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Is it possible that all solutions of a homogeneous system of twelve linear equations in fourteen variables are multiples of one fixed nonzero solution? Discuss. Consider the system as Ax = 0, where A is a 12x14 matrix. Choose the correct answer below. O A. Yes. Since A has at most 12 pivot positions, rank As 12. By the Rank Theorem, dim Nul A = 14- rank Az 2. Since there is at least one free variable in the system, all solutions are multiples of one fixed nonzero solution. O B. No. Since A has 12 pivot positions, rank A= 12. By the Rank Theorem, dim Nul A= 12- rank A= 0. Since Nul A= 0, it is impossible to find a single vector in Nul A that spans Nul A. OC. No. Since A has at most 12 pivot positions, rank As 12. By the Rank Theorem, dim Nul A= 14 - rank Az 2. Thus, it is impossible to find a single vector in Nul A that spans Nul A O D. Yes. Since A has 12 pivot positions, rank A = 12. By the Rank Theorem, dim Nul A= 12- rank A=0. Thus, all solutions are multiples of one fixed nonzero solution.

Is it possible that all solutions of a homogeneous system of twelve linear equations in fourteen variables are multiples of one fixed nonzero solution? Discuss. Consider the system as Ax = 0, where A is a 12x14 matrix. Choose the correct answer below. O A. Yes. Since A has at most 12 pivot positions, rank As 12. By the Rank Theorem, dim Nul A = 14- rank Az 2. Since there is at least one free variable in the system, all solutions are multiples of one fixed nonzero solution. O B. No. Since A has 12 pivot positions, rank A= 12. By the Rank Theorem, dim Nul A= 12- rank A= 0. Since Nul A= 0, it is impossible to find a single vector in Nul A that spans Nul A. OC. No. Since A has at most 12 pivot positions, rank As 12. By the Rank Theorem, dim Nul A= 14 - rank Az 2. Thus, it is impossible to find a single vector in Nul A that spans Nul A O D. Yes. Since A has 12 pivot positions, rank A = 12. By the Rank Theorem, dim Nul A= 12- rank A=0. Thus, all solutions are multiples of one fixed nonzero solution.

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Is it possible that all solutions of a homogeneous system of twelve linear equations in fourteen variables are multiples of one fixed nonzero solution? Discuss. Consider the system as Ax = 0, where A is a 12x 14 matrix. Choose the correct answer below. O A. Yes. Since A has at most 12 pivot positions, rank As 12. By the Rank Theorem, dim Nul A = 14- rank Az 2. Since there is at least one free variable in the system, all solutions are multiples of one fixed nonzero solution. O B. No. Since A has 12 pivot positions, rank A= 12. By the Rank Theorem, dim Nul A = 12 - rank A= 0. Since Nul A = 0, it is impossible to find a single vector in Nul A that spans Nul A. OC. No. Since A has at most 12 pivot positions, rank As 12. By the Rank Theorem, Nul A= 14 - rank Az 2. Thus, it is impossible to find a single vector in Nul A that spans Nul A. O D. Yes. Since A has 12 pivot positions, rank A= 12. By the Rank Theorem, dim Nul A= 12- rank A= 0. Thus, all solutions are multiples of one fixed nonzero solution.
Transcribed Image Text:Is it possible that all solutions of a homogeneous system of twelve linear equations in fourteen variables are multiples of one fixed nonzero solution? Discuss. Consider the system as Ax = 0, where A is a 12x 14 matrix. Choose the correct answer below. O A. Yes. Since A has at most 12 pivot positions, rank As 12. By the Rank Theorem, dim Nul A = 14- rank Az 2. Since there is at least one free variable in the system, all solutions are multiples of one fixed nonzero solution. O B. No. Since A has 12 pivot positions, rank A= 12. By the Rank Theorem, dim Nul A = 12 - rank A= 0. Since Nul A = 0, it is impossible to find a single vector in Nul A that spans Nul A. OC. No. Since A has at most 12 pivot positions, rank As 12. By the Rank Theorem, Nul A= 14 - rank Az 2. Thus, it is impossible to find a single vector in Nul A that spans Nul A. O D. Yes. Since A has 12 pivot positions, rank A= 12. By the Rank Theorem, dim Nul A= 12- rank A= 0. Thus, all solutions are multiples of one fixed nonzero solution.
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