University:
Stanford University
Course:
MATH 131P | Partial Differential Equations
Academic year:

2022
Views:

24

Pages:

16
Author:

MathPaladin

Questions and Answers Sheet 1 Systems of Equations Question #1 Find the complete solution of the system, or show that the system has no solution. x + y + 2z = 6  2 x + 5 z = 12  x + 2 y + 3z = 9  Answer: Question #2 Find the general solution of the following systems of difference equations: a) 𝑥𝑛−1 = 3𝑥𝑛 𝑦𝑛−1 = 𝑥𝑛 + 2𝑦𝑛 Answer: Question #3 Find the general solution of the following systems of difference equations: d) 𝑥𝑛+1 = 3𝑥𝑛 − 𝑦𝑛 𝑦𝑛+1 = −𝑥𝑛 + 2𝑦𝑛 − 𝑧𝑛 𝑧𝑛+1 = −𝑦𝑛 + 3𝑧𝑛 Answer: Question #4 Which of the following systems of linear equations is inconsistent? A) 𝑥1 − 2𝑥2 − 𝑥3 = 1 2𝑥1 − 3𝑥2 + 𝑥3 = 6 3𝑥1 − 5𝑥2 = 7 𝑥1 + 5𝑥3 = 9 B) 𝑥1 + 𝑥3 = 0 𝑥2 + 𝑥4 = 0 𝑥1 + 2𝑥2 + 3𝑥3 + 4𝑥4 = 0 C) 2𝑥1 − 𝑥2 + 𝑥3 = −4 2𝑥1 + 𝑥2 + 3𝑥3 = 0 𝑥1 + 𝑥2 + 2𝑥3 = 3 D) All the systems are consistent E) 𝑥1 + 2𝑥2 − 𝑥3 = −1 2𝑥1 + 2𝑥2 + 𝑥3 = 1 3𝑥1 + 5𝑥2 − 2𝑥3 = −1 Answer: Question #5 Solve the following systems of equations by Cramers rule a) a11 x1 + a12 x2 = b1 where 𝑎11 𝑎22 − 𝑎12 𝑎21 ≠ 0  a21 x1 + a22 x2 = b2 2 x1 + x2 − 3x3 = 5  b)  x1 − 2 x2 + x3 = 0 3x + 4 x − 2 x = 0 2 3  1 2 x1 + x2 − 3x3 = 1  c)  x1 − 2 x2 + x3 = 0 3x + 4 x − 2 x = −5 2 3  1 Answer: Question #6 Solve the systems of linear and quadratic equations y = x 2 − 4 y = −3 (-1,4) and (1,4) (-1,-3) and (1,-3) (-2,3) and (2,3) Answer: Question #7 Consider the three systems of linear equations: A) 4𝑥1 + 6𝑥2 + 2𝑥3 + 8𝑥4 = 2 6𝑥1 + 9𝑥2 + 16𝑥4 = 4 2𝑥1 + 3𝑥2 − 2𝑥3 + 7𝑥4 = 2 B) 4𝑥1 + 6𝑥2 + 2𝑥3 + 8𝑥4 = 2 2𝑥1 + 3𝑥2 + 4𝑥3 + 𝑥4 =0 2𝑥1 + 3𝑥2 − 2𝑥3 + 7𝑥4 = 2 C) 2𝑥1 + 3𝑥2 + 𝑥3 + 4𝑥4 = 1 3𝑥3 − 3𝑥4 = −1 −3𝑥3 + 3𝑥4 = 1 Select one: a)none of the others b) (A) and (B) are equivalent c) (A) and (C) are equivalent d) (A), (B) and (C) are equivalent e) (B) and (C) are equivalent Answer: Question #8 Consider the three systems of linear equations: A) 4𝑥1 + 6𝑥2 + 2𝑥3 + 8𝑥4 = 2 6𝑥1 + 9𝑥2 + 16𝑥4 = 4 2𝑥1 + 3𝑥2 − 2𝑥3 + 7𝑥4 = 2 B) 4𝑥1 + 6𝑥2 + 2𝑥3 + 8𝑥4 = 2 2𝑥1 + 3𝑥2 + 4𝑥3 + 𝑥4 = 0 2𝑥1 + 3𝑥2 − 2𝑥3 + 7𝑥4 = 2 C) 2𝑥1 + 3𝑥2 + 𝑥3 + 4𝑥4 = 1 3𝑥3 − 3𝑥4 = −1 −3𝑥3 + 3𝑥4 = 1 Select one: a)none of the others b) (A) ,(B) and (C) are equivalent c) (A) and (C) are equivalent d) (B) and (C) are equivalent e) (A) and (B) are equivalent Answer: Question #9 Solve (a) π𝑥 = 2 (b) 3𝑥 + 2 = 5𝑥 + 7 − 2𝑥 (c) 6𝑥 + 2 − 4𝑥 = 5 + 2𝑥 − 3 Answer: Question #10 Using a proof by contraposition, prove that if x and y are real numbers where 𝑥+𝑦 ≥2 then, 𝑥 ≥ 1 or 𝑦 ≥ 1 Answer: Let us assume that 𝑥 < 1 and 𝑦 < 1 Let us add the two inequalities 𝑥 < 1 and 𝑦 < 1: 𝑥+𝑦 <1+1 Evaluate the sum: 𝑥+𝑦<2 We have then shown that ''If 𝑥 < 1 and 𝑦 < 1, then 𝑥 + 𝑦 < 2′ ' is true, which implies that its contrapositive ''If 𝑥 + 𝑦 ≥ 2 hen 𝑥 ≥ 1 or 𝑦 ≥ 1''. is also true Question #11 Solve the system of equations: 2 x + 6 y = −12  2 x − 5 = 10 Answer: 2 x + 6 y = −12  2 x − 5 = 10 2 6   x   −12 In matrix form    =    Ax = b 2 −5  y   10  The argument of matrix is: 2 6 | −12 2 6 | −12  [ A | b] =  R2 → R2 − R1    2 −5 |10  0 −11| 22 2 6 | −12 2 0 | 0  [ A | b] − 1/11R2  R → R − 6 R 1 1 2  0 1| −2 0 1| −2    1 0 | 0  1/ 2R1   x = 0 and y = −2 0 1| −2 Question #12 Solve the system of equations: 3x − 8 + 72 = 0  7 x = 4 y − 8 Answer: The system of equation is : 3𝑥 − 8𝑦 + 72 = 0 ⇒ 3𝑥 − 8𝑦 = −72 and 7𝑥 = 4𝑦 − 8 ⇒ 7𝑥 − 4𝑦 = −8 In matrix form , 3 −8  x   −72 7 −4  y  =  −8       𝑅2 − 𝑅2 − 2𝑅1 3 −8  x   −72 1 12   y  = 136       𝑅1 → 𝑅1 − 3𝑅2 0 −44  x   −480 1 12   y  =  136       ⇒ −44𝑦 = −480 ⇒ 𝑦 = 480 −44 ⇒𝑦= 120 11 and 𝑥 + 12𝑦 = 136 ⇒ 𝑥 = 136 − 12𝑦 ⇒ 𝑥 = 136 − 12 ( 120 11 ) ⇒ 𝑥 = 136 − 1440 11 ⇒𝑥= 1496−1440 So, 𝑥 = 56 11 11 ⇒𝑥= and 𝑦 = 120 11 56 11

Questions and Answers #1 Systems of Equations

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