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












0














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






















  • 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 '18 at 2:35
















0














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






















  • 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 '18 at 2:35














0












0








0


0





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













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




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 '18 at 2:11









Future Math person

972717




972717












  • 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 '18 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 '18 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 '18 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 '18 at 2:35










2 Answers
2






active

oldest

votes


















1














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 '18 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 '18 at 11:39



















0














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 '18 at 2:49










  • @DonAntonio Thanks for fix. Put it in, with note.
    – coffeemath
    Nov 19 '18 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 '18 at 4:46










  • Yes, no a gives infinitely many solutions. the only possible value 4 of a gives no solutions.
    – coffeemath
    Nov 19 '18 at 8:07











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














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 '18 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 '18 at 11:39
















1














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 '18 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 '18 at 11:39














1












1








1






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 '18 at 2:43









DonAntonio

177k1491225




177k1491225












  • 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 11:39











0














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 '18 at 2:49










  • @DonAntonio Thanks for fix. Put it in, with note.
    – coffeemath
    Nov 19 '18 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 '18 at 4:46










  • Yes, no a gives infinitely many solutions. the only possible value 4 of a gives no solutions.
    – coffeemath
    Nov 19 '18 at 8:07
















0














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 '18 at 2:49










  • @DonAntonio Thanks for fix. Put it in, with note.
    – coffeemath
    Nov 19 '18 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 '18 at 4:46










  • