How to find values that make a matrix solvable and unsolvable?












0












$begingroup$


I need to find two more $b$'s other than $b=(2,5,7)$, such that the equation can be solved and two more such that the equation can't be solved. $$ubegin{bmatrix}1\2\3end{bmatrix}+vbegin{bmatrix}1\0\1end{bmatrix}+wbegin{bmatrix}1\3\4end{bmatrix}=b$$
How do I find those values? What should I do to find both, values of b that make the system solvable and values that make the system unsolvable?



I tried to set $b$ as $(b_1,b_2,b_3)$, but I ended up with this matrix:
$$left[begin{array}{l}1&1&1&b_1\0&2&-1&2b_1-b_2\0&0&0&b_1-b_3+b_2end{array}right]$$
Sorry if it is something obvious, I'm knew with linear algebra.






linear-algebra matrices







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$endgroup$












  • $begingroup$
    You're almost there. The last step is to use the row echelon form matrix you computed to the see if you can pick values of $b_1,b_2,b_3$ so that you do or don't have solutions. For instance, if $b_1=b_2=0$ and $b_3=1$ can you have a solution?
    $endgroup$
    – tch
    Jan 26 at 0:32










  • $begingroup$
    So I just need to give random values to$ b_1$,$ b_2$ and $b_3$? And if $b_1=b_2$ and $b_3=1$, I think that there is no solution, because in the last row there will be a 0=-1, right?
    $endgroup$
    – davidllerenav
    Jan 26 at 0:43










  • $begingroup$
    Exactly! The problem is asking you to find some values of $b$ where there is a solution, and some where there isn't.
    $endgroup$
    – tch
    Jan 26 at 0:46
















0












$begingroup$


I need to find two more $b$'s other than $b=(2,5,7)$, such that the equation can be solved and two more such that the equation can't be solved. $$ubegin{bmatrix}1\2\3end{bmatrix}+vbegin{bmatrix}1\0\1end{bmatrix}+wbegin{bmatrix}1\3\4end{bmatrix}=b$$
How do I find those values? What should I do to find both, values of b that make the system solvable and values that make the system unsolvable?



I tried to set $b$ as $(b_1,b_2,b_3)$, but I ended up with this matrix:
$$left[begin{array}{l}1&1&1&b_1\0&2&-1&2b_1-b_2\0&0&0&b_1-b_3+b_2end{array}right]$$
Sorry if it is something obvious, I'm knew with linear algebra.






linear-algebra matrices







share|cite|improve this question









$endgroup$












  • $begingroup$
    You're almost there. The last step is to use the row echelon form matrix you computed to the see if you can pick values of $b_1,b_2,b_3$ so that you do or don't have solutions. For instance, if $b_1=b_2=0$ and $b_3=1$ can you have a solution?
    $endgroup$
    – tch
    Jan 26 at 0:32










  • $begingroup$
    So I just need to give random values to$ b_1$,$ b_2$ and $b_3$? And if $b_1=b_2$ and $b_3=1$, I think that there is no solution, because in the last row there will be a 0=-1, right?
    $endgroup$
    – davidllerenav
    Jan 26 at 0:43










  • $begingroup$
    Exactly! The problem is asking you to find some values of $b$ where there is a solution, and some where there isn't.
    $endgroup$
    – tch
    Jan 26 at 0:46














0












0








0





$begingroup$


I need to find two more $b$'s other than $b=(2,5,7)$, such that the equation can be solved and two more such that the equation can't be solved. $$ubegin{bmatrix}1\2\3end{bmatrix}+vbegin{bmatrix}1\0\1end{bmatrix}+wbegin{bmatrix}1\3\4end{bmatrix}=b$$
How do I find those values? What should I do to find both, values of b that make the system solvable and values that make the system unsolvable?



I tried to set $b$ as $(b_1,b_2,b_3)$, but I ended up with this matrix:
$$left[begin{array}{l}1&1&1&b_1\0&2&-1&2b_1-b_2\0&0&0&b_1-b_3+b_2end{array}right]$$
Sorry if it is something obvious, I'm knew with linear algebra.






linear-algebra matrices







share|cite|improve this question









$endgroup$




I need to find two more $b$'s other than $b=(2,5,7)$, such that the equation can be solved and two more such that the equation can't be solved. $$ubegin{bmatrix}1\2\3end{bmatrix}+vbegin{bmatrix}1\0\1end{bmatrix}+wbegin{bmatrix}1\3\4end{bmatrix}=b$$
How do I find those values? What should I do to find both, values of b that make the system solvable and values that make the system unsolvable?



