What are the values of $k$ that males $x_1,x_2$ solutions to the linear system $AX=B$.












3












$begingroup$


Supose that $x_1$ and $x_2$ are solution to the linear system $ AX=B $ ,
where $B$ is not equal zero then $3x_1-kx_2$ is a solution also if $k = ?$



How to find the value of $A$ ?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Assuming $Aneq0$, $Ax_1=B=Ax_2implies A(x_1-x_2)=0$. So $x_1=x_2$.
    $endgroup$
    – Yadati Kiran
    Dec 17 '18 at 6:49








  • 1




    $begingroup$
    @YadatiKiran: I suspect $A$ is supposed to be a matrix, in which case $A(x_1-x_2) = 0$ does not necessarily imply $x_1 = x_2$. Of course, this isn't clear from the question.
    $endgroup$
    – JimmyK4542
    Dec 17 '18 at 7:22
















3












$begingroup$


Supose that $x_1$ and $x_2$ are solution to the linear system $ AX=B $ ,
where $B$ is not equal zero then $3x_1-kx_2$ is a solution also if $k = ?$



How to find the value of $A$ ?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Assuming $Aneq0$, $Ax_1=B=Ax_2implies A(x_1-x_2)=0$. So $x_1=x_2$.
    $endgroup$
    – Yadati Kiran
    Dec 17 '18 at 6:49








  • 1




    $begingroup$
    @YadatiKiran: I suspect $A$ is supposed to be a matrix, in which case $A(x_1-x_2) = 0$ does not necessarily imply $x_1 = x_2$. Of course, this isn't clear from the question.
    $endgroup$
    – JimmyK4542
    Dec 17 '18 at 7:22














3












3








3





$begingroup$


Supose that $x_1$ and $x_2$ are solution to the linear system $ AX=B $ ,
where $B$ is not equal zero then $3x_1-kx_2$ is a solution also if $k = ?$



How to find the value of $A$ ?










share|cite|improve this question











$endgroup$




Supose that $x_1$ and $x_2$ are solution to the linear system $ AX=B $ ,
where $B$ is not equal zero then $3x_1-kx_2$ is a solution also if $k = ?$



How to find the value of $A$ ?







linear-algebra systems-of-equations






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share|cite|improve this question













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share|cite|improve this question








edited Dec 17 '18 at 5:57









dmtri

1,5082521




1,5082521










asked Dec 17 '18 at 5:53









faisalfaisal

274




274












  • $begingroup$
    Assuming $Aneq0$, $Ax_1=B=Ax_2implies A(x_1-x_2)=0$. So $x_1=x_2$.
    $endgroup$
    – Yadati Kiran
    Dec 17 '18 at 6:49








  • 1




    $begingroup$
    @YadatiKiran: I suspect $A$ is supposed to be a matrix, in which case $A(x_1-x_2) = 0$ does not necessarily imply $x_1 = x_2$. Of course, this isn't clear from the question.
    $endgroup$
    – JimmyK4542
    Dec 17 '18 at 7:22


















  • $begingroup$
    Assuming $Aneq0$, $Ax_1=B=Ax_2implies A(x_1-x_2)=0$. So $x_1=x_2$.
    $endgroup$
    – Yadati Kiran
    Dec 17 '18 at 6:49








  • 1




    $begingroup$
    @YadatiKiran: I suspect $A$ is supposed to be a matrix, in which case $A(x_1-x_2) = 0$ does not necessarily imply $x_1 = x_2$. Of course, this isn't clear from the question.
    $endgroup$
    – JimmyK4542
    Dec 17 '18 at 7:22
















$begingroup$
Assuming $Aneq0$, $Ax_1=B=Ax_2implies A(x_1-x_2)=0$. So $x_1=x_2$.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 6:49






$begingroup$
Assuming $Aneq0$, $Ax_1=B=Ax_2implies A(x_1-x_2)=0$. So $x_1=x_2$.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 6:49






1




1




$begingroup$
@YadatiKiran: I suspect $A$ is supposed to be a matrix, in which case $A(x_1-x_2) = 0$ does not necessarily imply $x_1 = x_2$. Of course, this isn't clear from the question.
$endgroup$
– JimmyK4542
Dec 17 '18 at 7:22




$begingroup$
@YadatiKiran: I suspect $A$ is supposed to be a matrix, in which case $A(x_1-x_2) = 0$ does not necessarily imply $x_1 = x_2$. Of course, this isn't clear from the question.
$endgroup$
– JimmyK4542
Dec 17 '18 at 7:22










1 Answer
1






active

oldest

votes


















2












$begingroup$

Hints: The fact that $x_1$ and $x_2$ are solutions to $AX = B$ means that $Ax_1 = B$ and $Ax_2 = B$.



