obtain matrix $ A$ if $X$ and $b$ given












1












$begingroup$


enter image description here



for the right and detailed answer refer to user9077 answer






linear-algebra matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    OP, is $c$ the lower-left entry in $A$?
    $endgroup$
    – Mike
    Dec 11 '18 at 17:34










  • $begingroup$
    The letter "c" is being used twice -- once as the free parameter in expressing the general solution, and once as the lower-left entry in "c". To clear up confusion, perhaps replace the "c" in the general solution with something like "u".
    $endgroup$
    – John Hughes
    Dec 11 '18 at 18:14
















1












$begingroup$


enter image description here



for the right and detailed answer refer to user9077 answer






linear-algebra matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    OP, is $c$ the lower-left entry in $A$?
    $endgroup$
    – Mike
    Dec 11 '18 at 17:34










  • $begingroup$
    The letter "c" is being used twice -- once as the free parameter in expressing the general solution, and once as the lower-left entry in "c". To clear up confusion, perhaps replace the "c" in the general solution with something like "u".
    $endgroup$
    – John Hughes
    Dec 11 '18 at 18:14














1












1








1





$begingroup$


enter image description here



for the right and detailed answer refer to user9077 answer






linear-algebra matrices







share|cite|improve this question











$endgroup$




enter image description here



for the right and detailed answer refer to user9077 answer






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








edited Dec 12 '18 at 17:43







green life

















asked Dec 11 '18 at 17:17









green lifegreen life

186




186












  • $begingroup$
    OP, is $c$ the lower-left entry in $A$?
    $endgroup$
    – Mike
    Dec 11 '18 at 17:34










  • $begingroup$
    The letter "c" is being used twice -- once as the free parameter in expressing the general solution, and once as the lower-left entry in "c". To clear up confusion, perhaps replace the "c" in the general solution with something like "u".
    $endgroup$
    – John Hughes
    Dec 11 '18 at 18:14


















  • $begingroup$
    OP, is $c$ the lower-left entry in $A$?
    $endgroup$
    – Mike
    Dec 11 '18 at 17:34










  • $begingroup$
    The letter "c" is being used twice -- once as the free parameter in expressing the general solution, and once as the lower-left entry in "c". To clear up confusion, perhaps replace the "c" in the general solution with something like "u".
    $endgroup$
    – John Hughes
    Dec 11 '18 at 18:14
















$begingroup$
OP, is $c$ the lower-left entry in $A$?
$endgroup$
– Mike
Dec 11 '18 at 17:34




$begingroup$
OP, is $c$ the lower-left entry in $A$?
$endgroup$
– Mike
Dec 11 '18 at 17:34












$begingroup$
The letter "c" is being used twice -- once as the free parameter in expressing the general solution, and once as the lower-left entry in "c". To clear up confusion, perhaps replace the "c" in the general solution with something like "u".
$endgroup$
– John Hughes
Dec 11 '18 at 18:14




$begingroup$
The letter "c" is being used twice -- once as the free parameter in expressing the general solution, and once as the lower-left entry in "c". To clear up confusion, perhaps replace the "c" in the general solution with something like "u".
$endgroup$
– John Hughes
Dec 11 '18 at 18:14










3 Answers
3






active

oldest

votes


















1












$begingroup$

Since $begin{pmatrix}1\0end{pmatrix}+cbegin{pmatrix}0\1end{pmatrix}$ are all solutions to $Ax=begin{pmatrix}1\3end{pmatrix}$, then we have
$$
begin{pmatrix}1\3end{pmatrix}=Ax=Abegin{pmatrix}1\0end{pmatrix}+cAbegin{pmatrix}0\1end{pmatrix}.
$$



The above is true for any $c$. This can only happen when $Abegin{pmatrix}0\1end{pmatrix}=begin{pmatrix}0\0end{pmatrix}$. From this we know that the second column of $A$ is $begin{pmatrix}0\0end{pmatrix}$.



We also have $Abegin{pmatrix}1\0end{pmatrix}=begin{pmatrix}1\3end{pmatrix}$. So the first column of $A$ is $begin{pmatrix}1\3end{pmatrix}$. Therefore
$$A=begin{pmatrix}1&0\3&0end{pmatrix}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This does not answer the OP's question, which was "Can you confirm that my solution is right?"
    $endgroup$
    – John Hughes
    Dec 11 '18 at 22:51





















2












$begingroup$

Yes, your work is correct, although you might want to explain why you can write
$$
pmatrix{a & b \ c & d}pmatrix{0 \ 1} = pmatrix{0\0},
$$
even if it seems obvious to you.