I tried to set $b$ as $(b_1,b_2,b_3)$, but I ended up with this matrix:
$$left[begin{array}{l}1&1&1&b_1\0&2&-1&2b_1-b_2\0&0&0&b_1-b_3+b_2end{array}right]$$
Sorry if it is something obvious, I'm knew with linear algebra.






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 25 at 23:44









davidllerenavdavidllerenav

1888




1888












  • $begingroup$
    You're almost there. The last step is to use the row echelon form matrix you computed to the see if you can pick values of $b_1,b_2,b_3$ so that you do or don't have solutions. For instance, if $b_1=b_2=0$ and $b_3=1$ can you have a solution?
    $endgroup$
    – tch
    Jan 26 at 0:32










  • $begingroup$
    So I just need to give random values to$ b_1$,$ b_2$ and $b_3$? And if $b_1=b_2$ and $b_3=1$, I think that there is no solution, because in the last row there will be a 0=-1, right?
    $endgroup$
    – davidllerenav
    Jan 26 at 0:43










  • $begingroup$
    Exactly! The problem is asking you to find some values of $b$ where there is a solution, and some where there isn't.
    $endgroup$
    – tch
    Jan 26 at 0:46


















  • $begingroup$
    You're almost there. The last step is to use the row echelon form matrix you computed to the see if you can pick values of $b_1,b_2,b_3$ so that you do or don't have solutions. For instance, if $b_1=b_2=0$ and $b_3=1$ can you have a solution?
    $endgroup$
    – tch
    Jan 26 at 0:32










  • $begingroup$
    So I just need to give random values to$ b_1$,$ b_2$ and $b_3$? And if $b_1=b_2$ and $b_3=1$, I think that there is no solution, because in the last row there will be a 0=-1, right?
    $endgroup$
    – davidllerenav
    Jan 26 at 0:43










  • $begingroup$
    Exactly! The problem is asking you to find some values of $b$ where there is a solution, and some where there isn't.
    $endgroup$
    – tch
    Jan 26 at 0:46
















$begingroup$
You're almost there. The last step is to use the row echelon form matrix you computed to the see if you can pick values of $b_1,b_2,b_3$ so that you do or don't have solutions. For instance, if $b_1=b_2=0$ and $b_3=1$ can you have a solution?
$endgroup$
– tch
Jan 26 at 0:32




$begingroup$
You're almost there. The last step is to use the row echelon form matrix you computed to the see if you can pick values of $b_1,b_2,b_3$ so that you do or don't have solutions. For instance, if $b_1=b_2=0$ and $b_3=1$ can you have a solution?
$endgroup$
– tch
Jan 26 at 0:32












$begingroup$
So I just need to give random values to$ b_1$,$ b_2$ and $b_3$? And if $b_1=b_2$ and $b_3=1$, I think that there is no solution, because in the last row there will be a 0=-1, right?
$endgroup$
– davidllerenav
Jan 26 at 0:43




$begingroup$
So I just need to give random values to$ b_1$,$ b_2$ and $b_3$? And if $b_1=b_2$ and $b_3=1$, I think that there is no solution, because in the last row there will be a 0=-1, right?
$endgroup$
– davidllerenav
Jan 26 at 0:43












$begingroup$
Exactly! The problem is asking you to find some values of $b$ where there is a solution, and some where there isn't.
$endgroup$
– tch
Jan 26 at 0:46




$begingroup$
Exactly! The problem is asking you to find some values of $b$ where there is a solution, and some where there isn't.
$endgroup$
– tch
Jan 26 at 0:46










3 Answers
3






active

oldest

votes


















1












$begingroup$

After you apply your row reduction, look at the last row of your matrix.



