Having trouble solving for 3 constants in a system of 3 equations












0












$begingroup$


I'm trying to find out if a set of vectors is linearly independent or not, and the way I'm doing it is by setting the vectors to equal [0,0,0] when the 3 vectors are multiplied by a constant.



So for example: $c_1$[7, -14, 6] + $c_2$[-10, 15, $15over14$] + $c_3$[-1, 0 ,3] = [0,0,0].



Then I solve and see if there's a way to have this happen without all the constant's equaling 0.



But for some reason I'm having trouble coming up with how to use substitution here, I'm sure it's simple I'm just drawing a blank. If I solve for one variable, it's just equal to 0 so I can't use it to do anything and I'm confused.



One idea that I had was maybe just setting a constant, say $c_1$ to equal a random number and then use that?



Can I get help? Don't worry it's not an assignment question, it's a textbook question, the answer is $c_1$ $=$ $3over7$, $c_2$ $=$ $2over5$, $c_3$ $=$ $-1$.



(Also how do I know that I'll find an answer or not. And if there is one set of answers will there ALWAYS be an infinite amount of answers?)



Thanks!






linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hint: Consider the equations for each of the $3$ coordinates separately, e.g., for the first one, it's $7c_1 - 10c_2 - c_3 = 0$. Putting them together, you then have a set of $3$ linear equations in $3$ unknowns.
    $endgroup$
    – John Omielan
    Jan 14 at 1:17












  • $begingroup$
    Hi, sorry I should have mentioned I already did that. I just don't know how to use substitution for a set of 3 linear equations in 3 unknowns when there's a 0 on the right side for all of them. Because if I solve for, say c1, it'll end up equaling 0, no matter what coefficient goes in front of it
    $endgroup$
    – ming
    Jan 14 at 1:36










  • $begingroup$
    Thanks for the extra info. Actually, you don't usually use substitution to solve for a set of $3$ linear equations in $3$ unknowns as you don't yet know what to substitute. Also, it's incorrect to say that if you end up solving for $c_1$ that it'll end up always equaling $0$. The Wikipedia article System of linear equations does a quite job of explaining several methods to solve these types of equations. You should read this & pick one of the methods here, with "Elimination of Variables" being my choice for you to try.
    $endgroup$
    – John Omielan
    Jan 14 at 1:44










  • $begingroup$
    One other thing about the Wikipedia article. Your case is a special one of being a "Homogenous System". You may wish to read that section (which is further down) as well.
    $endgroup$
    – John Omielan
    Jan 14 at 1:46










  • $begingroup$
    Thanks I'll check it out. Do you know the answer to my last question in brackets though?
    $endgroup$
    – ming
    Jan 14 at 1:49
















0












$begingroup$


I'm trying to find out if a set of vectors is linearly independent or not, and the way I'm doing it is by setting the vectors to equal [0,0,0] when the 3 vectors are multiplied by a constant.



So for example: $c_1$[7, -14, 6] + $c_2$[-10, 15, $15over14$] + $c_3$[-1, 0 ,3] = [0,0,0].



Then I solve and see if there's a way to have this happen without all the constant's equaling 0.



But for some reason I'm having trouble coming up with how to use substitution here, I'm sure it's simple I'm just drawing a blank. If I solve for one variable, it's just equal to 0 so I can't use it to do anything and I'm confused.



One idea that I had was maybe just setting a constant, say $c_1$ to equal a random number and then use that?



Can I get help? Don't worry it's not an assignment question, it's a textbook question, the answer is $c_1$ $=$ $3over7$, $c_2$ $=$ $2over5$, $c_3$ $=$ $-1$.



(Also how do I know that I'll find an answer or not. And if there is one set of answers will there ALWAYS be an infinite amount of answers?)



