TASK MATHEMATICS for ECONOMIC APPLICATIONS 3/09/2019
I M 1) Determine all the roots of the equation B "' œ !% . Two possible procedures for the solution.
From B "' œ B % B % œ !% # # we get:
B % œ ! Ê B œ „# % Ê B œ „ # and
B % œ ! Ê B œ „# % œ „ "% Ê B œ „ #3. Or:
From B "' œ !% we get B œ% "' and so, since "' œ "' †cos! 3sin! we get:
% "' œ% "' †cos! 5 # sin! 5 # œ # †cos5 sin5
%1 3 %1 1# 3 1#
for ! Ÿ 5 Ÿ $ to get:
5 œ ! À # †cos! 3sin!œ # à 5 œ " À # †cos 3sin œ #3 à
# #
1 1
5 œ # À # †cos1 sin1 à 5 œ $ À # †cos 1 sin 1 .
3 œ # $ 3 $ œ #3
# #
I M 2) Given the matrix œ determine the value of the parameter for which the5
" " !
! " !
" ! 5
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 - ˆœ ! we get " " 5
" " !
! " !
" ! 5
œ œ !
-
-
-
- - - to get the
eigenvalues -" œ-# œ " and -$ œ5. So the matrix admits the imaginary unit as an eigenva-3 lue if and only if 5 œ 3. So we get œ . To check if the matrix is diagonalizable
" " !
! " !
" ! 3
or not we have to study only the multiple eigenvalue - œ " and its geometric multiplicity. Since
Rank Rank , we get and so the
" †ˆ œ œ # 7 œ $ # œ " 7 œ #
! " !
! ! !
" ! 3 " "
1 +
"
matrix is not a diagonalizable one.
I M 3) Given the linear system , check for the existence and
B #B B B œ "
B B 7 B B œ #
#B $B #B 5B œ $
" # $ %
" # $ %
" # $ %
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:
" # " " l " " # " " l "
" " 7 " l # ! " 7 " ! l "
# $ # 5 l $ ! ! $ 7 5 # l !
Ä
having used elementary operations: V Ã V V# # " and V Ã V V V$ $ # ". So:
if 7 Á $ or 5 Á #: Rank œRank ˜l œ $ and the system has ∞%$ œ ∞" solutions;
if 7 œ $ and 5 œ #: Rank œRank ˜l œ # and the system has ∞%# œ ∞# solutions.
I M 4) Given the two orthogonal vectors —" œ "ß "ß ! and —# œ "ß "ß ! , find a third vec- tor —$ orthogonal to —" and —#, so as to create a basis for ‘$. Then find the coordinates of the vector ˜ œ "ß "ß " in this basis.
If —$ œ Bß Cß D , since we need —"†—$ œ—#†—$ œ ! we get:
"ß "ß ! † Bß Cß D œ B C œ !
"ß "ß ! † Bß Cß D œ B C œ ! whose solution is B œ !ß C œ !ß a D. If we choose D œ "
we get —$ œ !ß !ß " and so the basis is —œ"ß "ß ! à "ß "ß ! à !ß !ß " .
To find the coordinates of the vector ˜ œ "ß "ß " in this basis we must solve the system:
" " ! B " B B œ " B œ "
" " ! B " B B œ " B œ !
! ! " B " B œ " B œ "
† œ Ê Ê
" " # "
# " # #
$ $ $
.
II M 1) Given 0 Bß C œ B / C / C B, determine all the directions @œcosαßsinα for which it results W@0 !ß ! œ W#@ @0 !ß ! .
0 Bß C œ B / C C /B is a twice differentiable function a Bß C − ‘#. So:
H 0@ !ß ! œ f0 !ß ! † @ and H 0#@ß@ !ß ! œ @ †‡ !ß ! † @X . We get f0 B ß C œ / C C /Bà B / C /B Ê f0 !ß ! œ "ß "; so W@0 !ß ! œ "ß " † cos αßsinαœcosαsin .α
From f0 B œ / à B / we get Bß C œ / and so:
/ B /
ß C C / / C / /
/
C C C
C C
B B B B
‡ B
‡ !ß ! œ ! ! ÊH 0 !ß ! œ † ! ! † œ !
