ll touch ?4:32 PM@ 86%docs.google.com18-18*Let W = {(a,b, c, d)| 5a – b = 0} be a subspace of R*. Then a basis for W isO {(1,5,0,0),(0,0,0,1)}{(1,5,0,0),(0,0,1,0),(0,0,0,1)}{(5,1,0,0),(0,0,1,0),(0,0,0,1)}O {(1,5,0,0)}Let k be a real number. By using Cramer's rule, the solution of the following system is:( 2х — 2у — 012x – 3y = 2k

Basis of a vector space is a vector which has every element writen in a unique way a finite linear…

Answered: Let W = {(a, b,c, d)| 5a – b = 0} be a…

Let W = {(a, b,c, d)| 5a – b = 0} be a subspace of R*. Then a basis for W is O {(1,5,0,0),(0,0,0,1)} O {(1,5,0,0),(0,0,1,0),(0,0,0,1)} O {(5,1,0,0),(0,0,1,0),(0,0,0,1)} O {(1,5,0,0)}

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN:9781305658004

Author:Ron Larson

Publisher:Ron Larson

Chapter6: Linear Transformations

Section6.1: Introduction To Linear Transformations

Problem 82E

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Linear Algebra

In mathematics, problems can be solved by the arithmetic operators like addition, subtraction, multiplication, and division. In algebra, we represent the numbers by any letter called a variable. If these variables are of degree 1, then it is studied under linear algebra. The first-degree equations involving algebraic expressions are called linear equations. That is, if you represent it by drawing a graph you will get a straight line.

Question

ll touch ?
4:32 PM
@ 86%
docs.google.com
18
-18
*
Let W = {(a,b, c, d)| 5a – b = 0} be a subspace of R*. Then a basis for W is
O {(1,5,0,0),(0,0,0,1)}
{(1,5,0,0),(0,0,1,0),(0,0,0,1)}
{(5,1,0,0),(0,0,1,0),(0,0,0,1)}
O {(1,5,0,0)}
Let k be a real number. By using Cramer's rule, the solution of the following system is:
( 2х — 2у — 0
12x – 3y = 2k

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