Also: the "MathJax" that I typed to produce this answer was exactly this:



$$

pmatrix{a & b \ c & d}pmatrix{0 \ 1} = 

pmatrix{0\0},

$$


You should, if you're going to stick around here, learn how to format things like this.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks for confirming i will clarify that for who want to know
    $endgroup$
    – green life
    Dec 11 '18 at 17:28












  • $begingroup$
    and yes i need to learn how to enter the math codes i will learn it soon i promise
    $endgroup$
    – green life
    Dec 11 '18 at 17:34



















0












$begingroup$

It is basically correct, but you can do it more easily.



If $a_1$ and $a_2$ are the columns of $A$, then you know that
$$
1a_1+ca_2=begin{bmatrix} 1 \ 3 end{bmatrix}
$$
for every $c$; in particular this holds for $c=0$ and $c=1$, so
begin{cases}
a_1=begin{bmatrix} 1 \ 3 end{bmatrix} \[4px]
a_1+a_2=begin{bmatrix} 1 \ 3 end{bmatrix}
end{cases}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This does not answer the OP's question, which was "Can you confirm that my solution is right?"
    $endgroup$
    – John Hughes
    Dec 11 '18 at 22:52










  • $begingroup$
    @JohnHughes The OP's solution is not available unless one goes back in the post history. Anyway, this is a confirmation that the solution is correct.
    $endgroup$
    – egreg
    Dec 11 '18 at 22:59











Your Answer





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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Since $begin{pmatrix}1\0end{pmatrix}+cbegin{pmatrix}0\1end{pmatrix}$ are all solutions to $Ax=begin{pmatrix}1\3end{pmatrix}$, then we have
$$
begin{pmatrix}1\3end{pmatrix}=Ax=Abegin{pmatrix}1\0end{pmatrix}+cAbegin{pmatrix}0\1end{pmatrix}.
$$



The above is true for any $c$. This can only happen when $Abegin{pmatrix}0\1end{pmatrix}=begin{pmatrix}0\0end{pmatrix}$. From this we know that the second column of $A$ is $begin{pmatrix}0\0end{pmatrix}$.



We also have $Abegin{pmatrix}1\0end{pmatrix}=begin{pmatrix}1\3end{pmatrix}$. So the first column of $A$ is $begin{pmatrix}1\3end{pmatrix}$. Therefore
$$A=begin{pmatrix}1&0\3&0end{pmatrix}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This does not answer the OP's question, which was "Can you confirm that my solution is right?"
    $endgroup$
    – John Hughes
    Dec 11 '18 at 22:51


















1












$begingroup$

Since $begin{pmatrix}1\0end{pmatrix}+cbegin{pmatrix}0\1end{pmatrix}$ are all solutions to $Ax=begin{pmatrix}1\3end{pmatrix}$, then we have
$$
begin{pmatrix}1\3end{pmatrix}=Ax=Abegin{pmatrix}1\0end{pmatrix}+cAbegin{pmatrix}0\1end{pmatrix}.
$$



The above is true for any $c$. This can only happen when $Abegin{pmatrix}0\1end{pmatrix}=begin{pmatrix}0\0end{pmatrix}$. From this we know that the second column of $A$ is $begin{pmatrix}0\0end{pmatrix}$.



We also have $Abegin{pmatrix}1\0end{pmatrix}=begin{pmatrix}1\3end{pmatrix}$. So the first column of $A$ is $begin{pmatrix}1\3end{pmatrix}$. Therefore
$$A=begin{pmatrix}1&0\3&0end{pmatrix}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This does not answer the OP's question, which was "Can you confirm that my solution is right?"
    $endgroup$
    – John Hughes
    Dec 11 '18 at 22:51
















1












1








1





$begingroup$

Since $begin{pmatrix}1\0end{pmatrix}+cbegin{pmatrix}0\1end{pmatrix}$ are all solutions to $Ax=begin{pmatrix}1\3end{pmatrix}$, then we have
$$
begin{pmatrix}1\3end{pmatrix}=Ax=Abegin{pmatrix}1\0end{pmatrix}+cAbegin{pmatrix}0\1end{pmatrix}.
$$



The above is true for any $c$. This can only happen when $Abegin{pmatrix}0\1end{pmatrix}=begin{pmatrix}0\0end{pmatrix}$. From this we know that the second column of $A$ is $begin{pmatrix}0\0end{pmatrix}$.



We also have $Abegin{pmatrix}1\0end{pmatrix}=begin{pmatrix}1\3end{pmatrix}$. So the first column of $A$ is $begin{pmatrix}1\3end{pmatrix}$. Therefore
$$A=begin{pmatrix}1&0\3&0end{pmatrix}.$$






share|cite|improve this answer









$endgroup$



Since $begin{pmatrix}1\0end{pmatrix}+cbegin{pmatrix}0\1end{pmatrix}$ are all solutions to $Ax=begin{pmatrix}1\3end{pmatrix}$, then we have
$$
begin{pmatrix}1\3end{pmatrix}=Ax=Abegin{pmatrix}1\0end{pmatrix}+cAbegin{pmatrix}0\1end{pmatrix}.
$$



The above is true for any $c$. This can only happen when $Abegin{pmatrix}0\1end{pmatrix}=begin{pmatrix}0\0end{pmatrix}$. From this we know that the second column of $A$ is $begin{pmatrix}0\0end{pmatrix}$.