! ! ! !
#@ß@ cos sin cos
α α sinα
α .
Finally W@ W# α α α α α α
0 !ß ! œ @ @0 !ß ! œ ! œ œ œ
% %
&
Ê cos sin Êcos sin Ê 1 or 1 .
II M 2) Given the system and the point P , de-
0 Bß Cß D œ BCD B C D œ !
1 Bß Cß D œ /BC /CD œ ! ! œ "ß "ß "
termine at least one implicit function that can be defined with it and then calculate the derivatives of such function at the proper point.
0 "ß "ß " œ " " " " œ !
1 "ß "ß " œ " " œ ! and the system is satisfied.
From ` 0 ß 1 we get:
` Bß Cß D œ CD " BD " BC "
/ / / /
BC BC CD CD
` 0 ß 1
` Bß Cß D "ß "ß " œ ! # ! # ! œ # Á !
" # " # "
and since it is possible to define an implicit function B Ä C B ß D B . For its derivatives we get:
.C ! .D #
.B œ œ # œ !à .B œ œ # œ "
! ! # !
" " # "
# ! # !
# " # "
.
II M 3) S Max/min
.c. :
olve the problem: .
u 0 Bß C œ #B C C Ÿ B Ÿ "
#
#
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:
ABß Cß- -"ß #œ#B C # -"C B # -#B ". 1) case -" œ !ß-# œ ! À
A A
wB wC
œ # !
œ #C œ ! Á Ÿ B C# B Ÿ "
: no solution.
2) case -" Á !ß-# œ ! À
A -
A - -
wB "
wC " "
œ # œ !
œ #C # C œ #C " œ ! C œ B#
B Ÿ "
from which we get two systems:
B œ ! C œ !
! !ß !
- -
" œ # "
À B Ÿ " true
and since the point may be a minimum point;
while the second system is impossible.
- -
"
"
œ # œ "
C œ B# B Ÿ "
3) case -" œ !ß-# Á! À
A -
A
- -
wB #
wC
# #
œ # œ ! œ #C œ !
Ê ! "ß
œ "
C œ ! B œ "
C
B
# Ÿ B
œ #
! Ÿ " À
! true
. Since the point may be a maximum
point.
4) case -" Á!ß-# Á! À
A - -
A -
- - - -
- -
wB " #
wC "
" # " #
" "
œ # œ !
œ #C # C œ ! Ê
œ # œ #
#C " œ ! #C " œ !
B œ C# B œ " B œ "
B œ " C œ " C œ "
and .
- -
-
- -
- -
" #
"
"
#
" #
œ #
" œ !
Ê ! ! "ß
œ "
œ B œ "
B œ "
"
C œ " "
C œ "
. Since and the point is nor a maxi-
mum nor a minimum point.
- -
-
- -
- -
" #
"
"
#
" #
œ #
" œ !
Ê ! ! "ß
œ "
œ B œ "
B œ "
"
C œ " "
C œ " . Since and the point is nor a ma- ximum nor a minimum point.
So, from Weierstrass Theorem, "ß! is the maximum point with 0 "ß ! œ# while !ß! is the minimum point with 0 !ß ! œ Þ!
II M 4) Given the function 0 Bß C œ B $BC C $ # 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 Ê 0 œ $B $C œ $ B C œ ! Ê
0 œ #C $B œ !
B B œ B B œ ! C œ B
0 Bß C œ
w # #
Bw C
# $ $
# #
$
#
and so we get two stationary points: T œ" !à ! and T œ# $ *à
# % Þ For the second order conditions we construct the Hessian matrix: ‡Bß C œ 'B $.
$ #
‡ !à ! œ ! $ ‡ œ * ! !à !
$ # . Since # the point is a saddle point.
‡ ‡
$ * ‡
# % œ * $ T
$ #
à
. Since " the point is a minimum point.
# œ * ! # ! #
œ ") * ! ;