It shows you that



$0u + 0v + 0 w = b_1 - b_3 + b_2$



This is only possible if $b_1 - b_3 + b_2 = 0$



So if you choose $b_1, b_2, b_3$ that satisfy this condition, your matrix has a solution (in this case infinite), otherwise it has no solutions.



for example if $b = (2, 5, 7)$ as you suggested, then it is solvable because $2 - 7 + 5 = 0$



Since you are new to linear algebra I did not mention determinants, dimensions, ranks etc. You don't need them in this case, just recall how a matrix represents a system of equations.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! After reading you answer I was able to find the solution.
    $endgroup$
    – davidllerenav
    Jan 27 at 0:34



















1












$begingroup$

Hint:



Use the criterion for the system to be (not) solvable:




$Ax=b$ has a solution if and only if $operatorname{rank}A$ is the same as the rank of the augmented matrix $(A|b)$.




So, as $operatorname{rank}A=2$, you have to find $b$ such that $(A|b)$ has the maximum rank ($3$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry, what does $rank$ mean?
    $endgroup$
    – davidllerenav
    Jan 26 at 0:41










  • $begingroup$
    The number of independent rows or columns. It can be computed via row reduction or with the maximum size of a non-zero subdeterminant in the matrix.
    $endgroup$
    – Bernard
    Jan 26 at 0:45



















1












$begingroup$

You can compute
$$det left(begin{matrix}
1 & 1 & 1 \
2 & 0 & 3 \
3 & 1 & 4
end{matrix}
right) = 9+2-3-8=0$$
which means the three vectors $(1,2,3), (1,0,1), (1,3,4)$ are linearly dependent. However the first two are linearly independent, so they generate a plane in $mathbb{R}^3$ which contains $(1,3,4)$. The plane is
$$t(1,2,3)+s(1,0,1) = (t+s,2t,3t+s)$$
which you can write in Cartesian coordinates as follows: define $x,y,z$ so that
$$begin{cases}
x = t+s \
y = 2t \
z = 3t+s.
end{cases}
$$
Hence $x+y-z=0$, and this is the Cartesian equation of your plane. So every $b=(x,y,z)$ satisfying this equation works. Notice that your $b = (2,5,7)$ works, and also $b=(0,0,0)$ is a trivial choice - which just means that $(1,3,4)$ can be expressed as a linear combination of the first two vectors, as we know. The $b$'s which do not lie on this plane do not work, e.g. $b=(2,5,8)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I didn't understand that well what you said at the beginning, but I understood that, I could choose any other $b$ that is a multiple of $b=(2,5,7)$ then it would lie on the same plane, right?
    $endgroup$
    – davidllerenav
    Jan 26 at 1:13










  • $begingroup$
    Right. My point at the beginning is essentially that $(1,3,4)$ can be expressed as $alpha(1,2,3)+beta(1,0,1)$, so your starting equation can be rewritten as $b = c(1,2,3)+d(1,0,1)$. Then $b$ lies in the plane generated by $(1,2,3)$ and $(1,0,1)$.
    $endgroup$
    – Gibbs
    Jan 26 at 1:14












  • $begingroup$
    Ok, I get it. How did you realize that the three vectors where linearly dependent?
    $endgroup$
    – davidllerenav
    Jan 26 at 2:35










  • $begingroup$
    I just tried to compute the determinant of the matrix they form, as I showed. It turns out to be zero, so the three vectors are linearly dependent.
    $endgroup$
    – Gibbs
    Jan 26 at 10:08











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

After you apply your row reduction, look at the last row of your matrix.



It shows you that



$0u + 0v + 0 w = b_1 - b_3 + b_2$



This is only possible if $b_1 - b_3 + b_2 = 0$



So if you choose $b_1, b_2, b_3$ that satisfy this condition, your matrix has a solution (in this case infinite), otherwise it has no solutions.



for example if $b = (2, 5, 7)$ as you suggested, then it is solvable because $2 - 7 + 5 = 0$



Since you are new to linear algebra I did not mention determinants, dimensions, ranks etc. You don't need them in this case, just recall how a matrix represents a system of equations.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! After reading you answer I was able to find the solution.
    $endgroup$
    – davidllerenav
    Jan 27 at 0:34
















1












$begingroup$

After you apply your row reduction, look at the last row of your matrix.



