Value of a such that the system has no solution/infinitely many solutions

Multi tool use
Multi tool use











up vote
0
down vote

favorite












I have this system of equations here:



$Ax=b$, where:



$A=$$
left(
begin{array}{ccc}
1 & 2 & 3 \
1 & 3 & 4\
1 & a & 5\
end{array}
right)
$



$b=$$
left(
begin{array}{c}
a\
3\
3\
end{array}
right)
$



I have to find the values of $a$ in which the system has no solution and the value of a in which the system has infinitely many solutions.



I wrote my augmented matrix as:



$
left(
begin{array}{cccc}
1 & 2 & 3 & a \
1 & 3 & 4 & 3\
1 & a & 5 & 3\
end{array}
right)
$



I then attempted to row reduce and then do cases but after a certain point, I am stuck. So far, I've row reduced down to:



$
left(
begin{array}{cccc}
1 & 2 & 3 & a \
0 & 1 & 1 & 3-a\
0 & a-2 & 2 & 3-a\
end{array}
right)
$



I'm not really sure what to do after this point though. I should probably row reduce more but I'm stuck on how I would show the conditions for each case of a. If anyone could guide me in the right direction, that would be much appreciated!










share|cite|improve this question






















  • Here you can use the 1 in row 2 column 2 to zero out the rest of column 2, still without assumptions on a. Then see what it becomes.
    – coffeemath
    Nov 19 at 2:35















up vote
0
down vote

favorite












I have this system of equations here:



$Ax=b$, where:



$A=$$
left(
begin{array}{ccc}
1 & 2 & 3 \
1 & 3 & 4\
1 & a & 5\
end{array}
right)
$



$b=$$
left(
begin{array}{c}
a\
3\
3\
end{array}
right)
$



I have to find the values of $a$ in which the system has no solution and the value of a in which the system has infinitely many solutions.



I wrote my augmented matrix as:



$
left(
begin{array}{cccc}
1 & 2 & 3 & a \
1 & 3 & 4 & 3\
1 & a & 5 & 3\
end{array}
right)
$



I then attempted to row reduce and then do cases but after a certain point, I am stuck. So far, I've row reduced down to:



$
left(
begin{array}{cccc}
1 & 2 & 3 & a \
0 & 1 & 1 & 3-a\
0 & a-2 & 2 & 3-a\
end{array}
right)
$



I'm not really sure what to do after this point though. I should probably row reduce more but I'm stuck on how I would show the conditions for each case of a. If anyone could guide me in the right direction, that would be much appreciated!










share|cite|improve this question






















  • Here you can use the 1 in row 2 column 2 to zero out the rest of column 2, still without assumptions on a. Then see what it becomes.
    – coffeemath
    Nov 19 at 2:35













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have this system of equations here:



$Ax=b$, where:



$A=$$
left(
begin{array}{ccc}
1 & 2 & 3 \
1 & 3 & 4\
1 & a & 5\
end{array}
right)
$



$b=$$
left(
begin{array}{c}
a\
3\
3\
end{array}
right)
$



I have to find the values of $a$ in which the system has no solution and the value of a in which the system has infinitely many solutions.



I wrote my augmented matrix as:



$
left(
begin{array}{cccc}
1 & 2 & 3 & a \
1 & 3 & 4 & 3\
1 & a & 5 & 3\
end{array}
right)
$



I then attempted to row reduce and then do cases but after a certain point, I am stuck. So far, I've row reduced down to:



$
left(
begin{array}{cccc}
1 & 2 & 3 & a \
0 & 1 & 1 & 3-a\
0 & a-2 & 2 & 3-a\
end{array}
right)
$



I'm not really sure what to do after this point though. I should probably row reduce more but I'm stuck on how I would show the conditions for each case of a. If anyone could guide me in the right direction, that would be much appreciated!










share|cite|improve this question













I have this system of equations here:



$Ax=b$, where:



$A=$$
left(
begin{array}{ccc}
1 & 2 & 3 \
1 & 3 & 4\
1 & a & 5\
end{array}
right)
$



$b=$$
left(
begin{array}{c}
a\
3\
3\
end{array}
right)
$



I have to find the values of $a$ in which the system has no solution and the value of a in which the system has infinitely many solutions.



