Determine whether the set of vectors in P2 is linearly dependent or linearly independent
0
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I am getting conflicting answers to what the book has, am I setting this up correctly? I did row reduced echelon form in the calculator, I am unsure what to do from here or if I am doing it properly
linear-algebra
asked Dec 18 '18 at 8:43
isuckatprogrammingisuckatprogramming
256
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add a comment |
0
$begingroup$
I am getting conflicting answers to what the book has, am I setting this up correctly? I did row reduced echelon form in the calculator, I am unsure what to do from here or if I am doing it properly
linear-algebra
asked Dec 18 '18 at 8:43
isuckatprogrammingisuckatprogramming
256
$endgroup$
add a comment |
0
0
0
$begingroup$
I am getting conflicting answers to what the book has, am I setting this up correctly? I did row reduced echelon form in the calculator, I am unsure what to do from here or if I am doing it properly
linear-algebra
asked Dec 18 '18 at 8:43
isuckatprogrammingisuckatprogramming
256
$endgroup$
I am getting conflicting answers to what the book has, am I setting this up correctly? I did row reduced echelon form in the calculator, I am unsure what to do from here or if I am doing it properly
linear-algebra
linear-algebra
asked Dec 18 '18 at 8:43
isuckatprogrammingisuckatprogramming
256
asked Dec 18 '18 at 8:43
isuckatprogrammingisuckatprogramming
256
asked Dec 18 '18 at 8:43
isuckatprogrammingisuckatprogramming
256
asked Dec 18 '18 at 8:43
isuckatprogrammingisuckatprogramming
256
asked Dec 18 '18 at 8:43
isuckatprogrammingisuckatprogramming
256
256
add a comment |
add a comment |
1 Answer
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$3(2-x)-(6-7x+x^{2})-(4x-x^{2})=0$, so the polynomials are linearly dependent.
answered Dec 18 '18 at 8:46
Kavi Rama MurthyKavi Rama Murthy
63k42362
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$begingroup$
can you show how you got that?
$endgroup$
– isuckatprogramming
Dec 18 '18 at 8:48
1
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Just by inspection!
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 8:49
$begingroup$
can you show the step by step work?
$endgroup$
– isuckatprogramming
Dec 18 '18 at 8:51
1
$begingroup$
You can first take a linear combination of the second and third so that the $x^{2}$ term gets cancelled and then see if what you get is a multiple of the first polynomial. [n this case just adding the second and the third removes the second degree term and you will be left with $3$ times the first polynomial.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 8:52
$begingroup$
@isuckatprogramming Alternatively, you can first take a linear combination of the first and second polynomials so that the constant term is cancelled, and see that what you get is a multiple of the third. There are many ways to go here. I don't suggest starting with the first-degree term, though.
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– Arthur
Dec 18 '18 at 8:54
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1 Answer
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1 Answer
1
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1
$begingroup$
$3(2-x)-(6-7x+x^{2})-(4x-x^{2})=0$, so the polynomials are linearly dependent.
answered Dec 18 '18 at 8:46
Kavi Rama MurthyKavi Rama Murthy
63k42362
$endgroup$
$begingroup$
can you show how you got that?
$endgroup$
– isuckatprogramming
Dec 18 '18 at 8:48
1
$begingroup$
Just by inspection!
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 8:49
$begingroup$
can you show the step by step work?
$endgroup$
– isuckatprogramming
Dec 18 '18 at 8:51
1
$begingroup$
You can first take a linear combination of the second and third so that the $x^{2}$ term gets cancelled and then see if what you get is a multiple of the first polynomial. [n this case just adding the second and the third removes the second degree term and you will be left with $3$ times the first polynomial.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 8:52
$begingroup$
@isuckatprogramming Alternatively, you can first take a linear combination of the first and second polynomials so that the constant term is cancelled, and see that what you get is a multiple of the third. There are many ways to go here. I don't suggest starting with the first-degree term, though.
$endgroup$
– Arthur
Dec 18 '18 at 8:54
|
show 1 more comment
1
$begingroup$
$3(2-x)-(6-7x+x^{2})-(4x-x^{2})=0$, so the polynomials are linearly dependent.
answered Dec 18 '18 at 8:46
Kavi Rama MurthyKavi Rama Murthy
63k42362
$endgroup$
$begingroup$
can you show how you got that?
$endgroup$
– isuckatprogramming
Dec 18 '18 at 8:48
1
$begingroup$
Just by inspection!
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 8:49
$begingroup$
can you show the step by step work?