In order for $3x_1-kx_2$ to be a solution to $AX=B$, we would need $A(3x_1-kx_2) = B$.



We can use linearity to simplify $A(3x_1-kx_2) = 3Ax_1-kAx_2 = cdots$.



Can you figure out what value of $k$ makes $A(3x_1-kx_2) = B$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    do you mean that $3x_1−kx_2$ must be equal 0 ?
    $endgroup$
    – faisal
    Dec 17 '18 at 7:21












  • $begingroup$
    No I don't mean that at all. Note that if $y$ is a solution to the equation $AX = B$, then $Ay = B$. So if $3x_1-kx_2$ is a solution to the equation $AX = B$, then $A(3x_1-kx_2) = B$.
    $endgroup$
    – JimmyK4542
    Dec 17 '18 at 7:24










  • $begingroup$
    then what is the values of k that makes $x_1 and x_2$ solutions to $AX=B$ could you explain this more please?
    $endgroup$
    – faisal
    Dec 17 '18 at 7:26












  • $begingroup$
    I think the problem given by the first two lines of the lines of the body of your post is saying that $x_1$ and $x_2$ are solutions to $AX = B$, and asking you to find a value of $k$ such that $3x_1-kx_2$ is also a solution to $AX = B$. If that's not the problem you are trying to solve, you might want to edit your question.
    $endgroup$
    – JimmyK4542
    Dec 17 '18 at 7:29










  • $begingroup$
    @JimmyK4542: I do admit that the later part of my comment is not true as $A$ can be singular. How is $A$ linear then? Since you said in the comment "I suspect $A$ is supposed to be a matrix" (just curious).
    $endgroup$
    – Yadati Kiran
    Dec 17 '18 at 7:36













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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

Hints: The fact that $x_1$ and $x_2$ are solutions to $AX = B$ means that $Ax_1 = B$ and $Ax_2 = B$.



In order for $3x_1-kx_2$ to be a solution to $AX=B$, we would need $A(3x_1-kx_2) = B$.



We can use linearity to simplify $A(3x_1-kx_2) = 3Ax_1-kAx_2 = cdots$.



Can you figure out what value of $k$ makes $A(3x_1-kx_2) = B$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    do you mean that $3x_1−kx_2$ must be equal 0 ?
    $endgroup$
    – faisal
    Dec 17 '18 at 7:21












  • $begingroup$
    No I don't mean that at all. Note that if $y$ is a solution to the equation $AX = B$, then $Ay = B$. So if $3x_1-kx_2$ is a solution to the equation $AX = B$, then $A(3x_1-kx_2) = B$.
    $endgroup$
    – JimmyK4542
    Dec 17 '18 at 7:24










  • $begingroup$
    then what is the values of k that makes $x_1 and x_2$ solutions to $AX=B$ could you explain this more please?
    $endgroup$
    – faisal
    Dec 17 '18 at 7:26












  • $begingroup$
    I think the problem given by the first two lines of the lines of the body of your post is saying that $x_1$ and $x_2$ are solutions to $AX = B$, and asking you to find a value of $k$ such that $3x_1-kx_2$ is also a solution to $AX = B$. If that's not the problem you are trying to solve, you might want to edit your question.
    $endgroup$
    – JimmyK4542
    Dec 17 '18 at 7:29










  • $begingroup$
    @JimmyK4542: I do admit that the later part of my comment is not true as $A$ can be singular. How is $A$ linear then? Since you said in the comment "I suspect $A$ is supposed to be a matrix" (just curious).
    $endgroup$
    – Yadati Kiran
    Dec 17 '18 at 7:36


















2












$begingroup$

Hints: The fact that $x_1$ and $x_2$ are solutions to $AX = B$ means that $Ax_1 = B$ and $Ax_2 = B$.



