The set of all polynomials of degree 6 under the standard addition and scalar multiplication operations is not a vector space because * We can find two polynomials P(x) and Q(x) for which P(x) Q(x)#Q(x)·P(x) We can find a polynomial P(x) for which 1 P(x)=P(x) It is not closed under addition. We can find a polynomial P(x) such that (c+d)P(x)#cP(x)+dP(x). [1 2 31 Let A = |1 3 9. Then the adjoint of A is given by: li 5 7
The set of all polynomials of degree 6 under the standard addition and scalar multiplication operations is not a vector space because * We can find two polynomials P(x) and Q(x) for which P(x) Q(x)#Q(x)·P(x) We can find a polynomial P(x) for which 1 P(x)=P(x) It is not closed under addition. We can find a polynomial P(x) such that (c+d)P(x)#cP(x)+dP(x). [1 2 31 Let A = |1 3 9. Then the adjoint of A is given by: li 5 7
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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Question
![5:16 M P
A docs.google.com/forms/d/e,
15
S is a basis for Rº
the above is true
The set of all polynomials of degree 6 under the
standard addition and scalar multiplication operations
is not a vector space because *
We can find two polynomials P(x) and Q(x)
for which P(x)-Q(x)#Q(x)·P(x)
We can find a polynomial P(x) for which
1-P(x)=P(x)
It is not closed under addition.
We can find a polynomial P(x) such that
(c+d)P(x)#cP(x)+dP(x).
[1 2 3]
3 9. Then the adjoint of A is given by:
li 5 7
Let A = |1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf13947d-bd88-4500-8649-4df845aa0f69%2Faa36574a-166f-48fa-91cf-c5382728d103%2Fcql36g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:5:16 M P
A docs.google.com/forms/d/e,
15
S is a basis for Rº
the above is true
The set of all polynomials of degree 6 under the
standard addition and scalar multiplication operations
is not a vector space because *
We can find two polynomials P(x) and Q(x)
for which P(x)-Q(x)#Q(x)·P(x)
We can find a polynomial P(x) for which
1-P(x)=P(x)
It is not closed under addition.
We can find a polynomial P(x) such that
(c+d)P(x)#cP(x)+dP(x).
[1 2 3]
3 9. Then the adjoint of A is given by:
li 5 7
Let A = |1
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