TASK MATHEMATICS for ECONOMIC APPLICATIONS 5/07/2019
I M 1) If D œ / and D œ / cos$ 3sin$ , calculate D † D .
% %
" "$ 31 # " #
1 1
From D œ /" "$ 31 œ /"† /$ 31 œ /"cos$1 3sin$1œ
œ /"cos1 3sin1œ /".
So D † D œ / † cos sin † / cos$ 3sin$ œ
% %
" # " 1 3 1 1 1
œ / † / $ $ œ " † ( 3 (
% % % %
"
cos1 1 sin1 1 cos 1 sin 1.
3
So D † D œ" # " †cos sin .
( # ( #
)1 5 #1 3 )1 5 #1 ß ! Ÿ 5 Ÿ "
For 5 œ ! À ( 3 ( , for 5 œ " À "& 3 "& Þ
) ) ) )
cos 1 sin 1 cos 1 sin 1
I M 2) Given the matrix œ find an orthogonal matrix which diagona-
! " ! !
" ! ! !
! ! ! "
! ! " !
lizes .
The matrix is a symmetric one, and so it is surely diagonalizable by an orthogonal matrix.
From
- ˆ
- -
- - -
- -
- -
œ œ œ
" ! ! " " ! !
" ! ! " ! !
! ! " ! ! "
! ! " ! ! "
œ " " œ
"
- -
-
- -
- -
! ! ! !
! " ! "
! " ! "
œ " -- - # " " -"-# " œ - "- "-# " œ !
for -" œ-# œ " and -$ œ -% œ ".
For finding the corresponding eigenvectors we solve two systems.
" †ˆ †—œ Ê † œ Ê
" " ! ! B !
" " ! ! B !
! ! " " B !
! ! " " B !
"
#
$
%
Ê B B œ ! Ê B œ B B ß B ß B ß B
B B œ ! B œ B
" # " #
$ % $ % and so the eigenvectors " " $ $ from which we get two orthogonal eigenvectors —" œ "ß "ß !ß ! and —# œ !ß !ß "ß " .
" †ˆ †—œ Ê † œ Ê
" " ! ! B !
" " ! ! B !
! ! " " B !
! ! " " B !
"
#
$
%
Ê B B œ ! Ê B œ B B ß B ß B ß B
B B œ ! B œ B
" # " #
$ % $ % and so the eigenvectors " " $ $ from which we get two orthogonal eigenvectors —$ œ "ß "ß !ß ! and —% œ !ß !ß "ß " .
Using the corresponding unit vectors as columns we find the orthogonal matrix:
” œ
! !
! !
! !
! !
" "
# #
" "
# #
" "
# #
" "
# #
.
I M 3) The matrix œ + + admits the eigenvectors "ß " and "ß " both correspon-
+ +
"" "#
#" ##
ding to the eigenvalue - œ ". Find the matrix . The matrix satisfies À
† " œ " † " œ " Ê + + † " œ " Ê
" " " + + " "
""#" "### Ê + + œ " Ê + œ " +
+ + œ " + œ " +
""#" "### "##" ""## . The matrix satisfies also À
† " œ " † " œ " Ê + + † " œ " Ê
" " " + + " "
""#" "###
Ê + + œ " Ê + " + œ " Ê #+ œ # Ê + œ "
+ + œ " " + + œ " #+ œ # + œ "
""#" "### "" ## ""## ##"" ""## .
So + œ ! and " !.
+"# œ ! œ ! "
#"
I M 4) Given the vector subspace of – ‘$ having base –" œ "ß !ß " à –# œ "ß "ß ! , check the values for the parameter for which the vector 5 —œ "ß "ß 5 belongs to .–
Our problem is like checking for the existence of solutions of a linear system having this augmented matrix: . To get Rank Rank so to satisfy
˜l œ œ ˜l
" " l "
! " l "
" ! l 5
Rouchè-Capelli Theorem, since Rank œ #, we simply need to get Det ˜l œ ! and so:
" " " " " " " " "
! " " ! " " ! " "
" ! 5 ! " 5 " ! ! 5
œ œ œ 5 œ ! .
So the vector —œ "ß "ß 5 belongs to only if – 5 œ !.
II M 1) Given the function 0 Bß C œ B / C C /B and the unit vector @œcosαßsinα, find the values of for which α H 0 "#@ß@ ß " œ /.
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 ! ;
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 ! H 0 " œ †‡ †
! /
/
Ê #@ß@ ß " cos sin "ß " cos œ
α α sinα
α
œ / œ / œ
cos sin cos cos sin / cos
sin sin
α α α α α α
α α
† ! † †
! /
œ /cos#α /sin#αœ /cos# œ /α forÀ
cos# œ " Ê # œ Ê œ or # œ $ Ê œ
# #
α α 1 α 1 α 1 α $1
.
II M 2) Given the equation 0 Bß Cß D œ B / CD C /BD œ !, satisfied at T œ !ß !ß ! , verify that it is possible to define an implicit function Bß D Ä C Bß D and then calculate the deriva- tives of such function at C œ !.
From f0 B ß Cß D œ /CD C /BDàB /CD /BDà B /CD C /BD we getÀ
f0!ß !ß ! œ " à "à ! and since 0Cw!ß !ß !œ " Á ! it is possible to define an implicit function Bß D Ä C Bß D . For its derivatives we have:
`C " `C !
`B œ " œ "à `D œ " œ !.
II M 3) S
Max/min u.c.:
olve the problem: .
0 Bß C œ B C
$C B $ Ÿ !
!
!
#
Ÿ B Ÿ C
The objective function of the problem is a continuous function, the feasible region is a triangleX in the first quadrant of the real plan, i.e. a compact set, and so surely exist maximum and mini- mum values.
It is not convenient to use Kuhn-Tucker's conditions. We check for free maximum or minimum points and then we study the problem in the boundary points.
In the first quadrant, since B ! and C !, we get 0 Bß C œ B C ! # , and so all the points belonging to the axes B œ ! and C œ !, where 0 Bß ! œ 0 !ß C œ ! , are minimum points.
For free maximum or minimum points we get:
0
0Bw # Ê a B
Cw
œ C œ !
œ #BC œ ! C œ ! , points just studied.
It remains only to study the points satisfying $C B $ œ ! Ê B œ $ $C Þ Substituting we get: 0 $ $Cß C œ $ $C C œ $C $C # # $ and deriving we get:
0 C œ 'C *C œ $C # $C ! ! Ÿ C Ÿ #
$
w # for .
So the point "à# is the maximum point, with 0 "à #œ % while all the points Bß ! and
$ $ *
!ß C are minimum points (red lines)
II M 4) Given the function 0 Bß Cß D œ B %B C D &B #C $ # # # nalyze the nature of itsa stationary points.
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 Ê
0 œ $B )B &
0 œ #C # œ ! 0 œ #D
B C œ "
D œ 0 Bß Cß D œ
œ ! œ !
œ B œ "
!
w #
Bw C Dw
&
$ or
and so we get two statio-
nary points: T œ" & and T œ"
$à "à ! "à "à ! Þ For the second order conditions we construct
the Hessian matrix: ‡ .
Bß Cß D œ
'B ) ! !
! # !
! ! #
‡
‡
‡
‡
&
$à "à ! œ T
# ! !
! # ! œ % !
! ! # œ ) !
. Since "# the point is a minimum point.
$
"
œ # !
‡ ‡
‡
"à "à ! œ T
# ! !
! # !
! ! #
. Since " the point is a saddle point.
" œ # ! #
œ # !
