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 $A$ ?
linear-algebra systems-of-equations
edited Dec 17 '18 at 5:57
dmtri
1,5082521
asked Dec 17 '18 at 5:53
faisalfaisal
274
$endgroup$
add a comment |
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 $A$ ?
linear-algebra systems-of-equations
edited Dec 17 '18 at 5:57
dmtri
1,5082521
asked Dec 17 '18 at 5:53
faisalfaisal
274
$endgroup$
$begingroup$
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
add a comment |
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 $A$ ?
linear-algebra systems-of-equations
edited Dec 17 '18 at 5:57
dmtri
1,5082521
asked Dec 17 '18 at 5:53
faisalfaisal
274
$endgroup$
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
linear-algebra systems-of-equations
edited Dec 17 '18 at 5:57
dmtri
1,5082521
asked Dec 17 '18 at 5:53
faisalfaisal
274
edited Dec 17 '18 at 5:57
dmtri
1,5082521
asked Dec 17 '18 at 5:53
faisalfaisal
274
edited Dec 17 '18 at 5:57
dmtri
1,5082521
edited Dec 17 '18 at 5:57
dmtri
1,5082521
edited Dec 17 '18 at 5:57
dmtri
1,5082521
1,5082521
asked Dec 17 '18 at 5:53
faisalfaisal
274
asked Dec 17 '18 at 5:53
faisalfaisal
274
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
add a comment |
$begingroup$
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
add a comment |
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 $A(3x_1-kx_2) = B$?
answered Dec 17 '18 at 7:18
JimmyK4542JimmyK4542
41.1k245106
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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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 $A(3x_1-kx_2) = B$?
answered Dec 17 '18 at 7:18
JimmyK4542JimmyK4542
41.1k245106
$endgroup$
$begingroup$
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
add a comment |
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 $A(3x_1-kx_2) = B$?
answered Dec 17 '18 at 7:18
JimmyK4542JimmyK4542
41.1k245106
$endgroup$
$begingroup$
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
add a comment |
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 $A(3x_1-kx_2) = B$?
answered Dec 17 '18 at 7:18
JimmyK4542JimmyK4542
41.1k245106
$endgroup$
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 $A(3x_1-kx_2) = B$?
answered Dec 17 '18 at 7:18
JimmyK4542JimmyK4542
41.1k245106
answered Dec 17 '18 at 7:18
JimmyK4542JimmyK4542
41.1k245106
answered Dec 17 '18 at 7:18
JimmyK4542JimmyK4542
41.1k245106
answered Dec 17 '18 at 7:18
JimmyK4542JimmyK4542
41.1k245106
41.1k245106
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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