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

The basis for W is given as follows

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 %3D {(1,5,0,0),(0,0,0,1) {(5,1,0,0),(0,0,1,0),(0,0,0,1)} O (1,5,0,0) O {(1,5,0,0),(0,0,1,0),(0,0,0,1)}

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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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.

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

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