Prove that the nonzero row vectors of a matrix in row-echelon form are linearly independent. Let (2) A be an m xn matrix in row-echelon form. If the first column of A is not all zero and e,, ezmy ea, denote leading ones, then the nonzero row vectors r,, r, of A, have the form of r, = ---Select--- 2 = ---Select--- 3 = ---Select--- and so forth. Then, the equation c,r, + c,r, + .. + cr, = 0 implies which of the following equations? (Select all that apply.) O cge3n + cze3n + Cze3n = 0 O cje2m + Cze2m = 0 O cen + Cze2n + Cz@3n = 0 O cgeim + cze2m = 0 O cze3n = 0 O c,e1 = 0 You can conclude in turn that c, - C, = * - Ck and so the row vectors are linearly independent.
Prove that the nonzero row vectors of a matrix in row-echelon form are linearly independent. Let (2) A be an m xn matrix in row-echelon form. If the first column of A is not all zero and e,, ezmy ea, denote leading ones, then the nonzero row vectors r,, r, of A, have the form of r, = ---Select--- 2 = ---Select--- 3 = ---Select--- and so forth. Then, the equation c,r, + c,r, + .. + cr, = 0 implies which of the following equations? (Select all that apply.) O cge3n + cze3n + Cze3n = 0 O cje2m + Cze2m = 0 O cen + Cze2n + Cz@3n = 0 O cgeim + cze2m = 0 O cze3n = 0 O c,e1 = 0 You can conclude in turn that c, - C, = * - Ck and so the row vectors are linearly independent.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Prove that the nonzero row vectors of a matrix in row-echelon form are linearly independent.
Let
= A be an m x n matrix in row-echelon form. If the first column of A is not all zero and e,,, e,m, ean denote leading ones, then the nonzero row vectors r, .. r, of A, have the form of
r1
---Select---
r2
-Select---
---Select---
and so forth.
Then, the equation c,r, + c,r, + ... + cy = 0 implies which of the following equations? (Select all that apply.)
U cze3n + C2e3n + Cze3n = 0
C1@2m + Cz€2m
= 0
= 0
C1ein + Cz@2n + cze3n
Cje im
* Cze2m = 0
Cze 3n
= 0
Cje11
= 0
You can conclude in turn that c, = c,
Ck
and so the row vectors are linearly independent.
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