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Unformatted text preview: THE UNIVERSITY OF WESTERN ONTARIO
Department of Applied Mathematics London Ontario
Applied Mathematics 025b First Tutorial Test — January 31, 2008 60 min Name: __..— Section 2 Closed book. Only simple calculators are allowed. Print your name above
and student number on the page 2. Justify answers by showing sufﬁcient
work to get the full marks. Solve each problem in space provided for that
speciﬁc problem. Use margins or back of each sheet to do your rough work. Instructor — Dr. N. Kiriushcheva PART A. Mark the correct answer.
[4] 1.1. The system of equations
ar+by+cz =0
cm: + (:3; : lnc
C3: 4— (:14.— HI}; = yz
(a) is linear in {33. y. z} and linear in {(5, b}; {b) is linear in {or}. 3} and linear in {l}. (3}: ﬂ is nonlinear in {12. y. z} and linear in {(1. b}; l/
(d) is nonlinear in {r.y_. z} and linear in {he};
(e) none of above. 1.2. The solution of the following system of equations
.‘L' + y = 0 23—35124
53+6y2—2 a) isr=4andy=—4;
b] is.t=6andy=*6;
c) is .1: = 0 and y = 0; d] is a: = 2 and y = *2;
is none of the above. z—xx—xr—xra AR}: 0251’) Test. 1  January 31. 2008 I»? Student 7%: FOR INSTRUCTOR’S USE ONLY lﬂﬂﬁ
IlIE'olFIIIIEl [8} 2.1 The following equation[ a is a constant) ozr—5225,r+e (a) exactly One solution a. E R;
has inﬁnity many solutions regardless what a. is;
c) has no solution if n. : —5: [g has a unique solution if a aé :5. 12.2. If the augmented matrix of the system of linear equations is l 0 3 0 6 *3
0 l 2 .—3 0 8
0 U 1 U 0 3
U 0 U 0 l 0 hen there
[a] is a unique solution to this system;
@are inﬁnity many solutions with one free parameter;
(c) are inﬁnity many solutions with two free parameters;
(cl) are inﬁnity many solutions with inﬁnity many parameters;
(e) is no solution. = on: —— be? 1 (:1? —— d is a curve passing through the point i: : “1,2; : 3, then
(a) 4 : —3;r:3 + 32:2 — 3x + d
/(b —4=33n——325+3e ——d
3 : —64a —— 16b — 4e —— d (d) : —33n + 326; ~ 3c—— of
(e) none of the above. 2.4. The following equation shows one step in row reduction. 1 ml 2 4
l 0 1 6
 2 —3 5 4
reduced to
1 —l '2 4
0 if} y z ,5 0 ii ‘U U:
/Give the values of missing elements:
/[a)r=1,y:0,7—1u——lellw —4:
(b):t=0,y=—1,z:—2‘u=l=e=—l,w=2;
(c)i:=11. i}=—l=2=l,u.:l.L‘I—1,LL‘=2;
J @;r 1,2 2,12. ’ —1,e 1.111.! 4; [e none 1of the aboye. ARI 0251') Test. 1 — January 31. 2008 3 Figure l: [8] 3.1. Consider the circuit in Fig. l. The positive directions of the currents are indicated by the
arrows. W'hich equation represents the Kirchhoff ’5 current law for the circuit above?
[aJi1+ig+i3+11=i4 '31 +152 + £3 = 1:4. (C) '31] + 222 + 3.33 2 461 (cl) i1 — Big 2 U 3.2. Consider the circuit in Fig. 1. Which equation represents the Kirchhoff‘s voltage law for
the circuit? (a) 212 — 333 —— 1133 = U
(b) 2n + 3223 —— 11 = 0 C Bliglg=4t4
22i2—3gi3+11=0 (e) 31 +1'l2 + '33 = 31. \/ 3.3. The system with the augmented matrix (a) has exactly one solution for any n; (b) has exactly one solution if a 72$ 4; (c) has no solution if a = 4; @has infinity man",r solutions if a = 4; 2 (6) none of the above. /
3.4. Consider the homogeneous system ox+by=0
cas+dy=0 where o, b, c. and d are arbitrary numbers.
For what relation among a. b. c, and d this system has the only one solution?