In order for $3x_1-kx_2$ to be a solution to $AX=B$, we would need $A(3x_1-kx_2) = B$.



We can use linearity to simplify $A(3x_1-kx_2) = 3Ax_1-kAx_2 = cdots$.



Can you figure out what value of $k$ makes $A(3x_1-kx_2) = B$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    do you mean that $3x_1−kx_2$ must be equal 0 ?
    $endgroup$
    – faisal
    Dec 17 '18 at 7:21












  • $begingroup$
    No I don't mean that at all. Note that if $y$ is a solution to the equation $AX = B$, then $Ay = B$. So if $3x_1-kx_2$ is a solution to the equation $AX = B$, then $A(3x_1-kx_2) = B$.
    $endgroup$
    – JimmyK4542
    Dec 17 '18 at 7:24










  • $begingroup$
    then what is the values of k that makes $x_1 and x_2$ solutions to $AX=B$ could you explain this more please?
    $endgroup$
    – faisal
    Dec 17 '18 at 7:26












  • $begingroup$
    I think the problem given by the first two lines of the lines of the body of your post is saying that $x_1$ and $x_2$ are solutions to $AX = B$, and asking you to find a value of $k$ such that $3x_1-kx_2$ is also a solution to $AX = B$. If that's not the problem you are trying to solve, you might want to edit your question.
    $endgroup$
    – JimmyK4542
    Dec 17 '18 at 7:29










  • $begingroup$
    @JimmyK4542: I do admit that the later part of my comment is not true as $A$ can be singular. How is $A$ linear then? Since you said in the comment "I suspect $A$ is supposed to be a matrix" (just curious).
    $endgroup$
    – Yadati Kiran
    Dec 17 '18 at 7:36
















2












2








2





$begingroup$

Hints: The fact that $x_1$ and $x_2$ are solutions to $AX = B$ means that $Ax_1 = B$ and $Ax_2 = B$.



In order for $3x_1-kx_2$ to be a solution to $AX=B$, we would need $A(3x_1-kx_2) = B$.



We can use linearity to simplify $A(3x_1-kx_2) = 3Ax_1-kAx_2 = cdots$.



Can you figure out what value of $k$ makes $A(3x_1-kx_2) = B$?






share|cite|improve this answer









$endgroup$



Hints: The fact that $x_1$ and $x_2$ are solutions to $AX = B$ means that $Ax_1 = B$ and $Ax_2 = B$.



In order for $3x_1-kx_2$ to be a solution to $AX=B$, we would need $A(3x_1-kx_2) = B$.



We can use linearity to simplify $A(3x_1-kx_2) = 3Ax_1-kAx_2 = cdots$.



Can you figure out what value of $k$ makes $A(3x_1-kx_2) = B$?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 17 '18 at 7:18









JimmyK4542JimmyK4542

41.1k245107




41.1k245107












  • $begingroup$
    do you mean that $3x_1−kx_2$ must be equal 0 ?
    $endgroup$
    – faisal
    Dec 17 '18 at 7:21












  • $begingroup$
    No I don't mean that at all. Note that if $y$ is a solution to the equation $AX = B$, then $Ay = B$. So if $3x_1-kx_2$ is a solution to the equation $AX = B$, then $A(3x_1-kx_2) = B$.
    $endgroup$
    – JimmyK4542
    Dec 17 '18 at 7:24










  • $begingroup$
    then what is the values of k that makes $x_1 and x_2$ solutions to $AX=B$ could you explain this more please?
    $endgroup$
    – faisal
    Dec 17 '18 at 7:26












  • $begingroup$
    I think the problem given by the first two lines of the lines of the body of your post is saying that $x_1$ and $x_2$ are solutions to $AX = B$, and asking you to find a value of $k$ such that $3x_1-kx_2$ is also a solution to $AX = B$. If that's not the problem you are trying to solve, you might want to edit your question.
    $endgroup$
    – JimmyK4542
    Dec 17 '18 at 7:29










  • $begingroup$
    @JimmyK4542: I do admit that the later part of my comment is not true as $A$ can be singular. How is $A$ linear then? Since you said in the comment "I suspect $A$ is supposed to be a matrix" (just curious).
    $endgroup$
    – Yadati Kiran
    Dec 17 '18 at 7:36




















