TASK MATHEMATICS for ECONOMIC APPLICATIONS 3/09/2019
I M 1) Determine all the roots of the equation B ο "' Ε !% . Two possible procedures for the solution.
From B ο "' Ε B ο % B ο % Ε !% ο # οο # ο we get:
B ο % Ε ! Γ B Ε β# ο% Γ B Ε β # and
B ο % Ε ! Γ B Ε β# οο % Ε βοο "ο% Γ B Ε β #3. Or:
From B ο "' Ε !% we get B Εο% "' and so, since "' Ε "' β ο‘cos! ο 3sin!ο’ we get:
ο% "' Εο% "' β οcosο! ο 5 # ο sinο! ο 5 # οο Ε # β οcosο5 ο sinο5 οο
%1 ο 3 %1 1# ο 3 1#
for ! ΕΈ 5 ΕΈ $ to get:
5 Ε ! Γ # β ο‘cos!ο 3sin!ο’Ε # Γ 5 Ε " Γ # β οcos ο 3sin οΕ #3 Γ
# #
1 1
5 Ε # Γ # β ο‘cos1 sin1ο’ Γ 5 Ε $ Γ # β οcos 1 sin 1ο .
ο 3 Ε ο # $ ο 3 $ Ε ο #3
# #
I M 2) Given the matrix ο Ε determine the value of the parameter for which the5
" " !
! " !
" ! 5
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
matrix admits the imaginary unit as an eigenvalue. For this value of , find all the eigenvalues3 5 of the matrix and check if it is diagonalizable or not.
From ο«οο- Λο«Ε ! we get " " 5
" " !
! " !
" ! 5
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο ο’ ο‘ ο’ο‘ ο’ο‘ ο’
ο
ο
Ε ο ο ο Ε !
-
-
-
- - - to get the
eigenvalues -" Ε-# Ε " and -$ Ε5. So the matrix admits the imaginary unit as an eigenva-3 lue if and only if 5 Ε 3. So we get Ε . To check if the matrix is diagonalizable
" " !
! " !
" ! 3
ο
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
or not we have to study only the multiple eigenvalue - Ε " and its geometric multiplicity. Since
Rankο¬ ο¬ Rank , we get and so the
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
οο " β Λ Ε Ε # 7 Ε $ ο # Ε " ο 7 Ε #
! " !
! ! !
" ! 3 ο " "
1 +
"
matrix is not a diagonalizable one.
I M 3) Given the linear system , check for the existence and
ο
οο
B ο #B ο B ο B Ε "
B ο B ο 7 B ο B Ε #
#B ο $B ο #B ο 5B Ε $
" # $ %
" # $ %
" # $ %
number of its solutions on varying the parameters and .7 5
To apply Rouchè-Capelli theorem we study the Rank of the matrix and the Rank of the augmen- ted matrix:
ο£ ο£ ο£ ο£
ο£ ο£ ο£ ο£
ο£ ο£ ο£ ο£
ο£ ο£ ο£ ο£
ο£ ο£ ο£ ο£
ο£ ο£ ο£ ο£
" # ο " " l " " # ο " " l "
" " 7 " l # ! ο " 7 ο " ! l "
# $ # 5 l $ ! ! $ ο 7 5 ο # l !
Γ
having used elementary operations: V Γ V ο V# # " and V Γ V ο V ο V$ $ # ". So:
if 7 Γ $ or 5 Γ #: Rankο‘ ο’ο ΕRankο‘ο Λl ο’Ε $ and the system has β%ο$ Ε β" solutions;
if 7 Ε $ and 5 Ε #: Rankο‘ ο’ο ΕRankο‘ο Λl ο’Ε # and the system has β%ο# Ε β# solutions.
I M 4) Given the two orthogonal vectors β" Ε "Γ "Γ !ο‘ ο’ and β# Ε "Γ ο "Γ !ο‘ ο’, find a third vec- tor β$ orthogonal to β" and β#, so as to create a basis for β$. Then find the coordinates of the vector Λ Ε "Γ "Γ "ο‘ ο’ in this basis.
If β$ Ε BΓ CΓ Dο‘ ο’, since we need β"β β$ Εβ#β β$ Ε ! we get:
οο‘ ο’ ο‘ ο’
ο‘"Γ "Γ ! β BΓ CΓ D Ε B ο C Ε !ο’ ο‘ ο’
"Γ ο "Γ ! β BΓ CΓ D Ε B ο C Ε ! whose solution is B Ε !Γ C Ε !Γ a D. If we choose D Ε "
we get β$ Ε !Γ !Γ "ο‘ ο’ and so the basis is βΕο₯ο‘"Γ "Γ ! Γ "Γ ο "Γ ! Γ !Γ !Γ "ο’ ο‘ ο’ ο‘ ο’ο¦.