(a) for any values of a, E}. c. and d: [b for ad # be;
for a?) 2 col: d) for o = andh = d;
(0) there ‘ no values of a, b, c, and at such that this system has a solution. WW1 Lb) 9% ARE 0213b rfest 1 January 31. 2008 J; PART B. Show all steps of your calculations. {13‘ rufr 50} 4.1. Find the equation of the parabola passing through the points (1.4). (—1, —2)'1 and (2,4). %NMow ax1+b7+ﬁ=%' Mﬁiaotb+oza iii a \I\ a
l_—',‘2‘}§ Q—b+Q:—Q § \ "k i, *3 737%”. ’2. O (l ’2.
{ZJWS'QG +ab+c=4 a 3‘ i L‘ “We 7” 0 ”‘ —"\ 311 a Li mbH—ra {an
i 20 2 9. .3; Qatar2:119 LE1) “9193 [0 0 ’3 ‘13 are ([233) E35/CIQ
E1: Q?‘C‘=im§0:‘—‘l:‘I
W57 b hoc.~:Ut+1—D=3 3' 'Lﬁnpmwboﬁﬂ Lb giw'n Ll‘dm'qh H: —cxa+3xt::i_ a 4.2. Suppose that the augmented matrix for a system of linear equations has been reduced by
row operations to the given row echelon form:
2
5
0 l 0 1
D D 1
U 0 U (a) Find a system of linear equations corresponding to the augmented matrix. Assume
that the variables are named 2:15 r2: r3. (b) Solve the s f tem by back substitution. 3.) "XM‘TBZ by.)
733:5 / LE2)
Oui+m7+mo=0 (E3) 2 EQ$13'5
L‘ 7:? ft‘D—‘Ig 2; 3,53
1““.5 E3 Q ‘13:; [JOW_ PQU‘CGJTLJW “‘7 ’1an VCLIM) D I 3 (76‘7“. '
'lx331z/‘Wnﬂ £015 MiG/lulu ow: Fosmbﬂa} "F 3, ’ AM 0251: Test 1 Jarmary '31. 2008. ".251 [10; 5. Solve the system of Linear equations below 131—532ng =0
2I51+JJ2+4173 =3.
3131—333 =0 (a) using. Gaussian Elimination and back substitution:
{i} (b) using GaussJ ordan Elimination. 5—1) \4 i Q i —\ i O i —i l {E1}
“"— 1 \ l 3 59129. 0 3 1 3 i7 0 3 't (C?)
'5 0 —\_0 23—32! 0 133—4 '0 125% 0 0 3 (E3? gimp A
$397331
127:) 313:”;«13
 3—H AlVI 02513 Test 1 January 31. 2008 6 [1 0] 6. Determine if the following system is eonsistent using Gaussian Elimination .JEll 103:3 =5
3331+l‘24Lt33 = —1
4$l+32+6113=1 a 3
(on) Find the row Ecalfeﬁm form of its augmented matrix;
[T] (b) Interpret the solution (no solution. unique solution or inﬁnitely many solu tions).
0 ‘0 5 i 010 5 i o :0 5l ’1.+f013=5 rs
\ L\ —\ $114“ :3: l 42% FM: § 0 132.! as 37 72..3Uf3:*!é (.
l. 6 l R‘bﬂz’lﬂl O l “(ill _lq ﬁgmﬁﬁ Cl C) O ‘3 Ox‘AO‘Ijriﬂﬂgﬂal REF E3 e m M J
{van3&1 O {Jaw Til vMQH Q2 mod} 3 . n 4'3 Baden/m i: 'm [email protected]’n$,1...§o7ULrbt lo ANI 0.25b Test 1 Janlmgy 3L 2008 I L_ Bk :7] IR: A ;; "i I. 5:11 ' {1? Figtme ‘2': [10] 7'. Determine the currents I]; I2, and ['3 in the Fig. 2 using K'irchhoff’s rules. A. J54 + 11 : L3 LX152! ) . , ’ HQ .
. ‘  i '_ . . 5r _« _ 1 .
(7; L J :3 1" i .+ J; i Mum. GU.» ibOMﬁ. <10 $511.1 ’F "" “any f r ‘ “I all“ > ...
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