It shows you that



$0u + 0v + 0 w = b_1 - b_3 + b_2$



This is only possible if $b_1 - b_3 + b_2 = 0$



So if you choose $b_1, b_2, b_3$ that satisfy this condition, your matrix has a solution (in this case infinite), otherwise it has no solutions.



for example if $b = (2, 5, 7)$ as you suggested, then it is solvable because $2 - 7 + 5 = 0$



Since you are new to linear algebra I did not mention determinants, dimensions, ranks etc. You don't need them in this case, just recall how a matrix represents a system of equations.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! After reading you answer I was able to find the solution.
    $endgroup$
    – davidllerenav
    Jan 27 at 0:34














1












1








1





$begingroup$

After you apply your row reduction, look at the last row of your matrix.



It shows you that



$0u + 0v + 0 w = b_1 - b_3 + b_2$



This is only possible if $b_1 - b_3 + b_2 = 0$



So if you choose $b_1, b_2, b_3$ that satisfy this condition, your matrix has a solution (in this case infinite), otherwise it has no solutions.



for example if $b = (2, 5, 7)$ as you suggested, then it is solvable because $2 - 7 + 5 = 0$



Since you are new to linear algebra I did not mention determinants, dimensions, ranks etc. You don't need them in this case, just recall how a matrix represents a system of equations.






share|cite|improve this answer











$endgroup$



After you apply your row reduction, look at the last row of your matrix.



It shows you that



$0u + 0v + 0 w = b_1 - b_3 + b_2$



This is only possible if $b_1 - b_3 + b_2 = 0$



So if you choose $b_1, b_2, b_3$ that satisfy this condition, your matrix has a solution (in this case infinite), otherwise it has no solutions.



for example if $b = (2, 5, 7)$ as you suggested, then it is solvable because $2 - 7 + 5 = 0$



Since you are new to linear algebra I did not mention determinants, dimensions, ranks etc. You don't need them in this case, just recall how a matrix represents a system of equations.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 26 at 22:11

























answered Jan 26 at 22:05









kkshourykkshoury

812




812












  • $begingroup$
    Thanks! After reading you answer I was able to find the solution.
    $endgroup$
    – davidllerenav
    Jan 27 at 0:34


















  • $begingroup$
    Thanks! After reading you answer I was able to find the solution.
    $endgroup$
    – davidllerenav
    Jan 27 at 0:34
















$begingroup$
Thanks! After reading you answer I was able to find the solution.
$endgroup$
– davidllerenav
Jan 27 at 0:34




$begingroup$
Thanks! After reading you answer I was able to find the solution.
$endgroup$
– davidllerenav
Jan 27 at 0:34











1












$begingroup$

Hint:



Use the criterion for the system to be (not) solvable:




$Ax=b$ has a solution if and only if $operatorname{rank}A$ is the same as the rank of the augmented matrix $(A|b)$.




So, as $operatorname{rank}A=2$, you have to find $b$ such that $(A|b)$ has the maximum rank ($3$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry, what does $rank$ mean?
    $endgroup$
    – davidllerenav
    Jan 26 at 0:41










  • $begingroup$
    The number of independent rows or columns. It can be computed via row reduction or with the maximum size of a non-zero subdeterminant in the matrix.
    $endgroup$
    – Bernard
    Jan 26 at 0:45
















1












$begingroup$

Hint:



Use the criterion for the system to be (not) solvable:




$Ax=b$ has a solution if and only if $operatorname{rank}A$ is the same as the rank of the augmented matrix $(A|b)$.




So, as $operatorname{rank}A=2$, you have to find $b$ such that $(A|b)$ has the maximum rank ($3$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry, what does $rank$ mean?
    $endgroup$
    – davidllerenav
    Jan 26 at 0:41










  • $begingroup$
    The number of independent rows or columns. It can be computed via row reduction or with the maximum size of a non-zero subdeterminant in the matrix.
    $endgroup$
    – Bernard
    Jan 26 at 0:45














1












1








1





$begingroup$

Hint:



Use the criterion for the system to be (not) solvable:




$Ax=b$ has a solution if and only if $operatorname{rank}A$ is the same as the rank of the augmented matrix $(A|b)$.




