Intermediate Test QUANTITATIVE METHODS for ECONOMIC APPLICATIONS 21/11/2019
I M 1) Given the polynomial equation B # # B 5 œ !, determine the value of for which5 this equation admits the solution B œ /1%3. Then calculate the square roots of the other solution.
From B œ /1%3 we get B œ cos1 sin1 ; from B # B 5 œ ! we get:
% 3 % œ ## 3 ## #
B œ # „ # %5 œ # %5 # %5 œ # Ê 5 œ "
# #
# # #
# „ œ # „ 3 # if .
So the second solution of the equation is B# œ # 3 # œcos( 3sin( .
# # % %
1 1
So B# œcos( 5 # 3sin( 5 # ß !Ÿ5 Ÿ" from which:
) # ) #
1 1 1 1
- œ" cos( sin( and - œ# cos"& sin"& .
)1 3 )1 )1 3 )1
I M 2) Given the matrix œ determine the value of the parameter such5
" 5 ! 5 " "
! " "
that the matrix admits the eigenvalue - œ ! and, for this value of , find one modal matrix5 which diagonalizes .
If the matrix admits the eigenvalue - œ ! surely the matrix is a singolar one, i.e. its determinant
is equal to zero. So iff .
œ œ " " " 5 5 ! œ 5 œ ! 5 œ !
" 5 ! 5 " "
! " "
#
So œ - ˆ œ !
" ! !
! " "
! " "
and so, from we get:
" ! !
! " "
! " "
œ " " " " œ " œ !
#
-
-
-
- - - - -# -
to get the eigenvalues -" œ !, -# œ " and -$ œ #.
Since the matrix is a symmetric one, the matrix is diagonalizable. Furthermore it has all distinct eigenvalues. To get the corresponding eigenvectors, we must solve three systems.
For -" œ ! we solve:
! †ˆ œ † œ Ê Ê
" ! ! B ! B œ !
! " " C ! C D œ !
! " " D ! C D œ !
B œ ! C œ D and so eigenvectors corresponding to -" œ ! are •" œ !ß Cß C Ê !ß "ß " . For -# œ " we solve:
" †ˆ œ † œ Ê
! ! ! B ! a B
! ! " C ! D œ !
! " ! D ! C œ !
and so eigenvectors corresponding to -# œ " are •# œ Bß !ß ! Ê "ß !ß ! .
For -$ œ # we solve:
# †ˆ œ † œ Ê Ê
" ! ! B ! B œ !
! " " C ! C D œ !
! " " D ! C D œ !
B œ ! D œ C and so eigenvectors corresponding to -$ œ # are •$ œ !ß Cß C Ê !ß "ß " .
Since the matrix is a symmetric one, the matrix is diagonalizable by an orthogonal matrix, and so we must transform the three vectors in unit vectors, to get the modal orthogonal matrix which is:
” œ
! " !
!
!
" "
# #
" "
# #
.
I M 3) If in the basis •œ"ß "ß ! à "ß "ß ! à #ß !ß " the vector has coordinates—
"ß #ß ", determine its coordinates in the basis –œ"ß "ß ! à "ß "ß ! à !ß #ß " .
If the vector has coordinates — "ß #ß " in the basis • œ"ß "ß ! à "ß "ß ! à #ß !ß " we get: —œ † œ . To determine the coordinates of in the other—
" " # " "
" " ! # "
! ! " " "
basis – œ"ß "ß ! à "ß "ß ! à !ß #ß " , we must solve the system:
" " ! B " B " " ! "
" " # C " C " " # "
! ! " D " D ! ! " "
† œ or calculate œ †
"
. Using the first method we get:
" " ! B " B C œ " B C œ " B œ "
" " # C " B C #D œ " B C œ " C œ !
! ! " D " D œ " D œ " D œ "
† œ Ê Ê Ê .
Using the second method we get: Adj( ) œ .
" " !
" " !
# # #
Then adj( ) , and since det , we finally obtain:
T œ œ #
" " #
" " #
! ! #
"
" " " "
# # # #
" " " "
# # # #
œ œ † œ
" "
" "
! ! " ! ! "
B " "
C " !
D " "
and so we get .
I M 4) Given a linear map 0 À‘$ Ä‘$, ˜œ —† , determine the matrix knowing that such linear map satisfies these three conditions:
a) 0 "ß #ß " œ "ß $ß $ à b) "ß "ß ! − Ker 0 à
c) "ß "ß " is an eigenvector of corresponding to the eigenvalue -œ ". From the first condition we get:
+ + + " " + #+ + œ "
+ + + # $ + #+ + œ $
+ + + " $ + #+ + œ $
† œ Ê
"" "# "$ "" "# "$
#" ## #$ #" ## $
$" $# $$ $" $# $$
2 ;
from the second condition we get:
+ + + " ! + + œ !
+ + + " ! + + œ !
+ + + ! ! + + œ !
† œ Ê
"" "# "$ "" "#
#" ## #$ #" ##
$" $# $$ $" $#
; from the third condition we get:
+ + + " " + + + œ "
+ + + " " + + + œ "
+ + + + + + œ "
† œ Ê
"" "# "$ "" "# "$
#" ## #$ #" ## #$
$" $# $$ 1 1 $" $# $$
. We have so obtained three systems:
+ #+ + œ " + œ !
+ + œ ! + œ !
+ + + œ " + œ "
Ê
"" "# "$ ""
"" "# "#
"" "# "$ "$
+ #+ + œ $ + œ #
+ + œ ! + œ #
+ + + œ " + œ "
Ê
#" ## #$ #"
#" ## ##
#" ## #$ #$
+ #+ + œ $ + œ #
+ + œ ! + œ #
+ + + œ " + œ "
Ê
$" $# $$ $"
$" $# $#
$" $# $$ $$
And finally œ
! ! "
# # "
# # "
.
I M 5) Given the linear system , check, depending on the para-
B #B #B B œ "
$B B #B œ "
B $B 7B 5 B œ !
" # $ %
" # %
" # $ %
meters and , the existence and the number of its solutions.7 5
To solve the problem we must apply Rouchè-Capelli theorem to the system, and so we study the Rank of the matrix and the Rank of the Augmented matrix ˜l .
By elementary operations on the rows V Ã V $V# # " and V Ã V V$ $ " we get:
" # # " l " " # # " l "
$ " ! # l " ! & ' " l #
" $ 7 5 l ! ! & 7 # 5 " l "
Ä ;
by another elementary operation on the rows V Ã V V$ $ # we get:
" # # " l " " # # " l "
! & ' " l # ! & ' " l #
! & 7 # 5 " l " ! ! 7 % 5 l "
Ä .
Using this last matrix we see that:
If 7 œ % and 5 œ ! it is Rank œ # Rank ˜l œ $ and so the system has no solu- tions;
if 7 Á % or 5 Á ! it is Rank œ $ œRank ˜l and so the system has ∞" solutions.
