check if the correspon-5 ding matrix is diagonalizable or not.
From we get
- ˆ œ ! œ œ
" " " " "
" " ! "
" 5 " 5 "
- -
- -
- - -
œ - - #- 5 -" - œ - - #- 5 " - œ
œ - - # # 5 "- œ !. The first eigenvalue is - œ !. To get multiple eigenva- lues there are two possibilities.
1) -# # 5 " œ !- for - œ !Ê5 " œ !Ê5 œ " Þ The characteristic polyno- mial becomes - - # #-œ !Ê- -# # œ ! Ê -" œ-# œ ! and -$ œ #.
For 5 œ " and - œ ! we get:
! †ˆ œ œ " Á ! ! †ˆ œ #
" " "
" "
" " "
" "
" !
! and since we get Rank
and so 7 œ $ # œ " 7 œ #1! +! and the matrix, for 5 œ " and - œ ! is not a diagona- lizable one.
2) The second degree polynomial -# # 5 "- may have a double root.
Solving we get: -œ "„" 5 " œ "„ 5 # and so we get the double root - œ " if 5 œ # ÞFor 5 œ # and -œ " we get:
" †ˆ œ œ " Á !
! " "
" "
" # !
! "
" "
" and since also this time we get:
Rank " †ˆœ # and so 7 œ $ # œ " 7 œ #1" +" and the matrix, for 5 œ # and - œ " is not a diagonalizable one.
I M 3) Consider the linear map 0 À‘% Ä‘%, ˜œ —† for which:
0 B ß B ß B ß B " # $ % œ B $B à #B #B à B #B à #B %B" # " # $ % $ %.
Determine the dimensions of the Kernel and of the Image of this linear map, and then find a basis for the Kernel.
From 0 B ß B ß B ß B " # $ % œ B $B à #B #B à B #B à #B %B" # " # $ % $ % we get:
œ
" $ ! !
# # ! !
! ! " #
! ! # %
. By elementary operations on the rows:
V Ã V #V# # " and V Ã V #V% % $ we get:
œ Ä
" $ ! ! " $ ! !
# # ! ! ! % ! !
! ! " # ! ! " #
! ! # % ! ! ! !
" $ !
! % !
!
. Since
! "
œ % Á ! we see that Rank œ $ and so Dim Imm œ $ and Dim Ker œ % $ œ ".
To find a basis for the Kernel we must solve the system:
—† œ Ê † œ Ê
" $ ! ! B ! B $B œ !
# # ! ! B ! #B #B œ !
! ! " # B ! B
! ! # % B !
" " #
# " #
$ $
%
#B œ !
#B %B œ !
Ê Ê
B $B œ !
%B œ ! B #B œ !
%
$ %
" #
#
$ %
Ê !ß !ß #5ß 5
B œ ! B œ ! B œ #B
"#
$ %
. So the vectors belonging to the Kernel are of the form and a basis for the Kernel may be the vector !ß !ß #ß ".
I M 4) Given the matrix œ " # determine at least a matrix , different from , that is
# "
similar to .
Similarity between matrices is satisfied if it exists a non singular matrix such that:
† œ † Ê œ † † œ # " œ # " œ " Á !
" " " "
" . If we choose it is
and so: œ # " Ê œ " " Ê œ " " and finally:
" " " # " #
‡ ‡ T
" œ " " Þ œ"† †
" #
So, from we get:
œ " " † " # † # " œ " " † % $ œ " !
" # # " " " " # & $ ' $
.
II M 1) Given 0 Bß C œ B C BC BC # # and @œcosαßsinα , determine at least two va- lues of for which it results α H 0 "@ ß " œ H 0 "@ß@# ß " .
0 Bß C œ B C BC BC # # is a twice differentiable function a Bß C − ‘#. So H 0 "@ ß " œ f0 " ß " † @ and H 0 "#@ß@ ß " œ @ †‡ "à " † @X .
It is f0 B ß Cœ#BC C C#à B B #BC Ê f0 "# ß " œ #ß # .