So, as $operatorname{rank}A=2$, you have to find $b$ such that $(A|b)$ has the maximum rank ($3$).






share|cite|improve this answer









$endgroup$



Hint:



Use the criterion for the system to be (not) solvable:




$Ax=b$ has a solution if and only if $operatorname{rank}A$ is the same as the rank of the augmented matrix $(A|b)$.




So, as $operatorname{rank}A=2$, you have to find $b$ such that $(A|b)$ has the maximum rank ($3$).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 26 at 0:35









BernardBernard

122k741116




122k741116












  • $begingroup$
    Sorry, what does $rank$ mean?
    $endgroup$
    – davidllerenav
    Jan 26 at 0:41










  • $begingroup$
    The number of independent rows or columns. It can be computed via row reduction or with the maximum size of a non-zero subdeterminant in the matrix.
    $endgroup$
    – Bernard
    Jan 26 at 0:45


















  • $begingroup$
    Sorry, what does $rank$ mean?
    $endgroup$
    – davidllerenav
    Jan 26 at 0:41










  • $begingroup$
    The number of independent rows or columns. It can be computed via row reduction or with the maximum size of a non-zero subdeterminant in the matrix.
    $endgroup$
    – Bernard
    Jan 26 at 0:45
















$begingroup$
Sorry, what does $rank$ mean?
$endgroup$
– davidllerenav
Jan 26 at 0:41




$begingroup$
Sorry, what does $rank$ mean?
$endgroup$
– davidllerenav
Jan 26 at 0:41












$begingroup$
The number of independent rows or columns. It can be computed via row reduction or with the maximum size of a non-zero subdeterminant in the matrix.
$endgroup$
– Bernard
Jan 26 at 0:45




$begingroup$
The number of independent rows or columns. It can be computed via row reduction or with the maximum size of a non-zero subdeterminant in the matrix.
$endgroup$
– Bernard
Jan 26 at 0:45











1












$begingroup$

You can compute
$$det left(begin{matrix}
1 & 1 & 1 \
2 & 0 & 3 \
3 & 1 & 4
end{matrix}
right) = 9+2-3-8=0$$
which means the three vectors $(1,2,3), (1,0,1), (1,3,4)$ are linearly dependent. However the first two are linearly independent, so they generate a plane in $mathbb{R}^3$ which contains $(1,3,4)$. The plane is
$$t(1,2,3)+s(1,0,1) = (t+s,2t,3t+s)$$
which you can write in Cartesian coordinates as follows: define $x,y,z$ so that
$$begin{cases}
x = t+s \
y = 2t \
z = 3t+s.
end{cases}
$$
Hence $x+y-z=0$, and this is the Cartesian equation of your plane. So every $b=(x,y,z)$ satisfying this equation works. Notice that your $b = (2,5,7)$ works, and also $b=(0,0,0)$ is a trivial choice - which just means that $(1,3,4)$ can be expressed as a linear combination of the first two vectors, as we know. The $b$'s which do not lie on this plane do not work, e.g. $b=(2,5,8)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I didn't understand that well what you said at the beginning, but I understood that, I could choose any other $b$ that is a multiple of $b=(2,5,7)$ then it would lie on the same plane, right?
    $endgroup$
    – davidllerenav
    Jan 26 at 1:13










  • $begingroup$
    Right. My point at the beginning is essentially that $(1,3,4)$ can be expressed as $alpha(1,2,3)+beta(1,0,1)$, so your starting equation can be rewritten as $b = c(1,2,3)+d(1,0,1)$. Then $b$ lies in the plane generated by $(1,2,3)$ and $(1,0,1)$.
    $endgroup$
    – Gibbs
    Jan 26 at 1:14












  • $begingroup$
    Ok, I get it. How did you realize that the three vectors where linearly dependent?
    $endgroup$
    – davidllerenav
    Jan 26 at 2:35










  • $begingroup$
    I just tried to compute the determinant of the matrix they form, as I showed. It turns out to be zero, so the three vectors are linearly dependent.
    $endgroup$
    – Gibbs
    Jan 26 at 10:08
