Thanks!






linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hint: Consider the equations for each of the $3$ coordinates separately, e.g., for the first one, it's $7c_1 - 10c_2 - c_3 = 0$. Putting them together, you then have a set of $3$ linear equations in $3$ unknowns.
    $endgroup$
    – John Omielan
    Jan 14 at 1:17












  • $begingroup$
    Hi, sorry I should have mentioned I already did that. I just don't know how to use substitution for a set of 3 linear equations in 3 unknowns when there's a 0 on the right side for all of them. Because if I solve for, say c1, it'll end up equaling 0, no matter what coefficient goes in front of it
    $endgroup$
    – ming
    Jan 14 at 1:36










  • $begingroup$
    Thanks for the extra info. Actually, you don't usually use substitution to solve for a set of $3$ linear equations in $3$ unknowns as you don't yet know what to substitute. Also, it's incorrect to say that if you end up solving for $c_1$ that it'll end up always equaling $0$. The Wikipedia article System of linear equations does a quite job of explaining several methods to solve these types of equations. You should read this & pick one of the methods here, with "Elimination of Variables" being my choice for you to try.
    $endgroup$
    – John Omielan
    Jan 14 at 1:44










  • $begingroup$
    One other thing about the Wikipedia article. Your case is a special one of being a "Homogenous System". You may wish to read that section (which is further down) as well.
    $endgroup$
    – John Omielan
    Jan 14 at 1:46










  • $begingroup$
    Thanks I'll check it out. Do you know the answer to my last question in brackets though?
    $endgroup$
    – ming
    Jan 14 at 1:49














0












0








0





$begingroup$


I'm trying to find out if a set of vectors is linearly independent or not, and the way I'm doing it is by setting the vectors to equal [0,0,0] when the 3 vectors are multiplied by a constant.



So for example: $c_1$[7, -14, 6] + $c_2$[-10, 15, $15over14$] + $c_3$[-1, 0 ,3] = [0,0,0].



Then I solve and see if there's a way to have this happen without all the constant's equaling 0.



But for some reason I'm having trouble coming up with how to use substitution here, I'm sure it's simple I'm just drawing a blank. If I solve for one variable, it's just equal to 0 so I can't use it to do anything and I'm confused.



One idea that I had was maybe just setting a constant, say $c_1$ to equal a random number and then use that?



Can I get help? Don't worry it's not an assignment question, it's a textbook question, the answer is $c_1$ $=$ $3over7$, $c_2$ $=$ $2over5$, $c_3$ $=$ $-1$.



(Also how do I know that I'll find an answer or not. And if there is one set of answers will there ALWAYS be an infinite amount of answers?)



Thanks!






linear-algebra







share|cite|improve this question









$endgroup$




I'm trying to find out if a set of vectors is linearly independent or not, and the way I'm doing it is by setting the vectors to equal [0,0,0] when the 3 vectors are multiplied by a constant.



So for example: $c_1$[7, -14, 6] + $c_2$[-10, 15, $15over14$] + $c_3$[-1, 0 ,3] = [0,0,0].



Then I solve and see if there's a way to have this happen without all the constant's equaling 0.



But for some reason I'm having trouble coming up with how to use substitution here, I'm sure it's simple I'm just drawing a blank. If I solve for one variable, it's just equal to 0 so I can't use it to do anything and I'm confused.



One idea that I had was maybe just setting a constant, say $c_1$ to equal a random number and then use that?



Can I get help? Don't worry it's not an assignment question, it's a textbook question, the answer is $c_1$ $=$ $3over7$, $c_2$ $=$ $2over5$, $c_3$ $=$ $-1$.



(Also how do I know that I'll find an answer or not. And if there is one set of answers will there ALWAYS be an infinite amount of answers?)



Thanks!






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 14 at 1:15









mingming

4606




4606












  • $begingroup$
    Hint: Consider the equations for each of the $3$ coordinates separately, e.g., for the first one, it's $7c_1 - 10c_2 - c_3 = 0$. Putting them together, you then have a set of $3$ linear equations in $3$ unknowns.
    $endgroup$
    – John Omielan
    Jan 14 at 1:17












  • $begingroup$
    Hi, sorry I should have mentioned I already did that. I just don't know how to use substitution for a set of 3 linear equations in 3 unknowns when there's a 0 on the right side for all of them. Because if I solve for, say c1, it'll end up equaling 0, no matter what coefficient goes in front of it
    $endgroup$
    – ming
    Jan 14 at 1:36










