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





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035528%2fobtain-matrix-a-if-x-and-b-given%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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

1,249612




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




















  • $begingroup$
    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

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


















  • $begingroup$
    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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035528%2fobtain-matrix-a-if-x-and-b-given%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
Read more