TASK MATHEMATICS for ECONOMIC APPLICATIONS 12/10/2019
I M 1) Calcolate " 3 % " 3'. Since
" 3 œ # † cos1 sin1, we get " 3 œ4 cos sin à
1 1
% 3 % % 3 œ %
" 3 œ # † cos ( sin( , we get " 3 œ ) cos #" sin#"
%1 3 %1 ' #1 3 #1 œ
œ ) cos 1 sin1 Þ So " 3 " 3 œ $#3 Þ
# 3 # œ )3 % '
Since $#3 œ $#cos$ sin$ we get:
#1 3 #1
$#3 œ$# †cos$ 5 # sin$ 5 # Ÿ Ÿ
%1 #1 3 %1 #1 ß ! 5 "
from which we get:
- œ % # † % # " 3 " œ % %3
# #
" cos$ sin $ and
%1 3 %1 œ
- œ % # † % # " 3 " œ % %3 Þ
# #
# cos( sin (
%1 3 %1 œ
I M 2) Find the eigenvalues of the matrix œ .
! ! " "
! ! " "
" " ! !
" " ! !
From we get
- ˆ œ !
! " " ! " "
! " " ! " "
" " ! " " !
" " ! ! !
œ œ
- -
- -
- -
- - -
œ " œ
" " ! " "
" ! " "
! !
- -
- -
- - - -
œ " œ # # œ
# "
" !
! !
- - - -
-
-
-
# #
œ -- - # # # -# œ-% %-# œ- -# # % œ ! and so eigenvalues are:
-" œ-# œ !ß-$ œ #ß-% œ # .
I M 3) Given the linear map 0 À Ä , œ † , with œ , deter-
" " # "
" " 7 !
" " " 5
‘% ‘$ ˜ —
mine a basis for the Kernel and a basis for the Image of the linear map generated by , knowing that the Kernel and the Image have the same dimensionsÞÞ
From Sylvester Theorem, it is Dim Imm œRank and Dim Ker œ 8 Rank Þ Now 8 œ % and so Dim Imm œDim Ker œ % Rank iff Rank œ # Þ
By elementary operations on the rows and V Ã V V# # " and V Ã V V$ $ # we get:
" " # " " " # "
" " 7 ! ! # 7 # "
" " " 5 ! ! " 7 5
Ä from which we see that Rank œ # iff
7 œ " 5 œ ! Þ œ
" " # "
" " " !
" " " !
and So .
To find a basis for the Kernel we must solve the system:
—† œ Ê † œ Ê Ê
" " # " ! B B #B B œ !
" " " ! ! B B B œ !
" " " ! ! B B B œ !
B B B B
"
#
$
%
" # $ %
" # $
" # $
Ê B B #B #B B œ ! Ê B œ B B
B œ B B B œ B $B
"$ # " " # # % %$ " " # # .
So the vectors belonging to the Kernel are — œ B ß B ß " # B B B $B" #ß " # and a basis may be — œ"ß !ß "ß " à !ß "ß "ß $ .
To find a basis for the Image we must apply Rouchè-Capelli theorem to the system:
—† œ˜ Ê † œ Ê
" " # " C B B #B B œ C
" " " ! C B B B œ C
" " " ! C B B B œ C
B B B B
"
#
$
%
" " # $ % "
# " # $ #
$ " # $ $
and then we
should study the Rank of the matrix and the Rank of the augmented matrix:
" " # " l C
" " " ! l C
" " " ! l C
"
#
$
.
By elementary operations on the rows and V Ã V V# # " and V Ã V V$ $ # we get:
" " # " l C " " # " l C
" " " ! l C ! # " " l C C
" " " ! l C ! ! ! ! l C C
Ä
" "
# # "
$ $ #
from which we get:
C œ C$ # and so the vectors belonging to the Image are ˜ œC ß C ß C" # # and a basis may be
˜ œ"ß !ß ! à !ß "ß ".
I M 4) Given the matrix œ determine its inverse matrix.
" ! "
! " !
! ! "
Since the matrix has its unique inverse matrix, which
œ œ " † œ " Á !