  • $begingroup$
    Thanks for the extra info. Actually, you don't usually use substitution to solve for a set of $3$ linear equations in $3$ unknowns as you don't yet know what to substitute. Also, it's incorrect to say that if you end up solving for $c_1$ that it'll end up always equaling $0$. The Wikipedia article System of linear equations does a quite job of explaining several methods to solve these types of equations. You should read this & pick one of the methods here, with "Elimination of Variables" being my choice for you to try.
    $endgroup$
    – John Omielan
    Jan 14 at 1:44










  • $begingroup$
    One other thing about the Wikipedia article. Your case is a special one of being a "Homogenous System". You may wish to read that section (which is further down) as well.
    $endgroup$
    – John Omielan
    Jan 14 at 1:46










  • $begingroup$
    Thanks I'll check it out. Do you know the answer to my last question in brackets though?
    $endgroup$
    – ming
    Jan 14 at 1:49


















  • $begingroup$
    Hint: Consider the equations for each of the $3$ coordinates separately, e.g., for the first one, it's $7c_1 - 10c_2 - c_3 = 0$. Putting them together, you then have a set of $3$ linear equations in $3$ unknowns.
    $endgroup$
    – John Omielan
    Jan 14 at 1:17












  • $begingroup$
    Hi, sorry I should have mentioned I already did that. I just don't know how to use substitution for a set of 3 linear equations in 3 unknowns when there's a 0 on the right side for all of them. Because if I solve for, say c1, it'll end up equaling 0, no matter what coefficient goes in front of it
    $endgroup$
    – ming
    Jan 14 at 1:36










  • $begingroup$
    Thanks for the extra info. Actually, you don't usually use substitution to solve for a set of $3$ linear equations in $3$ unknowns as you don't yet know what to substitute. Also, it's incorrect to say that if you end up solving for $c_1$ that it'll end up always equaling $0$. The Wikipedia article System of linear equations does a quite job of explaining several methods to solve these types of equations. You should read this & pick one of the methods here, with "Elimination of Variables" being my choice for you to try.
    $endgroup$
    – John Omielan
    Jan 14 at 1:44










  • $begingroup$
    One other thing about the Wikipedia article. Your case is a special one of being a "Homogenous System". You may wish to read that section (which is further down) as well.
    $endgroup$
    – John Omielan
    Jan 14 at 1:46










  • $begingroup$
    Thanks I'll check it out. Do you know the answer to my last question in brackets though?
    $endgroup$
    – ming
    Jan 14 at 1:49
















$begingroup$
Hint: Consider the equations for each of the $3$ coordinates separately, e.g., for the first one, it's $7c_1 - 10c_2 - c_3 = 0$. Putting them together, you then have a set of $3$ linear equations in $3$ unknowns.
$endgroup$
– John Omielan
Jan 14 at 1:17






$begingroup$
Hint: Consider the equations for each of the $3$ coordinates separately, e.g., for the first one, it's $7c_1 - 10c_2 - c_3 = 0$. Putting them together, you then have a set of $3$ linear equations in $3$ unknowns.
$endgroup$
– John Omielan
Jan 14 at 1:17














$begingroup$
Hi, sorry I should have mentioned I already did that. I just don't know how to use substitution for a set of 3 linear equations in 3 unknowns when there's a 0 on the right side for all of them. Because if I solve for, say c1, it'll end up equaling 0, no matter what coefficient goes in front of it
$endgroup$
– ming
Jan 14 at 1:36




$begingroup$
Hi, sorry I should have mentioned I already did that. I just don't know how to use substitution for a set of 3 linear equations in 3 unknowns when there's a 0 on the right side for all of them. Because if I solve for, say c1, it'll end up equaling 0, no matter what coefficient goes in front of it
$endgroup$
– ming
Jan 14 at 1:36












$begingroup$
Thanks for the extra info. Actually, you don't usually use substitution to solve for a set of $3$ linear equations in $3$ unknowns as you don't yet know what to substitute. Also, it's incorrect to say that if you end up solving for $c_1$ that it'll end up always equaling $0$. The Wikipedia article System of linear equations does a quite job of explaining several methods to solve these types of equations. You should read this & pick one of the methods here, with "Elimination of Variables" being my choice for you to try.
$endgroup$
– John Omielan
Jan 14 at 1:44