1












$begingroup$

You can compute
$$det left(begin{matrix}
1 & 1 & 1 \
2 & 0 & 3 \
3 & 1 & 4
end{matrix}
right) = 9+2-3-8=0$$
which means the three vectors $(1,2,3), (1,0,1), (1,3,4)$ are linearly dependent. However the first two are linearly independent, so they generate a plane in $mathbb{R}^3$ which contains $(1,3,4)$. The plane is
$$t(1,2,3)+s(1,0,1) = (t+s,2t,3t+s)$$
which you can write in Cartesian coordinates as follows: define $x,y,z$ so that
$$begin{cases}
x = t+s \
y = 2t \
z = 3t+s.
end{cases}
$$
Hence $x+y-z=0$, and this is the Cartesian equation of your plane. So every $b=(x,y,z)$ satisfying this equation works. Notice that your $b = (2,5,7)$ works, and also $b=(0,0,0)$ is a trivial choice - which just means that $(1,3,4)$ can be expressed as a linear combination of the first two vectors, as we know. The $b$'s which do not lie on this plane do not work, e.g. $b=(2,5,8)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I didn't understand that well what you said at the beginning, but I understood that, I could choose any other $b$ that is a multiple of $b=(2,5,7)$ then it would lie on the same plane, right?
    $endgroup$
    – davidllerenav
    Jan 26 at 1:13










  • $begingroup$
    Right. My point at the beginning is essentially that $(1,3,4)$ can be expressed as $alpha(1,2,3)+beta(1,0,1)$, so your starting equation can be rewritten as $b = c(1,2,3)+d(1,0,1)$. Then $b$ lies in the plane generated by $(1,2,3)$ and $(1,0,1)$.
    $endgroup$
    – Gibbs
    Jan 26 at 1:14












  • $begingroup$
    Ok, I get it. How did you realize that the three vectors where linearly dependent?
    $endgroup$
    – davidllerenav
    Jan 26 at 2:35










  • $begingroup$
    I just tried to compute the determinant of the matrix they form, as I showed. It turns out to be zero, so the three vectors are linearly dependent.
    $endgroup$
    – Gibbs
    Jan 26 at 10:08














1












1








1





$begingroup$

You can compute
$$det left(begin{matrix}
1 & 1 & 1 \
2 & 0 & 3 \
3 & 1 & 4
end{matrix}
right) = 9+2-3-8=0$$
which means the three vectors $(1,2,3), (1,0,1), (1,3,4)$ are linearly dependent. However the first two are linearly independent, so they generate a plane in $mathbb{R}^3$ which contains $(1,3,4)$. The plane is
$$t(1,2,3)+s(1,0,1) = (t+s,2t,3t+s)$$
which you can write in Cartesian coordinates as follows: define $x,y,z$ so that
$$begin{cases}
x = t+s \
y = 2t \
z = 3t+s.
end{cases}
$$
Hence $x+y-z=0$, and this is the Cartesian equation of your plane. So every $b=(x,y,z)$ satisfying this equation works. Notice that your $b = (2,5,7)$ works, and also $b=(0,0,0)$ is a trivial choice - which just means that $(1,3,4)$ can be expressed as a linear combination of the first two vectors, as we know. The $b$'s which do not lie on this plane do not work, e.g. $b=(2,5,8)$.






share|cite|improve this answer









$endgroup$



You can compute
$$det left(begin{matrix}
1 & 1 & 1 \
2 & 0 & 3 \
3 & 1 & 4
end{matrix}
right) = 9+2-3-8=0$$
which means the three vectors $(1,2,3), (1,0,1), (1,3,4)$ are linearly dependent. However the first two are linearly independent, so they generate a plane in $mathbb{R}^3$ which contains $(1,3,4)$. The plane is
$$t(1,2,3)+s(1,0,1) = (t+s,2t,3t+s)$$
which you can write in Cartesian coordinates as follows: define $x,y,z$ so that
$$begin{cases}
x = t+s \
y = 2t \
z = 3t+s.
end{cases}
$$
Hence $x+y-z=0$, and this is the Cartesian equation of your plane. So every $b=(x,y,z)$ satisfying this equation works. Notice that your $b = (2,5,7)$ works, and also $b=(0,0,0)$ is a trivial choice - which just means that $(1,3,4)$ can be expressed as a linear combination of the first two vectors, as we know. The $b$'s which do not lie on this plane do not work, e.g. $b=(2,5,8)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 26 at 0:40