To find the coordinates of the vector ο’ο’ο’ο’ο’ο’ ο’ ο£ ο£ ο£ ο£ο’ ο£ ο£ ο£ ο£ο’ ο£ ο£ ο£ ο£ο’ ο£ ο£ ο£ ο£ο’ ο£ ο£ ο£ ο£ο’ ο£ ο£ ο£ ο£ Λ Ε "Γ "Γ "ο‘ ο’ in this basis we must solve the system:
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο’ ο’
ο ο
ο ο
ο ο
" " ! B " B ο B Ε " B Ε "
" ο " ! B " B ο B Ε " B Ε !
! ! " B " B Ε " B Ε "
β Ε Γ Γ
" " # "
# " # #
$ $ $
.
II M 1) Given 0 BΓ C Ε B / ο C /ο‘ ο’ C B, determine all the directions @Εο‘cosΞ±ΓsinΞ±ο’ for which it results W@0 !Γ ! Εο‘ ο’ W#@ @0 !Γ !ο‘ ο’.
0 BΓ C Ε B / οο‘ ο’ C C /B is a twice differentiable function a BΓ C βο‘ ο’ β#. So:
H 0@ ο‘ ο’!Γ ! Ε f0ο‘ ο’!Γ ! β @ and H 0#@Γ@ ο‘ ο’!Γ ! Ε @ β β‘ο‘ ο’!Γ ! β @X . We get f0 Bο‘ Γ Cο’ ο‘Ε / οC C /BΓ B / οC /Bο’ Γ f0ο‘ ο’ ο‘!Γ ! Ε "Γ ο "ο’; so W@0 !Γ ! Εο‘ ο’ ο‘"Γ ο " β cosο’ ο‘ Ξ±ΓsinΞ±ο’ΕcosΞ±οsin .Ξ±
From f0 B Ε / ο Γ B / ο we get BΓ C Ε ο / ο and so:
/ ο B /
ο‘ Γ Cο’ ο‘ C / / ο’ ο‘ ο’ C / /
/
C C C
C C
B B B B
β‘ οΎ B οΎ
β‘ο‘ ο’!Γ ! Ε ! ! ΓH 0ο‘ ο’ ο¬!Γ ! Ε ο¬β ! ! β Ε !
! ! ! !
οΎ οΎ #@Γ@ cos sin οΎ οΎ οΎcos οΎ
Ξ± Ξ± sinΞ±
Ξ± .
Finally W@ W# Ξ± Ξ± Ξ± Ξ± Ξ± Ξ±
0 !Γ ! Ε @ @0 !Γ ! ο Ε ! Ε Ε Ε
% %
&
ο‘ ο’ ο‘ ο’ Γ cos sin Γcos sin Γ 1 or 1 .
II M 2) Given the system ο ο‘ ο’ and the point P , de-
ο‘ ο’ ο‘ ο’
0 BΓ CΓ D Ε BCD ο B ο C ο D Ε !
1 BΓ CΓ D Ε /BοCο /CοD Ε ! ! Ε "Γ "Γ "
termine at least one implicit function that can be defined with it and then calculate the derivatives of such function at the proper point.
ο ο‘ ο’
ο‘ ο’
0 "Γ "Γ " Ε " ο " ο " ο " Ε !
1 "Γ "Γ " Ε " ο " Ε ! and the system is satisfied.
From ` 0 Γ 1 we get:
` BΓ CΓ D Ε CD ο " BD ο " BC ο "
/ ο / ο / /
ο‘ ο’
ο‘ ο’ οΎ BοC BοC CοD CοD οΎ
` 0 Γ 1
` BΓ CΓ D "Γ "Γ " Ε ! # ! # ! Ε # Γ !
" ο # " ο # "
ο‘ ο’
ο‘ ο’ο‘ ο’ οΎ οΎ and since οΊ οΊ it is possible to define an implicit function B Γ C B Γ D Bο‘ ο‘ ο’ ο‘ ο’ο’. For its derivatives we get:
.C ! .D #
.B Ε ο Ε ο # Ε !Γ .B Ε ο Ε ο # Ε ο "
! ! # !
" " ο # "
# ! # !
ο # " ο # "
οΊ οΊ οΊ οΊ
οΊ οΊ οΊ οΊ
.
II M 3) S Max/min
.c. :
olve the problem: .
οu 0 BΓ C Ε #B ο C C ΕΈ B ΕΈ "
ο‘ ο’ #
#
The objective function of the problem is a continuous function, the feasible region is a com-X pact set, and so surely exist maximum and minimum values.