I wrote my augmented matrix as:



$
left(
begin{array}{cccc}
1 & 2 & 3 & a \
1 & 3 & 4 & 3\
1 & a & 5 & 3\
end{array}
right)
$



I then attempted to row reduce and then do cases but after a certain point, I am stuck. So far, I've row reduced down to:



$
left(
begin{array}{cccc}
1 & 2 & 3 & a \
0 & 1 & 1 & 3-a\
0 & a-2 & 2 & 3-a\
end{array}
right)
$



I'm not really sure what to do after this point though. I should probably row reduce more but I'm stuck on how I would show the conditions for each case of a. If anyone could guide me in the right direction, that would be much appreciated!







linear-algebra systems-of-equations matrix-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 19 at 2:11









Future Math person

864717




864717












  • Here you can use the 1 in row 2 column 2 to zero out the rest of column 2, still without assumptions on a. Then see what it becomes.
    – coffeemath
    Nov 19 at 2:35


















  • Here you can use the 1 in row 2 column 2 to zero out the rest of column 2, still without assumptions on a. Then see what it becomes.
    – coffeemath
    Nov 19 at 2:35
















Here you can use the 1 in row 2 column 2 to zero out the rest of column 2, still without assumptions on a. Then see what it becomes.
– coffeemath
Nov 19 at 2:35




Here you can use the 1 in row 2 column 2 to zero out the rest of column 2, still without assumptions on a. Then see what it becomes.
– coffeemath
Nov 19 at 2:35










2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










Continue reducing!



$$left(
begin{array}{cccc}
1 & 2 & 3 & a \
0 & 1 & 1 & 3-a\
0 & a-2 & 2 & 3-a\
end{array}
right)longrightarrowleft(
begin{array}{cccc}
1 & 2 & 3 & a \
0 & 1 & 1 & 3-a\
0 & 0 & 4-a & (a-3)^2\
end{array}
right)$$



Can you prove that for $;a=4;$ you get an incongruent system (no solution)? What happens if $;aneq4;$ ?






share|cite|improve this answer





















  • I get that there is no possible value for a in which there are infinitely many solutions. I don't think it's possible to write one row as a multiple of another in any way.
    – Future Math person
    Nov 19 at 4:47










  • @FutureMathperson Right. if $;a=4;$ there is no solution, and $;aneq4;$ renders a coefficient matrix that is regular, which means unique solution.
    – DonAntonio
    Nov 19 at 11:39


















up vote
0
down vote













After subtracting $a-2$ times row 2 from row 3, the (3,3) entry is $2-(a-2)=4-a.$ That is zero iff $a=4.$ So if $a=4$ there will be 0 or infinitely many solutions. In row 3 column 4 it will be $(3-a)-(3-a)(a-2)=(3-a)[1-(a-2)]=(3-a)^2$ which isn't $0$ when $a=4.$ So if your calculation so far is right, no solutions.



Thanks to Don Antonio for fix.






share|cite|improve this answer



















  • 1




    In row $3$ column $4$ it must be $$(3-a)[1-a+2]=(3-a)^2=(a-3)^2$$
    – DonAntonio
    Nov 19 at 2:49










  • @DonAntonio Thanks for fix. Put it in, with note.
    – coffeemath
    Nov 19 at 3:11










  • After mulling this over for a bit, I don't think there is a case where we have infinitely many solutions actually. That would mean I could write one row as a multiple of another row, but that's impossible for any choice of $a$. Is that correct?
    – Future Math person
    Nov 19 at 4:46










