obtain matrix $ A$ if $X$ and $b$ given
$begingroup$
for the right and detailed answer refer to user9077 answer
linear-algebra matrices
asked Dec 11 '18 at 17:17
green lifegreen life
186
$endgroup$
add a comment |
$begingroup$
for the right and detailed answer refer to user9077 answer
linear-algebra matrices
asked Dec 11 '18 at 17:17
green lifegreen life
186
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
for the right and detailed answer refer to user9077 answer
linear-algebra matrices
asked Dec 11 '18 at 17:17
green lifegreen life
186
$endgroup$
for the right and detailed answer refer to user9077 answer
linear-algebra matrices
linear-algebra matrices
asked Dec 11 '18 at 17:17
green lifegreen life
186
asked Dec 11 '18 at 17:17
green lifegreen life
186
edited Dec 12 '18 at 17:43
green life
asked Dec 11 '18 at 17:17
green lifegreen life
186
asked Dec 11 '18 at 17:17
green lifegreen life
186
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
add a comment |
$begingroup$
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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
$$A=begin{pmatrix}1&0\3&0end{pmatrix}.$$
answered Dec 11 '18 at 21:36
user9077user9077
1,249612
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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}
answered Dec 11 '18 at 21:16
egregegreg
185k1486206
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$A=begin{pmatrix}1&0\3&0end{pmatrix}.$$
answered Dec 11 '18 at 21:36
user9077user9077
1,249612
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$A=begin{pmatrix}1&0\3&0end{pmatrix}.$$
answered Dec 11 '18 at 21:36
user9077user9077
1,249612
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$A=begin{pmatrix}1&0\3&0end{pmatrix}.$$
answered Dec 11 '18 at 21:36
user9077user9077
1,249612
$endgroup$
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
$$A=begin{pmatrix}1&0\3&0end{pmatrix}.$$
answered Dec 11 '18 at 21:36
user9077user9077
1,249612
answered Dec 11 '18 at 21:36
user9077user9077
1,249612
answered Dec 11 '18 at 21:36
user9077user9077
1,249612
answered Dec 11 '18 at 21:36
user9077user9077
1,249612
1,249612
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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}
answered Dec 11 '18 at 21:16
egregegreg
185k1486206
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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}
answered Dec 11 '18 at 21:16
egregegreg
185k1486206
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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}
answered Dec 11 '18 at 21:16
egregegreg
185k1486206
$endgroup$
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}
answered Dec 11 '18 at 21:16
egregegreg
185k1486206
edited Dec 11 '18 at 23:00
answered Dec 11 '18 at 21:16
egregegreg
185k1486206
answered Dec 11 '18 at 21:16
egregegreg
185k1486206
answered Dec 11 '18 at 21:16
egregegreg
185k1486206
185k1486206
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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