MATHEMATICS for ECONOMIC APPLICATIONS TASK 23/4/2020
I M 1) After having determined the complex number which is the solution of the equationD
D D
" 3 " 3 œ ", calculate its square roots D.
D D " " " 3 " 3
" 3 " 3 œ D " 3 " 3 œ D " 3 " 3 œ D #3 œ " Ê D œ " œ 3
# 3
.
Since 3 œcos $ 3sin $ we getÀ
# #
1 1
D œ cos$ # sin$ #
%1 5 #1 3 %1 5 #1 ß ! Ÿ 5 Ÿ "
.
For 5 œ ! D œ # 3 # à
# #
we get ! cos $ sin$
%1 3 %1 œ
for 5 œ " D œ # 3 #.
# #
we get " cos ( sin(
%1 3 %1 œ
I M 2) Given the matrix œ , since œ " is a multiple eigenvalue for ,
" ! !
" 5 " !
" 5 5
-
find the values of the real parameter and check, for such values, if is diagonalizable or5 not.
From - ˆœ ! we get:
1 .
" ! !
" 5 " !
" 5 5
5 5
" œ !
-
-
-
- - -
œ
To get - œ " as a multiple eigenvalue we have two possibilities:
5 " œ !Ê5 œ " and 5 " " œ !Ê5 œ # .
For 5 œ " the matrix is œ à
" ! ! ! ! !
" ! ! " " !
" " " " " !
and the matrix is
ˆ
"
since " " we get Rank
" " œ # Á ! "ˆ œ #Ê 7 œ $ # œ " 7 œ #1" +" and so the matrix is not a diagonalizable one.
For 5 œ # the matrix is œ à
" ! ! ! ! !
" " ! " ! !
" # # " # "
and the matrix is
ˆ
"
since " ! we get Rank
" # œ # Á ! "ˆ œ #Ê 7 œ $ # œ " 7 œ #1" +" and so the matrix is not a diagonalizable one.
I M 3) Since the linear system has the solution , calculate the
5B #B B œ #
#B 5B B œ #
$B B 5B œ $
"ß "ß "
" # $
" # $
" # $
value of the real parameter and then find the number of the solutions of the system.5
Substituting the solution we get:
5 # " œ #
# 5 " œ #
$ " 5 œ $
Ê 5 œ " and the system becomes:
B #B B œ # " " " " " "
#B B B œ # # " " ! & $ ! & $
$B B B œ $ $ " ! & ! !
" # $
" # $
" # $
. But
2 2 2
" # "
œ œ œ & .
From Cramer's Theorem, since the determinant of the matrix is different from zero the system has one and only one solution.
I M 4) Determine all the vectors orthogonal to the vector —" œ #ß "ß " and to the vector —# œ "ß "ß " .
Two vectors are orthogonal if their scalar product is equal to zero.
So a vector —œ Bß Cß D is orthogonal to the vector —" œ #ß "ß " ifÀ
Bß Cß D † #ß "ß " œ #B C D œ ! Ê D œ C #B . Vector becomes — Bß Cß C #B. Now the vector —œ Bß Cß C #B is orthogonal to the vector —# œ "ß "ß " ifÀ
Bß Cß C #B † "ß "ß " œ B C C #B œ #C B œ ! Ê B œ #C .
So all the vectors orthogonal to the vector —" œ #ß "ß " and to the vector —# œ "ß "ß " are the vectors — œ #Cß Cß $C œ 5 #ß "ß $ .
II M 1) Given the equation 0 Bß C œ / BC /BC œ ! satisfied at the point !ß ! , verify that an implicit function B Ä C B can be defined with it, and then calculate the first order deri- vative of this function.
The function 0 Bß C is a differentiable function a Bß C − ‘#. It also turns outÀ f0 B ß C œ /BC /BCà /BC /BC for which f0 !ß ! œ !à # .
Since 0Cw !ß ! œ # Á ! it is possible to define an implicit function B Ä C B . For its derivative we have: C ! œ ! œ ! Þ
#
w
II M 2) Solve the problem Max min . s v
Î 0 Bß C œ B C D Þ Þ B C D œ $# # #
The objective function of the problem is a continuous function, the constraint defines a feasi- ble region which is a compact set since it consists of only boundary points (the surface of aX sphere) and therefore surely maximum and minimum values exist.
For a problem with equality constraints we construct the Lagrangian function and apply to it the first and second order conditions. We have:
ABß Cß Dß-œ B C D-B C D $# # # . Applying the first order conditions we have:
f Ê Ê Ê
œ " # B œ ! œ " # C œ ! œ " # œ !
œ œ œ
A -
A -
A -
A -
Bß Cß Dß œ
D B C D œ $
B C D
œ $
wB wC wD
# # #
"
# "
" #
" # " "
% % %
- - -
-# -# -#
Ê Ê Ê ∪ œ
œ œ
œ œ œ
œ " œ "
œ " œ "
œ " œ "
B C D
œ $
B C D
œ
B B
C C
D D
œ œ
Þ
"
# "
" #
$ #
%
"
# "
" #
#" " "
- - - -
- - -
# - - -
# % # #
We have only two stationary points: T œ "ß "ß "" and T œ "ß "ß "# .
By Weierstrass' theorem, since 0 "ß "ß " œ $ and 0 "ß "ß " œ $ it obviously turns out that is the maximum point while is the minimum point.T" T#
If we want to apply the second order conditions we have to built the bordered Hessian matrix:
‡
Bß Cß Dß œ
! #B #C #D
#B # ! !
#C ! # !
#D ! ! #
- -
-
-
to calculate then, in the points and , theT" T#
minors ‡$ and ‡%. Since ‡
T" œ À
! # # #
# " ! !
# ! " !
# ! ! "
it is
‡$ T" œ ) ! and ‡% T" œ "# ! and so is the maximum point;T"
since ‡
T# œ À
! # # #
# " ! !
# ! " !
# ! ! "
it is
‡$ T# œ ) ! and ‡% T# œ "# ! and so is the minimum point.T#
II M 3) Given 0 Bß C œ B C and 1 Bß C œ B C determine for which values of parameter α it is W@0 "ß " œ W@1 "ß " , with @œcosαßsinα.
The functions 0 Bß C œ B C and 1 Bß C œ B C are polynomials and therefore are differen- tiable in any order a Bß C − ‘#.
So H 0@ "ß " œ f0 "ß " † @ and H 1@ "ß " œ f1 "ß " † @. Since:
f0 B ß C œ Cà BÊf0 "ß " œ "ß " and f1 B ß C œ "à "Êf1 "ß " œ"à ", to get W@0 "ß " œ W@1 "ß " it will be:
"ß " cosαßsinαœ " à "cosαßsinαÊcosαsinαœcosαsinαÊsinαœ ! and so αœ ! or αœ1.
II M 4) Given the function 0 Bß Cß D œ B C D # $ logD B / CD, determine the gradient vector of the function at T œ "ß #ß #! .
It is f BC D " à $B C D / à B C " /
D B D B
0 Bß Cß D œ # $ # # CD # $ CD and so:
f "ß #ß # " à #% / à ) " /
# "
0 œ $# ## ## œ $"à #$à "!.