" ! "
! " !
! ! "
" !
! "
is given by the transpose of the adjugate matrix adj( ), divided by the determinant of .
It is Adj( ) œ , then Adj( ) œ , and since det œ ", we
" ! ! " ! "
! " ! ! " !
" ! " ! ! "
T
finally obtain: " œ .
" ! "
! " !
! ! "
II M 1) Given the equation 0 Bß C œ B C BC B C œ " $ $ and the point P! œ "ß ! satisfy- ing it, determine first order derivative of a possible implicit function definable with it.
From f0Bß Cœ $B C C "à B $BC " # $ $ # we get f0 "ß ! œ "à ! and so it is possible to define only an implicit function C ÄB C : B ! œ ! œ !
"
w .
II M 2) Given the functions 0 Bß C œ B C # # and 1 Bß C œ B BC , the point T œ "ß "! and the unit vector @œcosαßsinα, find the values of for which α W@0 T ! œW@1 T ! .
0 Bß C œ B C # # and 1 Bß C œ B BC are differentiable function a Bß C − ‘#. So H 0@ "ß " œ f0 "ß " † @ œ H 1@ "ß " œ f1 "ß " † @. We get:
f0 B ß C œ #Bà #C Ê f0 "ß " œ #ß # , f1 B ß C œ " Cà B Ê f1 "ß " œ #ß " . So H 0@ "ß " œ f0 "ß " † cosαßsinαœ H 1@ "ß " œ f1 "ß " † cosαßsinαÊ
Ê #ß # †cosαßsinαœ #ß " † cosαßsinαÊ #cosα #sinαœ #cosαsinαÊ sinαœ ! Êαœ ! or αœ 1 .
II M 3) Determine maximum and minimum points for the function 0 Bß C œ B C # in the rec- tangle œBß C − ‘2: ! Ÿ B Ÿ # à " Ÿ C Ÿ !.
The objective function of the problem is a continuous function, the feasible region is a com- pact set, and so surely exist maximum and minimum values.
We don't use Kuhn-Tucker's conditions nor the Lagrangian function. Testing for free maximum and minimum points we get:
f Ê 0 œ #B œ !
0 œ " Á !
0 Bß C œ BCww and so we get no solutions.
Now we study our function on the points of the boundary of .
For B œ ! we get 0 !ß C œ C : a decreasing function;
for B œ # we get 0 #ß C œ % C : a decreasing function;
for C œ " we get 0 Bß " œ B " # : an increasing function for B !; for C œ ! we get 0 Bß ! œ B #: an increasing function for B !.
So, using proper arrows, we see that !ß ! is the minimum point, with 0 !ß ! œ ! while
#ß " is the maximum point, with 0 #ß " œ & .
II M 4) Given the function 0 Bß C œ B B C 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 #BC œ ! Ê
0 œ #B C #C œ !
B $B #C œ !
#C " B œ !
0 Bß C œ
w # #
Bw #
C
#
# from which we get:
B œ ! $B #C œ !
C œ !
B œ !
" B œ ! C œ !
$B #C œ !
" B œ !
or # or # or # # .
From the first system we get the point !ß ! ; the second system is impossible.
From the third system we get another time the point !ß ! , and from the last system we get
B œ " B œ "
C œ B# $# , an impossible system, and C œ B# $# , from which we get the two solutions
"ß$# and "ß $#.
For the second order conditions we construct the Hessian matrix:
‡Bß C œ 'B #C %BC
%BC # #B
#
# .
Since ‡ !ß ! œ ! ! we cannot use the leading principle minors. But 0 !ß ! œ ! and, in
! #
the orizontal axis we have 0 Bß ! œ B $, which is negative for B ! and positive for B !; so !ß ! is a saddle point.
Then ‡"ß and since ‡ , "ß is a sad-
$
# œ $ % œ #% !
% !
$
#
$
#
# $#
dle point.
Finally ‡"ß and since ‡ , "ß is ano-
$
# œ $ % œ #% !
% !
$
#
$
#
# $#
ther saddle point.