We also have $Abegin{pmatrix}1\0end{pmatrix}=begin{pmatrix}1\3end{pmatrix}$. So the first column of $A$ is $begin{pmatrix}1\3end{pmatrix}$. Therefore
$$A=begin{pmatrix}1&0\3&0end{pmatrix}.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 11 '18 at 21:36









user9077user9077

1,239612




1,239612












  • $begingroup$
    This does not answer the OP's question, which was "Can you confirm that my solution is right?"
    $endgroup$
    – John Hughes
    Dec 11 '18 at 22:51




















  • $begingroup$
    This does not answer the OP's question, which was "Can you confirm that my solution is right?"
    $endgroup$
    – John Hughes
    Dec 11 '18 at 22:51


















$begingroup$
This does not answer the OP's question, which was "Can you confirm that my solution is right?"
$endgroup$
– John Hughes
Dec 11 '18 at 22:51






$begingroup$
This does not answer the OP's question, which was "Can you confirm that my solution is right?"
$endgroup$
– John Hughes
Dec 11 '18 at 22:51













2












$begingroup$

Yes, your work is correct, although you might want to explain why you can write
$$
pmatrix{a & b \ c & d}pmatrix{0 \ 1} = pmatrix{0\0},
$$
even if it seems obvious to you.



Also: the "MathJax" that I typed to produce this answer was exactly this:



$$

pmatrix{a & b \ c & d}pmatrix{0 \ 1} = 

pmatrix{0\0},

$$


You should, if you're going to stick around here, learn how to format things like this.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks for confirming i will clarify that for who want to know
    $endgroup$
    – green life
    Dec 11 '18 at 17:28












  • $begingroup$
    and yes i need to learn how to enter the math codes i will learn it soon i promise
    $endgroup$
    – green life
    Dec 11 '18 at 17:34
















2












$begingroup$

Yes, your work is correct, although you might want to explain why you can write
$$
pmatrix{a & b \ c & d}pmatrix{0 \ 1} = pmatrix{0\0},
$$
even if it seems obvious to you.



Also: the "MathJax" that I typed to produce this answer was exactly this:



$$

pmatrix{a & b \ c & d}pmatrix{0 \ 1} = 

pmatrix{0\0},

$$


You should, if you're going to stick around here, learn how to format things like this.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks for confirming i will clarify that for who want to know
    $endgroup$
    – green life
    Dec 11 '18 at 17:28












  • $begingroup$
    and yes i need to learn how to enter the math codes i will learn it soon i promise
    $endgroup$
    – green life
    Dec 11 '18 at 17:34














2












2








2





$begingroup$

Yes, your work is correct, although you might want to explain why you can write
$$
pmatrix{a & b \ c & d}pmatrix{0 \ 1} = pmatrix{0\0},
$$
even if it seems obvious to you.



Also: the "MathJax" that I typed to produce this answer was exactly this:



$$

pmatrix{a & b \ c & d}pmatrix{0 \ 1} = 

pmatrix{0\0},

$$


You should, if you're going to stick around here, learn how to format things like this.






share|cite|improve this answer











$endgroup$



Yes, your work is correct, although you might want to explain why you can write
$$
pmatrix{a & b \ c & d}pmatrix{0 \ 1} = pmatrix{0\0},
$$
even if it seems obvious to you.



Also: the "MathJax" that I typed to produce this answer was exactly this:



$$

pmatrix{a & b \ c & d}pmatrix{0 \ 1} = 

pmatrix{0\0},

$$


You should, if you're going to stick around here, learn how to format things like this.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








answered Dec 11 '18 at 17:25


























community wiki





John Hughes













  • $begingroup$
    thanks for confirming i will clarify that for who want to know
    $endgroup$
    – green life
    Dec 11 '18 at 17:28












  • $begingroup$
    and yes i need to learn how to enter the math codes i will learn it soon i promise
    $endgroup$
    – green life
    Dec 11 '18 at 17:34


















  • $begingroup$
    thanks for confirming i will clarify that for who want to know
    $endgroup$
    – green life
    Dec 11 '18 at 17:28












  • $begingroup$
    and yes i need to learn how to enter the math codes i will learn it soon i promise
    $endgroup$
    – green life
    Dec 11 '18 at 17:34
















$begingroup$
thanks for confirming i will clarify that for who want to know
$endgroup$
– green life
Dec 11 '18 at 17:28






$begingroup$
thanks for confirming i will clarify that for who want to know
$endgroup$
– green life
Dec 11 '18 at 17:28














$begingroup$
and yes i need to learn how to enter the math codes i will learn it soon i promise
$endgroup$
– green life
Dec 11 '18 at 17:34




$begingroup$
and yes i need to learn how to enter the math codes i will learn it soon i promise
$endgroup$
– green life
Dec 11 '18 at 17:34











0












$begingroup$

It is basically correct, but you can do it more easily.