So we get: H 0 "@ ß " œ f0 " ß " † @ œ #ß # † cosαßsinαœ #cosαsinα. Since ‡Bß Cœ #C #B " #C it is ‡ "ß " œ # $ and so:
#B " #C #B $ #
H 0 " œ † # $ † œ #
$ #
@ß@# ß " cos sin cos # ' † # #
sin cos cos sin sin
α α α
α α α α α
.
Therefore H 0 "#@ß@ ß " œ # 'cosα†sinαœ # " $cosα†sinα.
Then having to be H 0 "@ ß " œ H 0 "@ß@# ß " it will be cosαsinαœ" $cosα†sinα and this equality is certainly verified at least for α and for α 1.
œ ! œ
#
II M 2) Solve the problem .
Max/min u.c.:
0 Bß C œ B C C Ÿ " B
" B Ÿ C
# #
#
We rewrite the problem as .
Max/min u.c.:
0 Bß C œ B C B C " Ÿ !
" B C Ÿ !
# #
#
The objective function of the problem is a continuous function, the feasible region is a com-X pact set and therefore there are certainly maximum and minimum values.
As can be seen from the figure, it is 0 Bß C ! a Bß C − , X . Using Kuhn-Tucker conditions, we form the Lagrangian function:
ABß Cß- -"ß #œB C # # -"B C " # -#" B C. 1) case -" œ !ß-# œ ! À
A A
wB wC
œ #B œ ! œ #C œ !
Ê
B œ ! C œ ! B C " Ÿ !
" B C Ÿ !
! ! " Ÿ !
" ! ! Ÿ ! À
# !à !
not satisfied
in fact  X.
2) case -" Á !ß-# œ ! À
A - -
A -
-
wB " "
wC "
"
œ #B # B œ #B " œ ! œ #C œ !
Ê
B œ ! C œ
C œ " B
" B Ÿ C
"
œ # !
" Ÿ "
# so !ß " is a possible Maximum
point; or:
A - -
A -
- -
wB " "
wC "
# #
" "
œ #B # B œ #B " œ ! œ #C œ !
Ê Ê
B œ " B œ
C œ C œ
C œ " B
" B Ÿ C œ " œ "
" B Ÿ C " B Ÿ C
#
" " "
# # #
" "
# #
œ
from which:
B œ C œ
Ê
"
#
"
#
" " " "
# # # #
-" œ " !
" Ÿ Ÿ À
" "
#ß
# satisfied
so is a possible Maximum point,
while , in fact .
B œ C œ
"
#
"
#
" "
# #
-" œ "
" Ÿ À
" "
#ß
# not satisfied
 X
3) case -" œ !ß-# Á! À
A -
A -
- -
wB #
wC #
#
œ #B œ ! œ #C œ !
C œ " B Ê Ê
B œ C œ
B œ C œ
œ
C Ÿ " B œ "
C Ÿ " B
" ! Ÿ " À
# #
- -
- -
#
#
# #
#
#
# #
"
"#
#
" "
# % vera
; from # ! it fol-
lows that the point " "
# #ß is a possible Minimum point.
4) case -" Á!ß-# Á! À
A - -
A - -
- -
- - -
wB " #
wC " #
# "
" # #
œ #B # B œ !
œ #C œ !
C œ " B
Ê Ê
B œ ! B œ !
C œ " C œ "
œ ! œ #
# œ ! œ !
C œ " B# À !ß " already studied,
and:
B œ " B œ "
C œ ! C œ !
# # œ ! œ # !
- - Ê -
- - -
" # "
" # œ ! # œ # !
so "ß ! is a possible Maximum point.
Let's study the objective function 0 Bß C œ B C # # on the points of the first constraint C œ " B#. Since:
0 Bß " B # œ B " B# ## œ B B "% # it is 0 Bß " Bw #œ %B #B$ from this we get 0 Bß " B œ #B #B " ! if " (in
#
w # #
B X it is ! Ÿ B Ÿ ". Therefore:
and so the point " "
#ß
# is, relative to the boundary points only, a Minimum point, contra- dicting the previous indication -" œ " ! which indicated it as a possible maximum point. So " " it is neither a maximum nor a minimum point.
