LinAlg Test 1 [Sep 07]

LinAlg Test 1 [Sep 07] - THE UNIVERSITY OF WESTERN ONTARIO...

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Unformatted text preview: THE UNIVERSITY OF WESTERN ONTARIO DEPARTMENT OF APPLIED MATHEMATICS Applied Mathematics 025a First Tutorial Test, September 27, 2007 Time: 60 minutes The test is closed book! You are allowed to use only simple calculators. Section 002, Professor Kiriushcheva SOLUTIONS PART A (20 marks). Mark the correct answer. 1. The system of equations or: + by + as = 1n 0 as: + acy = 0 cm+(a+b)mz=y (a) is linear in {55,352} and linear in {(2, b}; (b) is nonlinear in {:1:,y, z} and linear in {(2, b}; (c) is linear in {3,932} and linear in {13,0}; ((1) is nonlinear in {x,y,z} and linear in {b,c}; (e) none of above. Answer: b. 2. The solution of the following system of equations w+y=4 m—y=8 2m+3y26 (a)is:z:=4andy=0; (b)is:1:==6andy=-——2; (c)is:t==0andy=4; (d)ism=2andy=——2; (e) is none of the ab0ve. Answer: b. 3. The following equation (a is a constant) (a2 — 4) a: = 3 (a) has always exactly one solution (for all a); (b) has infinity many solutions regardless what a is: (c) has unique solution if a = 2; (d) has no solution if a. = :l:2 ; (6) can be solved only after a gets a numerical value. Answer: d. 4. If the augmented matrix of the system of linear equations is 10401l—3 01—20—1|1 00010|2 00000|1 then there (a) is a unique solution to this system; (b) are infinity many solutions with one free parameter; (c) are infinity many solutions with two free parameters; (d) are infinity many solutions with infinity many parameters; (e) is no solution. Answer: e. 5. Consider the following traffic flow. Which equation represents the junction rule for this system? (a) 40—x1+:r5-3:4 =0; (b) 60—332—335—22320; (c) 60—221+a:2 =0; ((1) 40+w3 +224 = 0; (e) none of the above. Answer: e. 6. Consider the circuit. The positive directions of the currents are indicated by the arrows. L" ; 1J1 . "9‘er L5 Ii; 2J7. 53 3-12. Which equatiOn represents the Kirchhofi’s loop rule for the circuit? (a) 2'1 + 22'2 + 3i3 = 4724; i1 + = 0; (c) 2&2 —~ 37;3 = 0: (d) 2'1 — 3% = 4n; (6)51 +i2 + is = '54. Answer: c. 7. Which equation represents the Kirchhoff ’s junction rule for the circuit above? (a) 2'1 + 2% + 3i3 = 42'4; 2-1 "" 3’53 == 0; (C) 2i2 + 47:4 + ’65 == 0: 7:1 — 37:3 = 4’64 + ’55; (3) i1 + ’52 “Fig = ’54. Answer: e. 8. If y(:c) = ax3+bm2+cx+d is a curve passing through the point a: =- 5, y = —3, then (a) 3 == —5:v3 + 53:2 —— 5.7: + d; (b) —3 = 5% + 526 + 5c + d; (c) -—5 = 27a+ 9b+3c+d; (d) 3 = -—53a + 52b — 5c+ d; (e) none of the above. Answer: b. 9. The following equation shows one step in row reduction. Give the values of missing elements. 1 —1 3 8 1 —1 3 8 2 -—1 4 11 reduces to O a: y z —-1 2 —4 ~11 0 'u u w (a) m: Ly: —2,z= —5,v= 1,u=—-1,w=—3; (b) 2:: 1,y=0,z=1,v= —1,u=1,w=——11; (c) 31: Ly: -—1,z=2,v= —1,u= 1,20: ——4; (d) :1:=0,y=-1,z=—5,u:1,u=——1,w=2; (e) none of the above. Answer: a. 10. The system with the augmented matrix [2 3 f 4] a 6 I 8 (a) has exactly one solution for any a; (b) has no solution for some values of a; (c) has infinity many solutions if {1 7E 4;; (d) has exactly one solution if a 75 4; (e) none of the above. Answer: (1. PART B. Show all steps of your calculations. 1. (10 marks). Solve the system of equations using Gaussian elimination. m1+3m2+a¢3—~$4 =0 31151 +132 + 311:3 = —2 21131 + 6932 + 2933 "- 2134 = 2 Solution Augmented matrix: R1 1 3 1 *1 0 R1 1 3 1 —1 0 R; 3 1 3 0 —2 R2—3R1 U v—8 0 3 ——2 R3 2 6 2 —-2 2 R3—2R1 0 0 0 0 2 Row Echelon Form: R1 1 3 1 —1 0 R1 1 3 1 ——1 0 s; 0 1 0 :83. 21; 122 0 1 o -?3 i R3 0 0 0 0 2 R3 0 0 0 0 2 Row 3 in the matrix shows that the system of equations has NO SOLUTION. 2. (10 marks). Find the equation of the parabola passing through the points (1! 0)! (—116): {MN-1(2) Solution Lety=ax2+bx+c. oPoint(l,0)—>a+b+c=0 oPoint(-1,6)—)a-b+c=6 oPoint(2,0)—)4a+2b+c=0 Augmented Matrix: R1 1 1 1 0 R1 1 1 1 0 122 [1—116] 112—121 0H2 0 6 R3 4 2 1 0 res—ml o —2 —3 o 121 1 1 1 0 R1 1 1 1 0 32 0—2 0 6] £3 010—3 33—182 0 0 —3 —6 :3; 001 2 Rnw3gives: c=2 Row2gives: b =-3 Rowlgives:a+b+c=O-+a—3+2=O—>a=1 Therefore the equation of the parabola is given by y = x2 — 3 x + 2 3. (10 marks). Find currents i1, i2, and 1'3 using Kirchhofl’s rules in the following circuit: Solution 0 Kirchhofi’s junction rule Junction A: 13 = 11 + 12 Junction B: II + 12 = I3 * Junctions A and B give the redundant equations, so we can choose either of them. a Kirchhofi’s loop rule LOOP 1: 13 +13 = 9 313:3 —}13=3 Loop 2: (-4) 12 + (-2) 13 + (-1) 13 = -13 412—313 :43 —>4Ig+313=13 Loop 3: (-4) 12 = 9 - 13 412 :4 +I2=1 * Loops 1+ 3 give the equation for the L00p 2. o Currents of ll, 12, and I3 II+I2“13=0—)I1=-12+13—)Il=-1+3 11:2 ...
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LinAlg Test 1 [Sep 07] - THE UNIVERSITY OF WESTERN ONTARIO...

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