  • Yes, no a gives infinitely many solutions. the only possible value 4 of a gives no solutions.
    – coffeemath
    Nov 19 at 8:07











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',
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%2f3004428%2fvalue-of-a-such-that-the-system-has-no-solution-infinitely-many-solutions%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Continue reducing!



$$left(
begin{array}{cccc}
1 & 2 & 3 & a \
0 & 1 & 1 & 3-a\
0 & a-2 & 2 & 3-a\
end{array}
right)longrightarrowleft(
begin{array}{cccc}
1 & 2 & 3 & a \
0 & 1 & 1 & 3-a\
0 & 0 & 4-a & (a-3)^2\
end{array}
right)$$



Can you prove that for $;a=4;$ you get an incongruent system (no solution)? What happens if $;aneq4;$ ?






share|cite|improve this answer





















  • I get that there is no possible value for a in which there are infinitely many solutions. I don't think it's possible to write one row as a multiple of another in any way.
    – Future Math person
    Nov 19 at 4:47










  • @FutureMathperson Right. if $;a=4;$ there is no solution, and $;aneq4;$ renders a coefficient matrix that is regular, which means unique solution.
    – DonAntonio
    Nov 19 at 11:39















up vote
1
down vote



accepted










Continue reducing!



$$left(
begin{array}{cccc}
1 & 2 & 3 & a \
0 & 1 & 1 & 3-a\
0 & a-2 & 2 & 3-a\
end{array}
right)longrightarrowleft(
begin{array}{cccc}
1 & 2 & 3 & a \
0 & 1 & 1 & 3-a\
0 & 0 & 4-a & (a-3)^2\
end{array}
right)$$



Can you prove that for $;a=4;$ you get an incongruent system (no solution)? What happens if $;aneq4;$ ?






share|cite|improve this answer





















  • I get that there is no possible value for a in which there are infinitely many solutions. I don't think it's possible to write one row as a multiple of another in any way.
    – Future Math person
    Nov 19 at 4:47










  • @FutureMathperson Right. if $;a=4;$ there is no solution, and $;aneq4;$ renders a coefficient matrix that is regular, which means unique solution.
    – DonAntonio
    Nov 19 at 11:39













up vote
1
down vote



accepted







up vote
1
down vote



accepted






Continue reducing!



$$left(
begin{array}{cccc}
1 & 2 & 3 & a \
0 & 1 & 1 & 3-a\
0 & a-2 & 2 & 3-a\
end{array}
right)longrightarrowleft(
begin{array}{cccc}
1 & 2 & 3 & a \
0 & 1 & 1 & 3-a\
0 & 0 & 4-a & (a-3)^2\
end{array}
right)$$



Can you prove that for $;a=4;$ you get an incongruent system (no solution)? What happens if $;aneq4;$ ?






share|cite|improve this answer












Continue reducing!



$$left(
begin{array}{cccc}
1 & 2 & 3 & a \
0 & 1 & 1 & 3-a\
0 & a-2 & 2 & 3-a\
end{array}
right)longrightarrowleft(
begin{array}{cccc}
1 & 2 & 3 & a \
0 & 1 & 1 & 3-a\
0 & 0 & 4-a & (a-3)^2\
end{array}
right)$$



Can you prove that for $;a=4;$ you get an incongruent system (no solution)? What happens if $;aneq4;$ ?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 19 at 2:43









DonAntonio

176k1491224




176k1491224












  • I get that there is no possible value for a in which there are infinitely many solutions. I don't think it's possible to write one row as a multiple of another in any way.
    – Future Math person
    Nov 19 at 4:47










  • @FutureMathperson Right. if $;a=4;$ there is no solution, and $;aneq4;$ renders a coefficient matrix that is regular, which means unique solution.
    – DonAntonio
    Nov 19 at 11:39


















  • I get that there is no possible value for a in which there are infinitely many solutions. I don't think it's possible to write one row as a multiple of another in any way.
    – Future Math person
    Nov 19 at 4:47