$begingroup$
Thanks for the extra info. Actually, you don't usually use substitution to solve for a set of $3$ linear equations in $3$ unknowns as you don't yet know what to substitute. Also, it's incorrect to say that if you end up solving for $c_1$ that it'll end up always equaling $0$. The Wikipedia article System of linear equations does a quite job of explaining several methods to solve these types of equations. You should read this & pick one of the methods here, with "Elimination of Variables" being my choice for you to try.
$endgroup$
– John Omielan
Jan 14 at 1:44












$begingroup$
One other thing about the Wikipedia article. Your case is a special one of being a "Homogenous System". You may wish to read that section (which is further down) as well.
$endgroup$
– John Omielan
Jan 14 at 1:46




$begingroup$
One other thing about the Wikipedia article. Your case is a special one of being a "Homogenous System". You may wish to read that section (which is further down) as well.
$endgroup$
– John Omielan
Jan 14 at 1:46












$begingroup$
Thanks I'll check it out. Do you know the answer to my last question in brackets though?
$endgroup$
– ming
Jan 14 at 1:49




$begingroup$
Thanks I'll check it out. Do you know the answer to my last question in brackets though?
$endgroup$
– ming
Jan 14 at 1:49










1 Answer
1






active

oldest

votes


















1












$begingroup$

Put the vectors into a matrix and row-reduce. Then read off the solutions.



There will be a unique solution, $(0,0,0)$, if the matrix is invertible. If not, there will be free variables, and infinitely many solutions.



So, if there is one nonzero solution, there will be infinitely many. For instance, multiplying by any nonzero scalar would give another nonzero solution. This is always true for a homogeneous system.



So, we get $left(begin{array}{rrr|r}7&-10&-1&0\-14&15&0&0\6&frac{15}{14}&3&0end{array}right)to left(begin{array}{rrr|r}1&-frac{10}7&-frac17&0\0&-5&-2&0\0&frac{135}{14}&frac{27}7&0end{array}right)toleft(begin{array}{rrr|r}1&-frac{10}7&-frac17&0\0&1&frac25&0\0&0&0&0end{array}right) $.



Now by back substitution we get $(frac 37,frac25,-1)$. This was done by setting $c_3=-1$. There are infinitely many solutions.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Wait why Is it $-3 over 7$ when my solution has it as positive?
    $endgroup$
    – ming
    Jan 14 at 3:34












  • $begingroup$
    Sorry. It's a typo...
    $endgroup$
    – Chris Custer
    Jan 14 at 3:39












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

Put the vectors into a matrix and row-reduce. Then read off the solutions.



There will be a unique solution, $(0,0,0)$, if the matrix is invertible. If not, there will be free variables, and infinitely many solutions.



So, if there is one nonzero solution, there will be infinitely many. For instance, multiplying by any nonzero scalar would give another nonzero solution. This is always true for a homogeneous system.



So, we get $left(begin{array}{rrr|r}7&-10&-1&0\-14&15&0&0\6&frac{15}{14}&3&0end{array}right)to left(begin{array}{rrr|r}1&-frac{10}7&-frac17&0\0&-5&-2&0\0&frac{135}{14}&frac{27}7&0end{array}right)toleft(begin{array}{rrr|r}1&-frac{10}7&-frac17&0\0&1&frac25&0\0&0&0&0end{array}right) $.



Now by back substitution we get $(frac 37,frac25,-1)$. This was done by setting $c_3=-1$. There are infinitely many solutions.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Wait why Is it $-3 over 7$ when my solution has it as positive?
    $endgroup$
    – ming
    Jan 14 at 3:34












  • $begingroup$
    Sorry. It's a typo...
    $endgroup$
    – Chris Custer
    Jan 14 at 3:39
















1












$begingroup$

Put the vectors into a matrix and row-reduce. Then read off the solutions.



There will be a unique solution, $(0,0,0)$, if the matrix is invertible. If not, there will be free variables, and infinitely many solutions.



So, if there is one nonzero solution, there will be infinitely many. For instance, multiplying by any nonzero scalar would give another nonzero solution. This is always true for a homogeneous system.



