University:
Florida Gateway College
Course:
MAC 2312 | Calculus II
Academic year:

2024
Views:

112

Pages:

3
Author:

Ruby Bates

Question 1 Use Cramer's rule to solve: 𝑥 1 + 2 𝑥 2 = 5 x 1 +2x 2 =5 − 𝑥 1 + 𝑥 2 = 1 −x 1 +x 2 =1 Answer The solution is computed using Cramer's rule: Matrix 𝐴 A: 𝐴 = [ 1 2 − 1 1 ] A=[ 1 −1 2 1 ] Determinant ∣ 𝐴 ∣ ∣A∣: ∣ 𝐴 ∣ = ( 1 ) ( 1 ) − ( − 1 ) ( 2 ) = 1 + 2 = 3 ∣A∣=(1)(1)−(−1)(2)=1+2=3 Matrix 𝐴 1 A 1 : Replace the first column of 𝐴 A with the constants vector 𝐵 = [ 5 , 1 ] B=[5,1]: 𝐴 1 = [ 5 2 1 1 ] A 1 =[ 5 1 2 1 ] Determinant ∣ 𝐴 1 ∣ ∣A 1 ∣: ∣ 𝐴 1 ∣ = ( 5 ) ( 1 ) − ( 1 ) ( 2 ) = 5 − 2 = 3 ∣A 1 ∣=(5)(1)−(1)(2)=5−2=3 Matrix 𝐴 2 A 2 : Replace the second column of 𝐴 A with 𝐵 B: 𝐴 2 = [ 1 5 − 1 1 ] A 2 =[ 1 −1 5 1 ] Determinant ∣ 𝐴 2 ∣ ∣A 2 ∣: ∣ 𝐴 2 ∣ = ( 1 ) ( 1 ) − ( − 1 ) ( 5 ) = 1 + 5 = 6 ∣A 2 ∣=(1)(1)−(−1)(5)=1+5=6 Solution: Using Cramer's rule: 𝑥 1 = ∣ 𝐴 1 ∣ ∣ 𝐴 ∣ = 3 3 = 1 , 𝑥 2 = ∣ 𝐴 2 ∣ ∣ 𝐴 ∣ = 6 3 = 2 x 1 = ∣A∣ ∣A 1 ∣ = 3 3 =1,x 2 = ∣A∣ ∣A 2 ∣ = 3 6 =2 Final Answer: 𝑥 1 = 1 , 𝑥 2 = 2 x 1 =1,x 2 =2 Question 2 What is a system of linear equations? Provide an example with your description. Answer A system of linear equations is a set of two or more linear equations with the same variables. The goal is to find the values of the variables that satisfy all the equations simultaneously. Example: 2 𝑥 − 3 𝑦 = 8 , 7 𝑦 = 23 2x−3y=8,7y=23 Here, we have two linear equations with two variables 𝑥 x and 𝑦 y. Question 3 A system of linear equations with more equations than unknowns is called an overdetermined system. Can it be consistent? Yes, if the extra equations are linear combinations of the other equations. Example: Equations: 𝑥 + 𝑦 = 1 , 𝑥 − 𝑦 = 0 , 2 𝑥 = 2 x+y=1,x−y=0,2x=2 The third equation 2 𝑥 = 2 2x=2 is a multiple of the first equation. The solution is: 𝑥 = 1 , 𝑦 = 0 x=1,y=0 Question 4 Solve the linear and quadratic system: 𝑦 + 3 𝑥 = 2 , 𝑦 = 𝑥 2 y+3x=2,y=x 2 Answer Substitute 𝑦 = 𝑥 2 y=x 2 into 𝑦 + 3 𝑥 = 2 y+3x=2: 𝑥 2 + 3 𝑥 = 2 x 2 +3x=2 Rearrange: 𝑥 2 + 3 𝑥 − 2 = 0 x 2 +3x−2=0 Solve using the quadratic formula: 𝑥 = − 𝑏 ± 𝑏 2 − 4 𝑎 𝑐 2 𝑎 = − 3 ± 3 2 − 4 ( 1 ) ( − 2 ) 2 ( 1 ) x= 2a −b± b 2 −4ac = 2(1) −3± 3 2 −4(1)(−2) 𝑥 = − 3 ± 17 2 x= 2 −3± 17 𝑥 = 3.562 and 𝑥 = − 0.562 x=3.562 and x=−0.562 Find 𝑦 y for both 𝑥 x: For 𝑥 = 3.562 x=3.562: 𝑦 = ( 3.562 ) 2 = 12.685 y=(3.562) 2 =12.685 For 𝑥 = − 0.562 x=−0.562: 𝑦 = ( − 0.562 ) 2 = 0.315 y=(−0.562) 2 =0.315 Intersection Points: ( − 0.562 , 0.315 ) , ( 3.562 , 12.685 ) (−0.562,0.315),(3.562,12.685) Graph: The graph shows the line 𝑦 + 3 𝑥 = 2 y+3x=2 and the parabola 𝑦 = 𝑥 2 y=x 2 intersecting at these points.