If $a_1$ and $a_2$ are the columns of $A$, then you know that
$$
1a_1+ca_2=begin{bmatrix} 1 \ 3 end{bmatrix}
$$
for every $c$; in particular this holds for $c=0$ and $c=1$, so
begin{cases}
a_1=begin{bmatrix} 1 \ 3 end{bmatrix} \[4px]
a_1+a_2=begin{bmatrix} 1 \ 3 end{bmatrix}
end{cases}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This does not answer the OP's question, which was "Can you confirm that my solution is right?"
    $endgroup$
    – John Hughes
    Dec 11 '18 at 22:52










  • $begingroup$
    @JohnHughes The OP's solution is not available unless one goes back in the post history. Anyway, this is a confirmation that the solution is correct.
    $endgroup$
    – egreg
    Dec 11 '18 at 22:59
















0












$begingroup$

It is basically correct, but you can do it more easily.



If $a_1$ and $a_2$ are the columns of $A$, then you know that
$$
1a_1+ca_2=begin{bmatrix} 1 \ 3 end{bmatrix}
$$
for every $c$; in particular this holds for $c=0$ and $c=1$, so
begin{cases}
a_1=begin{bmatrix} 1 \ 3 end{bmatrix} \[4px]
a_1+a_2=begin{bmatrix} 1 \ 3 end{bmatrix}
end{cases}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This does not answer the OP's question, which was "Can you confirm that my solution is right?"
    $endgroup$
    – John Hughes
    Dec 11 '18 at 22:52










  • $begingroup$
    @JohnHughes The OP's solution is not available unless one goes back in the post history. Anyway, this is a confirmation that the solution is correct.
    $endgroup$
    – egreg
    Dec 11 '18 at 22:59














0












0








0





$begingroup$

It is basically correct, but you can do it more easily.



If $a_1$ and $a_2$ are the columns of $A$, then you know that
$$
1a_1+ca_2=begin{bmatrix} 1 \ 3 end{bmatrix}
$$
for every $c$; in particular this holds for $c=0$ and $c=1$, so
begin{cases}
a_1=begin{bmatrix} 1 \ 3 end{bmatrix} \[4px]
a_1+a_2=begin{bmatrix} 1 \ 3 end{bmatrix}
end{cases}






share|cite|improve this answer











$endgroup$



It is basically correct, but you can do it more easily.



If $a_1$ and $a_2$ are the columns of $A$, then you know that
$$
1a_1+ca_2=begin{bmatrix} 1 \ 3 end{bmatrix}
$$
for every $c$; in particular this holds for $c=0$ and $c=1$, so
begin{cases}
a_1=begin{bmatrix} 1 \ 3 end{bmatrix} \[4px]
a_1+a_2=begin{bmatrix} 1 \ 3 end{bmatrix}
end{cases}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 11 '18 at 23:00

























answered Dec 11 '18 at 21:16









egregegreg

181k1485203




181k1485203












  • $begingroup$
    This does not answer the OP's question, which was "Can you confirm that my solution is right?"
    $endgroup$
    – John Hughes
    Dec 11 '18 at 22:52










  • $begingroup$
    @JohnHughes The OP's solution is not available unless one goes back in the post history. Anyway, this is a confirmation that the solution is correct.
    $endgroup$
    – egreg
    Dec 11 '18 at 22:59


















  • $begingroup$
    This does not answer the OP's question, which was "Can you confirm that my solution is right?"
    $endgroup$
    – John Hughes
    Dec 11 '18 at 22:52










  • $begingroup$
    @JohnHughes The OP's solution is not available unless one goes back in the post history. Anyway, this is a confirmation that the solution is correct.
    $endgroup$
    – egreg
    Dec 11 '18 at 22:59
















$begingroup$
This does not answer the OP's question, which was "Can you confirm that my solution is right?"
$endgroup$
– John Hughes
Dec 11 '18 at 22:52




$begingroup$
This does not answer the OP's question, which was "Can you confirm that my solution is right?"
$endgroup$
– John Hughes
Dec 11 '18 at 22:52












$begingroup$
@JohnHughes The OP's solution is not available unless one goes back in the post history. Anyway, this is a confirmation that the solution is correct.
$endgroup$
– egreg
Dec 11 '18 at 22:59




$begingroup$
@JohnHughes The OP's solution is not available unless one goes back in the post history. Anyway, this is a confirmation that the solution is correct.
$endgroup$
– egreg
Dec 11 '18 at 22:59


















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