sample problems econ 106
DESCRIPTION
Sample Problems + Solutions to Econ 106TRANSCRIPT
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Econ 106 Macroeconomics12nd Semester, AY 2014-2015
Problem Set 1 Solution Key
1. Given uT =5 2 3
; v
T
=3 1 9
; w
T
=7 5 8
; and x
T
=x1 x2 x3
;
write out the column vectors, u; v; w; and x; and nd
(a) uvT =
24 15 5 456 2 189 3 27
35(b) uwT =
24 35 25 4014 10 1621 15 24
35(c) xxT =
24 x21 x1x2 x1x3x2x1 x22 x2x3x3x1 x3x2 x
23
35(d) vTu = 44
(e) uT v =vTu
T= vTu = 44
(f) wTx = 7x1 + 5x2 + 8x3(g) uTu = 38
(h) xTx = x21 + x22 + x
23
2. Given w =
24 3216
35 ; x = x1x2
; y =
y1y2
; and z =
z1z2
:
(a) The following products are dened: xyT ; yT y; zzT ; ywT ; inner product of x and y (= xT y)
i. xyT =x1y1 x1y2x2y1 x2y2
ii. yT y = y21 + y
22 + y
23
iii. zzT =z21 z1z2z2z1 z
22
iv. ywT =
3y1 2y1 16y13y2 2y2 16y2
v. xT y = x1y1 + x2y2
3. Find AB and BA:
AB =
24 3 1 21 0 34 0 2
35| {z }
A
24 0 15 3101 15 7100 25 110
35| {z }
B
=
24 1 0 00 1 00 0 1
35 :
BA =
24 0 15 3101 15 7100 25 110
3524 3 1 21 0 34 0 2
35 =24 1 0 00 1 00 0 1
35 :1S. Daway
1
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4. Given
A =
a11 a12a21 a22
, B =
b11 b12b21 b22
; and C =
c11 c12 c13c21 c22 c23
;
(a) Find AT ; BT ; and CT :
AT =
a11 a21a12 a22
, BT =
b11 b21b12 b22
; and CT =
24 c11 c21c12 c22c13 c23
35 ;(b) Given
A = [aij ] ; B = [bij ] ; and C = [cji] ; i = 1; 2; :::;m and j = 1; 2; :::; n;
Show that
(a) i. (A+B)T = AT +BT
Let (A+B) = [aij + bij ] [zij ] :AT +BT = [aij ]
T+ [bij ]
T= [aji] + [bji] = [zji] = [zij ]
T= [aij + bij ]
T= (A+B)
T
ii. (AC)T = CTAT
2
-
(AC)T
om=
0BB@2664a11 a12 ::: a1na21 a22 ::: a2n: : : :am1 am2 ::: amn
37752664c11 c12 ::: c1oc21 c22 ::: c2o: : : :cn1 cn2 ::: cno
37751CCAT
=
2664a11c11 + a12c21 + :::a1ncn1 ::: a11c1m + a12c2m + :::a1ncnoa21c11 + a22c21 + :::a2ncn1 ::: a21c1m + a22c2m + :::a2ncno
: : :am1c11 + am2c21 + :::amncn1 ::: am1c1o + am2c2o + :::amncno
3775T
=
266666666664
nXk=1
a1kck1
nXk=1
a1kck2 :::nXk=1
a1kcko
nXk=1
a2kck1
nXk=1
a2kck2 :::
nXk=1
a2kcko
: : : :nXk=1
amkck1
nXk=1
amkck2 :::nXk=1
amkcko
377777777775
T
=
266666666664
nXk=1
a1kck1
nXk=1
a2kck1 :::
nXk=1
amkck1
nXk=1
a1kck2
nXk=1
a2kck2 :::nXk=1
amkck2
: : : :nXk=1
a1kcko
nXk=1
a2kcko :::
nXk=1
amkcko
377777777775
=
266666666664
nXk=1
ck1a1k
nXk=1
ck1a2k :::nXk=1
ck1amk
nXk=1
ck2a1k
nXk=1
ck2a2k :::nXk=1
ck2amk
: : : :nXk=1
ckoa1k
nXk=1
ckoa2k :::nXk=1
ckoamk
377777777775
=
2664c11a11 + c21a12 + :::cn1a1n ::: c11am1 + c21am2 + :::cn1amnc12a11 + c22a12 + :::cn2a1n ::: c12am1 + c22am2 + :::cn2amn
: : :c1oa11 + c2oa12 + :::cnoa1n ::: c1oam1 + c2oam2 + :::cnoamn
3775
=
2664c11 c21 ::: cn1c12 c22 ::: cn2: : : :c1o c2o ::: cno
37752664a11 a21 ::: am1a12 a22 ::: am2: : : :a1n a2n ::: amn
3775= CT
(on)AT
(nm):
5. Given
a b cd e fg h i
; nd the minors and cofactors of the elements a; b and f:3
-
ma = ei fh; mb = di fg; mf = ah bg
ca = (1)2 (ei fh) ; cb = (1)3 (di fg) ; cf = (1)5 (ah bg)
6. Given
A =
26641 2 0 92 3 4 61 6 0 10 5 0 8
3775 ; B =26642 7 0 15 6 4 80 0 9 01 3 1 4
3775Find the determinants of the following:
(a) det (A) = 5 (1)61 0 92 4 61 0 1
+ 8 (1)81 2 02 3 41 6 0
= 200 128 = 72(b) det (B) = 9 (1)6
2 7 15 6 81 3 4
= 9 (96 84 21) = 81
(c) det(A+B)T = det (A+B) =
3 9 0 107 9 8 141 6 9 11 8 1 12
3 (1)2
9 8 146 9 18 1 12
+9 (1)37 8 141 9 11 1 12
+10 (1)57 9 81 6 91 8 1
= 4683 49235060 = 5300
(d) det (AB) = det (A) det (B) = 5832
(e) det (AB)T = det (AB) = 5832
4