#ß
#
The same conclusion could be reached using the bordered Hessian matrix of the Lagrangian function ABß Cß-"œB C # # -"B C "# . It is for -" œ ":
‡ ‡
Bß C œ œ œ # † # # !
! #B " ! # "
#B !
" ! # # !
" ! #
# #-" Ê " " !
#ß
# ,
and this result shows us again " " as a minimum point.
#ß
#
Now let's study the objective function 0 Bß C œ B C # # on the points of the second constra- int C œ " B.
Since 0 Bß " B œ B " B # # œ #B #B "# it is 0 Bß " B œ %B #w from which 0 Bß " B œ # #B " ! "
#
w if B (in X it is ! Ÿ B Ÿ ". Therefore:
so we have confirmation that the point " "
# #ß s a minimum point, conclusion already en-i
sured by the Weierstrass Theorem, since the point " " was the only candidate found for
# #ß
a minimum point. Here too we could confirm using the bordered Hessian matrix of the Lagrangian function ABß Cß-#œB C # # -#" B C. It is:
‡ ‡
Bß C œ œ œ % !
! " " ! " "
" ! " !
" ! # " ! #
# Ê " " #
# #ß , a result
that shows us again " " .
# #ß as a minimum point
So " " " " "
# #ß s the minimum point withi 0 # #ß œ #; !ß " are "ß ! maximum points, with 0 !ß " œ 0 "ß ! œ " .
II M 3) Given the equation 0 Bß C œ B C BC #BC #B #C œ ! $ $ , satisfied at "ß " , verify that an implicit function B Ä C B can be defined with it and that such function has a stationary point. Then determine the nature of this stationary point.
Since f0Bß Cœ$B C C #C # B $BC #B ## $ à $ # it is f0 "ß " œ !ß % . It is therefore possible to define an implicit function for which .
B Ä C B C œ 4! œ !
w 1
So B œ 1 is a stationary point for the implicit function.
From ‡Bß C œ 'BC $B $C # we get ‡ œ .
$B $C # 'BC "ß "
# #
# # 6 4
4 6
From: C œ 0 #0 C 0 C C œ œ ! Þ
0 % #
' ) † ! ' † ! $
ww ww ww w ww w ww
BB BC CC #
Cw
#
, we get 1 From Cw " œ ! C " œ $ ! we get that B œ"
and ww # is a maximum point for the impli- cit function.
II M 4) Given the vectors —œ BCß # $C and ˜œ B %ß BC , determine if pairs Bß C exist for which the scalar product of the two vectors — ˜† œ 0 Bß C is maximum or mini- mum.
0 Bß C œ — ˜† œ BCß # $C † B %ß BC œ B C %BC #BC $BC # #. So 0 Bß C œ B C 'BC $BC # #. We apply first order conditions:
f Ê 0 œ #BC 'C $C œ C #B ' $C œ ! 0 œ B 'B 'BC œ B B ' 'C œ !
0 Bß C œ
w #
Bw #
C
and so we get four possi- ble solutions:
B œ ! B œ ! B œ ' B œ 'C '
C œ ! ∪ C œ # ∪ C œ ! ∪ "#C "# ' $C œ ! B œ #
Ê C œ #
$ . For the second order conditions we construct the Hessian matrix:
‡Bß C œ #C #B ' 'C
#B ' 'C 'B .
Since ‡ !ß ! œ ! ' it is ‡ œ $' ! !ß ! is a saddle point;
' !
#
Since ‡ !ß # œ % ' it is ‡ œ $' ! !ß # is a saddle point;
' !
#
Since ‡ 'ß !œ ! ' it is ‡ œ $' ! 'ß ! is a saddle point;
' $'
#
Since ‡ it is ‡ ‡ , so is
#ß # ‡ #ß
$ œ # œ !à œ "# !
# "# œ "' % œ "# !
% %
$ $ #
" " $
#
a minimum point.