  • @FutureMathperson Right. if $;a=4;$ there is no solution, and $;aneq4;$ renders a coefficient matrix that is regular, which means unique solution.
    – DonAntonio
    Nov 19 at 11:39
















I get that there is no possible value for a in which there are infinitely many solutions. I don't think it's possible to write one row as a multiple of another in any way.
– Future Math person
Nov 19 at 4:47




I get that there is no possible value for a in which there are infinitely many solutions. I don't think it's possible to write one row as a multiple of another in any way.
– Future Math person
Nov 19 at 4:47












@FutureMathperson Right. if $;a=4;$ there is no solution, and $;aneq4;$ renders a coefficient matrix that is regular, which means unique solution.
– DonAntonio
Nov 19 at 11:39




@FutureMathperson Right. if $;a=4;$ there is no solution, and $;aneq4;$ renders a coefficient matrix that is regular, which means unique solution.
– DonAntonio
Nov 19 at 11:39










up vote
0
down vote













After subtracting $a-2$ times row 2 from row 3, the (3,3) entry is $2-(a-2)=4-a.$ That is zero iff $a=4.$ So if $a=4$ there will be 0 or infinitely many solutions. In row 3 column 4 it will be $(3-a)-(3-a)(a-2)=(3-a)[1-(a-2)]=(3-a)^2$ which isn't $0$ when $a=4.$ So if your calculation so far is right, no solutions.



Thanks to Don Antonio for fix.






share|cite|improve this answer



















  • 1




    In row $3$ column $4$ it must be $$(3-a)[1-a+2]=(3-a)^2=(a-3)^2$$
    – DonAntonio
    Nov 19 at 2:49










  • @DonAntonio Thanks for fix. Put it in, with note.
    – coffeemath
    Nov 19 at 3:11










  • After mulling this over for a bit, I don't think there is a case where we have infinitely many solutions actually. That would mean I could write one row as a multiple of another row, but that's impossible for any choice of $a$. Is that correct?
    – Future Math person
    Nov 19 at 4:46










  • Yes, no a gives infinitely many solutions. the only possible value 4 of a gives no solutions.
    – coffeemath
    Nov 19 at 8:07















up vote
0
down vote













After subtracting $a-2$ times row 2 from row 3, the (3,3) entry is $2-(a-2)=4-a.$ That is zero iff $a=4.$ So if $a=4$ there will be 0 or infinitely many solutions. In row 3 column 4 it will be $(3-a)-(3-a)(a-2)=(3-a)[1-(a-2)]=(3-a)^2$ which isn't $0$ when $a=4.$ So if your calculation so far is right, no solutions.



Thanks to Don Antonio for fix.






share|cite|improve this answer



















  • 1




    In row $3$ column $4$ it must be $$(3-a)[1-a+2]=(3-a)^2=(a-3)^2$$
    – DonAntonio
    Nov 19 at 2:49










  • @DonAntonio Thanks for fix. Put it in, with note.
    – coffeemath
    Nov 19 at 3:11










  • After mulling this over for a bit, I don't think there is a case where we have infinitely many solutions actually. That would mean I could write one row as a multiple of another row, but that's impossible for any choice of $a$. Is that correct?
    – Future Math person
    Nov 19 at 4:46










  • Yes, no a gives infinitely many solutions. the only possible value 4 of a gives no solutions.
    – coffeemath
    Nov 19 at 8:07













up vote
0
down vote










up vote
0
down vote









After subtracting $a-2$ times row 2 from row 3, the (3,3) entry is $2-(a-2)=4-a.$ That is zero iff $a=4.$ So if $a=4$ there will be 0 or infinitely many solutions. In row 3 column 4 it will be $(3-a)-(3-a)(a-2)=(3-a)[1-(a-2)]=(3-a)^2$ which isn't $0$ when $a=4.$ So if your calculation so far is right, no solutions.



