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

2024
Views:

52

Pages:

3
Author:

Ruby Bates

Question 1 Given a homogeneous system of linear equations, if the system is overdetermined, what are the possibilities as to the number of solutions? Explain. Answer 1 If the system is overdetermined (more equations than variables): If there are more linearly independent equations than variables, the system is inconsistent (no solutions). If there are exactly 𝑛 n linearly independent equations (where 𝑛 n is the number of variables), the system has either no solution or a unique solution. If there are fewer linearly independent equations than variables, the system can be reduced to a consistent system with infinitely many solutions. Answer 2 A homogeneous system is overdetermined if the number of equations exceeds the number of variables ( 𝑚 > 𝑛 m>n). If all equations are linearly independent and 𝑚 > 𝑛 m>n: No solution exists. If the system is consistent: There may be either one unique solution or infinitely many solutions. Example: Consider 𝑥 + 𝑦 = 0 x+y=0, 2 𝑥 + 3 𝑦 = 0 2x+3y=0, 3 𝑥 + 2 𝑦 = 0 3x+2y=0: Here 𝑚 = 3 m=3, 𝑛 = 2 n=2, so the system is overdetermined. Solution: 𝑥 = 0 , 𝑦 = 0 x=0,y=0. Question 2 Suppose a nonhomogeneous system of nine linear equations in ten unknowns has a solution for all possible constants on the right-hand side of the equations. Is it possible to find two nonzero solutions of the associated homogeneous system that are not multiples of each other? Discuss. Answer From the rank-nullity theorem: Rank ( 𝐴 ) + dim ( Null ( 𝐴 ) ) = 𝑛 Rank(A)+dim(Null(A))=n Given Rank ( 𝐴 ) = 9 Rank(A)=9, 𝑛 = 10 n=10, so: dim ( Null ( 𝐴 ) ) = 10 − 9 = 1 dim(Null(A))=10−9=1 This means the null space is one-dimensional, so there is only one linearly independent solution in the null space. Hence, it is impossible to find two nonzero solutions that are not multiples of each other. Question 3 Is there any value of 𝑘 k for which this system of linear equations has infinitely many solutions? Equations: 𝑥 ( 𝑘 + 2 ) + 𝑦 ( 𝑘 − 1 ) + 𝑧 ( 𝑘 ) = 2 x(k+2)+y(k−1)+z(k)=2 𝑦 ( 𝑘 + 2 ) + 2 𝑧 = 0 y(k+2)+2z=0 𝑧 ( 𝑘 2 − 𝑘 − 2 ) = 𝑘 + 2 z(k 2 −k−2)=k+2 Answer Analyze the system: Third equation: 𝑧 ( 𝑘 2 − 𝑘 − 2 ) = 𝑘 + 2 z(k 2 −k−2)=k+2. Solve for 𝑘 k: If 𝑘 2 − 𝑘 − 2 = 0 k 2 −k−2=0: ( 𝑘 − 2 ) ( 𝑘 + 1 ) = 0 ⟹ 𝑘 = 2 or 𝑘 = − 1 (k−2)(k+1)=0⟹k=2 or k=−1 Cases: For 𝑘 = − 1 k=−1, the third equation becomes 𝑧 ( 0 ) = 1 z(0)=1, which is inconsistent (no solutions). For 𝑘 = 2 k=2, the rank of the system reduces to 2 (dependent rows), giving infinitely many solutions. Conclusion: The system has infinitely many solutions when 𝑘 = 2 k=2. Question 4 Find the equation of the line with slope − 3 −3 and passing through ( − 3 , 5 ) (−3,5). Write your equation in point-slope and slope-intercept forms. Answer Point-slope form: 𝑦 − 𝑦 1 = 𝑚 ( 𝑥 − 𝑥 1 ) y−y 1 =m(x−x 1 ) Substituting 𝑚 = − 3 m=−3, ( 𝑥 1 , 𝑦 1 ) = ( − 3 , 5 ) (x 1 ,y 1 )=(−3,5): 𝑦 − 5 = − 3 ( 𝑥 + 3 ) y−5=−3(x+3) Slope-intercept form: Simplify: 𝑦 − 5 = − 3 𝑥 − 9 y−5=−3x−9 𝑦 = − 3 𝑥 − 4 y=−3x−4 Final Answer: Point-slope form: 𝑦 − 5 = − 3 ( 𝑥 + 3 ) y−5=−3(x+3) Slope-intercept form: 𝑦 = − 3 𝑥 − 4 y=−3x−4