So, we get $left(begin{array}{rrr|r}7&-10&-1&0\-14&15&0&0\6&frac{15}{14}&3&0end{array}right)to left(begin{array}{rrr|r}1&-frac{10}7&-frac17&0\0&-5&-2&0\0&frac{135}{14}&frac{27}7&0end{array}right)toleft(begin{array}{rrr|r}1&-frac{10}7&-frac17&0\0&1&frac25&0\0&0&0&0end{array}right) $.



Now by back substitution we get $(frac 37,frac25,-1)$. This was done by setting $c_3=-1$. There are infinitely many solutions.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Wait why Is it $-3 over 7$ when my solution has it as positive?
    $endgroup$
    – ming
    Jan 14 at 3:34












  • $begingroup$
    Sorry. It's a typo...
    $endgroup$
    – Chris Custer
    Jan 14 at 3:39














1












1








1





$begingroup$

Put the vectors into a matrix and row-reduce. Then read off the solutions.



There will be a unique solution, $(0,0,0)$, if the matrix is invertible. If not, there will be free variables, and infinitely many solutions.



So, if there is one nonzero solution, there will be infinitely many. For instance, multiplying by any nonzero scalar would give another nonzero solution. This is always true for a homogeneous system.



So, we get $left(begin{array}{rrr|r}7&-10&-1&0\-14&15&0&0\6&frac{15}{14}&3&0end{array}right)to left(begin{array}{rrr|r}1&-frac{10}7&-frac17&0\0&-5&-2&0\0&frac{135}{14}&frac{27}7&0end{array}right)toleft(begin{array}{rrr|r}1&-frac{10}7&-frac17&0\0&1&frac25&0\0&0&0&0end{array}right) $.



Now by back substitution we get $(frac 37,frac25,-1)$. This was done by setting $c_3=-1$. There are infinitely many solutions.






share|cite|improve this answer











$endgroup$



Put the vectors into a matrix and row-reduce. Then read off the solutions.



There will be a unique solution, $(0,0,0)$, if the matrix is invertible. If not, there will be free variables, and infinitely many solutions.



So, if there is one nonzero solution, there will be infinitely many. For instance, multiplying by any nonzero scalar would give another nonzero solution. This is always true for a homogeneous system.



So, we get $left(begin{array}{rrr|r}7&-10&-1&0\-14&15&0&0\6&frac{15}{14}&3&0end{array}right)to left(begin{array}{rrr|r}1&-frac{10}7&-frac17&0\0&-5&-2&0\0&frac{135}{14}&frac{27}7&0end{array}right)toleft(begin{array}{rrr|r}1&-frac{10}7&-frac17&0\0&1&frac25&0\0&0&0&0end{array}right) $.



Now by back substitution we get $(frac 37,frac25,-1)$. This was done by setting $c_3=-1$. There are infinitely many solutions.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 14 at 3:40

























answered Jan 14 at 3:22









Chris CusterChris Custer

14.4k3827




14.4k3827












  • $begingroup$
    Wait why Is it $-3 over 7$ when my solution has it as positive?
    $endgroup$
    – ming
    Jan 14 at 3:34












  • $begingroup$
    Sorry. It's a typo...
    $endgroup$
    – Chris Custer
    Jan 14 at 3:39


















  • $begingroup$
    Wait why Is it $-3 over 7$ when my solution has it as positive?
    $endgroup$
    – ming
    Jan 14 at 3:34












  • $begingroup$
    Sorry. It's a typo...
    $endgroup$
    – Chris Custer
    Jan 14 at 3:39
















$begingroup$
Wait why Is it $-3 over 7$ when my solution has it as positive?
$endgroup$
– ming
Jan 14 at 3:34






$begingroup$
Wait why Is it $-3 over 7$ when my solution has it as positive?
$endgroup$
– ming
Jan 14 at 3:34














$begingroup$
Sorry. It's a typo...
$endgroup$
– Chris Custer
Jan 14 at 3:39




$begingroup$
Sorry. It's a typo...
$endgroup$
– Chris Custer
Jan 14 at 3:39


















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Cabo Verde

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  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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Gyllenstierna

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