GibbsGibbs

5,4103827




5,4103827












  • $begingroup$
    I didn't understand that well what you said at the beginning, but I understood that, I could choose any other $b$ that is a multiple of $b=(2,5,7)$ then it would lie on the same plane, right?
    $endgroup$
    – davidllerenav
    Jan 26 at 1:13










  • $begingroup$
    Right. My point at the beginning is essentially that $(1,3,4)$ can be expressed as $alpha(1,2,3)+beta(1,0,1)$, so your starting equation can be rewritten as $b = c(1,2,3)+d(1,0,1)$. Then $b$ lies in the plane generated by $(1,2,3)$ and $(1,0,1)$.
    $endgroup$
    – Gibbs
    Jan 26 at 1:14












  • $begingroup$
    Ok, I get it. How did you realize that the three vectors where linearly dependent?
    $endgroup$
    – davidllerenav
    Jan 26 at 2:35










  • $begingroup$
    I just tried to compute the determinant of the matrix they form, as I showed. It turns out to be zero, so the three vectors are linearly dependent.
    $endgroup$
    – Gibbs
    Jan 26 at 10:08


















  • $begingroup$
    I didn't understand that well what you said at the beginning, but I understood that, I could choose any other $b$ that is a multiple of $b=(2,5,7)$ then it would lie on the same plane, right?
    $endgroup$
    – davidllerenav
    Jan 26 at 1:13










  • $begingroup$
    Right. My point at the beginning is essentially that $(1,3,4)$ can be expressed as $alpha(1,2,3)+beta(1,0,1)$, so your starting equation can be rewritten as $b = c(1,2,3)+d(1,0,1)$. Then $b$ lies in the plane generated by $(1,2,3)$ and $(1,0,1)$.
    $endgroup$
    – Gibbs
    Jan 26 at 1:14












  • $begingroup$
    Ok, I get it. How did you realize that the three vectors where linearly dependent?
    $endgroup$
    – davidllerenav
    Jan 26 at 2:35










  • $begingroup$
    I just tried to compute the determinant of the matrix they form, as I showed. It turns out to be zero, so the three vectors are linearly dependent.
    $endgroup$
    – Gibbs
    Jan 26 at 10:08
















$begingroup$
I didn't understand that well what you said at the beginning, but I understood that, I could choose any other $b$ that is a multiple of $b=(2,5,7)$ then it would lie on the same plane, right?
$endgroup$
– davidllerenav
Jan 26 at 1:13




$begingroup$
I didn't understand that well what you said at the beginning, but I understood that, I could choose any other $b$ that is a multiple of $b=(2,5,7)$ then it would lie on the same plane, right?
$endgroup$
– davidllerenav
Jan 26 at 1:13












$begingroup$
Right. My point at the beginning is essentially that $(1,3,4)$ can be expressed as $alpha(1,2,3)+beta(1,0,1)$, so your starting equation can be rewritten as $b = c(1,2,3)+d(1,0,1)$. Then $b$ lies in the plane generated by $(1,2,3)$ and $(1,0,1)$.
$endgroup$
– Gibbs
Jan 26 at 1:14






$begingroup$
Right. My point at the beginning is essentially that $(1,3,4)$ can be expressed as $alpha(1,2,3)+beta(1,0,1)$, so your starting equation can be rewritten as $b = c(1,2,3)+d(1,0,1)$. Then $b$ lies in the plane generated by $(1,2,3)$ and $(1,0,1)$.
$endgroup$
– Gibbs
Jan 26 at 1:14














$begingroup$
Ok, I get it. How did you realize that the three vectors where linearly dependent?
$endgroup$
– davidllerenav
Jan 26 at 2:35




$begingroup$
Ok, I get it. How did you realize that the three vectors where linearly dependent?
$endgroup$
– davidllerenav
Jan 26 at 2:35












$begingroup$
I just tried to compute the determinant of the matrix they form, as I showed. It turns out to be zero, so the three vectors are linearly dependent.
$endgroup$
– Gibbs
Jan 26 at 10:08




$begingroup$
I just tried to compute the determinant of the matrix they form, as I showed. It turns out to be zero, so the three vectors are linearly dependent.
$endgroup$
– Gibbs
Jan 26 at 10:08


















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