$endgroup$
– isuckatprogramming
Dec 18 '18 at 8:51
1
$begingroup$
You can first take a linear combination of the second and third so that the $x^{2}$ term gets cancelled and then see if what you get is a multiple of the first polynomial. [n this case just adding the second and the third removes the second degree term and you will be left with $3$ times the first polynomial.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 8:52
$begingroup$
@isuckatprogramming Alternatively, you can first take a linear combination of the first and second polynomials so that the constant term is cancelled, and see that what you get is a multiple of the third. There are many ways to go here. I don't suggest starting with the first-degree term, though.
$endgroup$
– Arthur
Dec 18 '18 at 8:54
|
show 1 more comment
1
1
1
$begingroup$
$3(2-x)-(6-7x+x^{2})-(4x-x^{2})=0$, so the polynomials are linearly dependent.
answered Dec 18 '18 at 8:46
Kavi Rama MurthyKavi Rama Murthy
63k42362
$endgroup$
$3(2-x)-(6-7x+x^{2})-(4x-x^{2})=0$, so the polynomials are linearly dependent.
answered Dec 18 '18 at 8:46
Kavi Rama MurthyKavi Rama Murthy
63k42362
answered Dec 18 '18 at 8:46
Kavi Rama MurthyKavi Rama Murthy
63k42362
answered Dec 18 '18 at 8:46
Kavi Rama MurthyKavi Rama Murthy
63k42362
answered Dec 18 '18 at 8:46
Kavi Rama MurthyKavi Rama Murthy
63k42362
63k42362
$begingroup$
can you show how you got that?
$endgroup$
– isuckatprogramming
Dec 18 '18 at 8:48
1
$begingroup$
Just by inspection!
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 8:49
$begingroup$
can you show the step by step work?
$endgroup$
– isuckatprogramming
Dec 18 '18 at 8:51
1
$begingroup$
You can first take a linear combination of the second and third so that the $x^{2}$ term gets cancelled and then see if what you get is a multiple of the first polynomial. [n this case just adding the second and the third removes the second degree term and you will be left with $3$ times the first polynomial.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 8:52
$begingroup$
@isuckatprogramming Alternatively, you can first take a linear combination of the first and second polynomials so that the constant term is cancelled, and see that what you get is a multiple of the third. There are many ways to go here. I don't suggest starting with the first-degree term, though.
$endgroup$
– Arthur
Dec 18 '18 at 8:54
|
show 1 more comment
$begingroup$
can you show how you got that?
$endgroup$
– isuckatprogramming
Dec 18 '18 at 8:48
1
$begingroup$
Just by inspection!
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 8:49
$begingroup$
can you show the step by step work?
$endgroup$
– isuckatprogramming
Dec 18 '18 at 8:51
1
$begingroup$
You can first take a linear combination of the second and third so that the $x^{2}$ term gets cancelled and then see if what you get is a multiple of the first polynomial. [n this case just adding the second and the third removes the second degree term and you will be left with $3$ times the first polynomial.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 8:52
$begingroup$
@isuckatprogramming Alternatively, you can first take a linear combination of the first and second polynomials so that the constant term is cancelled, and see that what you get is a multiple of the third. There are many ways to go here. I don't suggest starting with the first-degree term, though.
$endgroup$
– Arthur
Dec 18 '18 at 8:54
$begingroup$
can you show how you got that?
$endgroup$
– isuckatprogramming
Dec 18 '18 at 8:48
$begingroup$
can you show how you got that?
$endgroup$
– isuckatprogramming
Dec 18 '18 at 8:48
1
1
$begingroup$
Just by inspection!
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 8:49
$begingroup$
Just by inspection!
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 8:49
$begingroup$
can you show the step by step work?
$endgroup$
– isuckatprogramming
Dec 18 '18 at 8:51
$begingroup$
can you show the step by step work?
$endgroup$
– isuckatprogramming
Dec 18 '18 at 8:51
1
1
$begingroup$
You can first take a linear combination of the second and third so that the $x^{2}$ term gets cancelled and then see if what you get is a multiple of the first polynomial. [n this case just adding the second and the third removes the second degree term and you will be left with $3$ times the first polynomial.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 8:52
$begingroup$
You can first take a linear combination of the second and third so that the $x^{2}$ term gets cancelled and then see if what you get is a multiple of the first polynomial. [n this case just adding the second and the third removes the second degree term and you will be left with $3$ times the first polynomial.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 8:52
$begingroup$
@isuckatprogramming Alternatively, you can first take a linear combination of the first and second polynomials so that the constant term is cancelled, and see that what you get is a multiple of the third. There are many ways to go here. I don't suggest starting with the first-degree term, though.
$endgroup$
– Arthur
Dec 18 '18 at 8:54
$begingroup$
@isuckatprogramming Alternatively, you can first take a linear combination of the first and second polynomials so that the constant term is cancelled, and see that what you get is a multiple of the third. There are many ways to go here. I don't suggest starting with the first-degree term, though.
$endgroup$
– Arthur
Dec 18 '18 at 8:54
|
show 1 more comment
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