Thanks to Don Antonio for fix.






share|cite|improve this answer














After subtracting $a-2$ times row 2 from row 3, the (3,3) entry is $2-(a-2)=4-a.$ That is zero iff $a=4.$ So if $a=4$ there will be 0 or infinitely many solutions. In row 3 column 4 it will be $(3-a)-(3-a)(a-2)=(3-a)[1-(a-2)]=(3-a)^2$ which isn't $0$ when $a=4.$ So if your calculation so far is right, no solutions.



Thanks to Don Antonio for fix.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 19 at 3:10

























answered Nov 19 at 2:44









coffeemath

2,1501413




2,1501413








  • 1




    In row $3$ column $4$ it must be $$(3-a)[1-a+2]=(3-a)^2=(a-3)^2$$
    – DonAntonio
    Nov 19 at 2:49










  • @DonAntonio Thanks for fix. Put it in, with note.
    – coffeemath
    Nov 19 at 3:11










  • After mulling this over for a bit, I don't think there is a case where we have infinitely many solutions actually. That would mean I could write one row as a multiple of another row, but that's impossible for any choice of $a$. Is that correct?
    – Future Math person
    Nov 19 at 4:46










  • Yes, no a gives infinitely many solutions. the only possible value 4 of a gives no solutions.
    – coffeemath
    Nov 19 at 8:07














  • 1




    In row $3$ column $4$ it must be $$(3-a)[1-a+2]=(3-a)^2=(a-3)^2$$
    – DonAntonio
    Nov 19 at 2:49










  • @DonAntonio Thanks for fix. Put it in, with note.
    – coffeemath
    Nov 19 at 3:11










  • After mulling this over for a bit, I don't think there is a case where we have infinitely many solutions actually. That would mean I could write one row as a multiple of another row, but that's impossible for any choice of $a$. Is that correct?
    – Future Math person
    Nov 19 at 4:46










  • Yes, no a gives infinitely many solutions. the only possible value 4 of a gives no solutions.
    – coffeemath
    Nov 19 at 8:07








1




1




In row $3$ column $4$ it must be $$(3-a)[1-a+2]=(3-a)^2=(a-3)^2$$
– DonAntonio
Nov 19 at 2:49




In row $3$ column $4$ it must be $$(3-a)[1-a+2]=(3-a)^2=(a-3)^2$$
– DonAntonio
Nov 19 at 2:49












@DonAntonio Thanks for fix. Put it in, with note.
– coffeemath
Nov 19 at 3:11




@DonAntonio Thanks for fix. Put it in, with note.
– coffeemath
Nov 19 at 3:11












After mulling this over for a bit, I don't think there is a case where we have infinitely many solutions actually. That would mean I could write one row as a multiple of another row, but that's impossible for any choice of $a$. Is that correct?
– Future Math person
Nov 19 at 4:46




After mulling this over for a bit, I don't think there is a case where we have infinitely many solutions actually. That would mean I could write one row as a multiple of another row, but that's impossible for any choice of $a$. Is that correct?
– Future Math person
Nov 19 at 4:46












Yes, no a gives infinitely many solutions. the only possible value 4 of a gives no solutions.
– coffeemath
Nov 19 at 8:07




Yes, no a gives infinitely many solutions. the only possible value 4 of a gives no solutions.
– coffeemath
Nov 19 at 8:07


















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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


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%2f3004428%2fvalue-of-a-such-that-the-system-has-no-solution-infinitely-many-solutions%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







cHdd,r zDQf CQ1HrU3Mw,na0,7 SQgHk7En5T8I 5PGot1
QY78GU7Y 1PAgo BV4HTK9Bk9tqx,gKn3,Sxyf7aY,b2jr x QDO,yr hQ bCz5,67i6tjpjMZ9Mbx1YDx

Popular posts from this blog

mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

How to change which sound is reproduced for terminal bell?

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