Using Kuhn-Tucker's conditions, we form the Lagrangian function:
Aο‘BΓ CΓ- -"Γ #ο’Ε#B ο C ο# -"οC ο B ο# ο -#ο‘B ο "ο’. 1) case -" Ε !Γ-# Ε ! Γ
οο
ο
οο
οο A A
wB wC
Ε # !
Ε ο #C Ε ! Γ ΕΈ B C# B ΕΈ "
: no solution.
2) case -" Γ !Γ-# Ε ! Γ
οο
ο
οο
οο
A -
A - -
wB "
wC " "
Ε # ο Ε !
Ε ο #C ο # C Ε ο #C " οο‘ ο’Ε ! C Ε B#
B ΕΈ "
from which we get two systems:
οο
ο
οο
οο B Ε ! C Ε !
ο ! !Γ !
- -
" Ε ο # "
Γ B ΕΈ " true
and since the point ο‘ ο’ may be a minimum point;
while the second system is impossible.
οο
ο
οο
οο - -
"
"
Ε ο # Ε ο "
C Ε B# B ΕΈ "
3) case -" Ε !Γ-# Γ! Γ
ο ο
ο ο
ο ο
ο ο
ο ο
ο ο
ο ο
A -
A
- -
wB #
wC
# #
Ε # ο Ε ! Ε ο #C Ε !
Γ ο ! "Γ
Ε "
C Ε ! B Ε "
C
B
# ΕΈ B
Ε #
! ΕΈ " Γ
! true
. Since the point ο‘ ο’ may be a maximum
point.
4) case -" Γ!Γ-# Γ! Γ
ο ο ο
ο ο ο
ο ο ο
ο ο ο
ο ο ο
ο ο ο
ο ο ο
A - -
A -
- - - -
- -
wB " #
wC "
" # " #
" "
Ε # ο ο Ε !
Ε ο #C ο # C Ε ! Γ
ο Ε ο # ο Ε ο #
ο #C " ο Ε ! ο #C " ο Ε !
B Ε C# B Ε " B Ε "
B Ε " C Ε " C Ε ο "
ο‘ ο’ and ο‘ ο’ .
ο ο
ο ο
ο ο
ο ο
ο ο
ο ο
ο ο
- -
-
- -
- -
" #
"
"
#
" #
ο Ε ο #
" ο Ε !
Γ ο ! ο ! "Γ
Ε ο "
Ε B Ε "
B Ε "
"
C Ε " "
C Ε "
. Since and the point ο‘ ο’ is nor a maxi-
mum nor a minimum point.
ο ο
ο ο
ο ο
ο ο
ο ο
ο ο
ο ο
- -
-
- -
- -
" #
"
"
#
" #
ο Ε ο #
" ο Ε !
Γ ο ! ο ! "Γ ο
Ε ο "
Ε B Ε "
B Ε "
"
C Ε ο " "
C Ε ο " . Since and the point ο‘ ο’ is nor a ma- ximum nor a minimum point.
So, from Weierstrass Theorem, ο‘ ο’"Γ! is the maximum point with 0 "Γο‘ ο’! Ε# while ο‘ ο’!Γ! is the minimum point with 0 !Γο‘ ο’! Ε Γ!
II M 4) Given the function 0 BΓ C Ε B ο $BC ο Cο‘ ο’ $ # nalyze the nature of its stationary points.a To nalyze the nature of the stationary points of the function we apply first and second order a conditions. For the first order conditions we pose:
f Γ 0 Ε $B ο $C Ε $ B ο C Ε ! Γ
0 Ε #C ο $B Ε !
B ο B Ε B B ο Ε ! C Ε B
0 BΓ C Εο‘ ο’ ο ο ο‘ ο’
ο ο ο
w # #
Bw C
# $ $
# #
$
#
and so we get two stationary points: T Ε" ο‘ ο’!Γ ! and T Ε# ο$ *Γ ο
# % Γ For the second order conditions we construct the Hessian matrix: β‘ο‘BΓ C Εο’ οΎ 'B ο $οΎ.
ο $ #
β‘ο‘ ο’!Γ ! ΕοΎ ! ο $οΎ ο« ο«β‘ Ε ο * ο ! ο‘ ο’!Γ !
ο $ # . Since # the point is a saddle point.
β‘ β‘
ο$ *ο β‘
# % Ε * ο $ T
ο $ #
Γ οΎ οΎ οο« ο«
. Since ο« ο«" the point is a minimum point.
# Ε * ο ! # ο ! #
Ε ") ο * ο ! ;