  • $begingroup$
    do you mean that $3x_1−kx_2$ must be equal 0 ?
    $endgroup$
    – faisal
    Dec 17 '18 at 7:21












  • $begingroup$
    No I don't mean that at all. Note that if $y$ is a solution to the equation $AX = B$, then $Ay = B$. So if $3x_1-kx_2$ is a solution to the equation $AX = B$, then $A(3x_1-kx_2) = B$.
    $endgroup$
    – JimmyK4542
    Dec 17 '18 at 7:24










  • $begingroup$
    then what is the values of k that makes $x_1 and x_2$ solutions to $AX=B$ could you explain this more please?
    $endgroup$
    – faisal
    Dec 17 '18 at 7:26












  • $begingroup$
    I think the problem given by the first two lines of the lines of the body of your post is saying that $x_1$ and $x_2$ are solutions to $AX = B$, and asking you to find a value of $k$ such that $3x_1-kx_2$ is also a solution to $AX = B$. If that's not the problem you are trying to solve, you might want to edit your question.
    $endgroup$
    – JimmyK4542
    Dec 17 '18 at 7:29










  • $begingroup$
    @JimmyK4542: I do admit that the later part of my comment is not true as $A$ can be singular. How is $A$ linear then? Since you said in the comment "I suspect $A$ is supposed to be a matrix" (just curious).
    $endgroup$
    – Yadati Kiran
    Dec 17 '18 at 7:36


















$begingroup$
do you mean that $3x_1−kx_2$ must be equal 0 ?
$endgroup$
– faisal
Dec 17 '18 at 7:21






$begingroup$
do you mean that $3x_1−kx_2$ must be equal 0 ?
$endgroup$
– faisal
Dec 17 '18 at 7:21














$begingroup$
No I don't mean that at all. Note that if $y$ is a solution to the equation $AX = B$, then $Ay = B$. So if $3x_1-kx_2$ is a solution to the equation $AX = B$, then $A(3x_1-kx_2) = B$.
$endgroup$
– JimmyK4542
Dec 17 '18 at 7:24




$begingroup$
No I don't mean that at all. Note that if $y$ is a solution to the equation $AX = B$, then $Ay = B$. So if $3x_1-kx_2$ is a solution to the equation $AX = B$, then $A(3x_1-kx_2) = B$.
$endgroup$
– JimmyK4542
Dec 17 '18 at 7:24












$begingroup$
then what is the values of k that makes $x_1 and x_2$ solutions to $AX=B$ could you explain this more please?
$endgroup$
– faisal
Dec 17 '18 at 7:26






$begingroup$
then what is the values of k that makes $x_1 and x_2$ solutions to $AX=B$ could you explain this more please?
$endgroup$
– faisal
Dec 17 '18 at 7:26














$begingroup$
I think the problem given by the first two lines of the lines of the body of your post is saying that $x_1$ and $x_2$ are solutions to $AX = B$, and asking you to find a value of $k$ such that $3x_1-kx_2$ is also a solution to $AX = B$. If that's not the problem you are trying to solve, you might want to edit your question.
$endgroup$
– JimmyK4542
Dec 17 '18 at 7:29




$begingroup$
I think the problem given by the first two lines of the lines of the body of your post is saying that $x_1$ and $x_2$ are solutions to $AX = B$, and asking you to find a value of $k$ such that $3x_1-kx_2$ is also a solution to $AX = B$. If that's not the problem you are trying to solve, you might want to edit your question.
$endgroup$
– JimmyK4542
Dec 17 '18 at 7:29












$begingroup$
@JimmyK4542: I do admit that the later part of my comment is not true as $A$ can be singular. How is $A$ linear then? Since you said in the comment "I suspect $A$ is supposed to be a matrix" (just curious).
$endgroup$
– Yadati Kiran
Dec 17 '18 at 7:36






$begingroup$
@JimmyK4542: I do admit that the later part of my comment is not true as $A$ can be singular. How is $A$ linear then? Since you said in the comment "I suspect $A$ is supposed to be a matrix" (just curious).
$endgroup$
– Yadati Kiran
Dec 17 '18 at 7:36




















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