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LinAlg Test1 [Jan 08]

LinAlg Test1 [Jan 08] - THE UNIVERSITY OF WESTERN ONTARIO...

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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 l-flflfi IlIE'olFIIIIE-l [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——le-llw —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? [aJ-i1+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) 21-2 — 333 —— 1133 = U (b) 2n + 3-223 —— 11 = 0 C Bliglg=4t4 22-i2—3g-i3+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 = and-h = 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. {Z-JWS'QG +ab+c=4 a 3‘ i L‘ “We 7”- 0 ”‘ -—"\ 311 a Li mbH—ra {an i 20 2 9. .3; Qatar-2:119 LE1) “91-93 [0 0 ’3 ‘13 are ([233) E35/CIQ E1: Q?‘C‘=im§0:‘|—‘l:-‘I W57 b- h-o-c.~:Ut+1—D=3 3' 'Lfinpmwbofifl Lb giw'n Ll‘dm'qh H: —cxa+3x-t::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—532-ng =0 2I51+JJ2+4173 =3. 3131—33-3 =0 (a) using. Gaussian Elimination and back substitution: {i} (b) using Gauss-J ordan Elimination. 5—1) \4 i Q i —\ i O i —i l {E1} “"— 1 \ l 3 591-29. 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 $397331 127:) 313:”;«13 - 3—H AlVI 02513 Test 1 January 31. 2008 6 [1 0] 6. Determine if the following system is eonsistent using Gaussian Elimination .JEl-l- 103:3 =5 3331+l‘2-4Lt33 = —1 4$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\ —\ $114“ :3: l 42% FM: § 0 1-32.! as 37 72..-3Uf3:*!é (. l. 6 l R‘bflz’lfll O l “(ill _lq figmfifi Cl C) O ‘3 Ox‘A-O‘Ijriflflgflal REF E3 e m M J {van-3&1 O {Jaw Til v-MQH Q2 mod} 3 . n 4'3 Baden/m i: 'm [email protected]’n$,1...§o7UL--rbt 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.» ibOM-fi. <10 $511.1 ’F "" “any f r ‘ “I all“ > ...
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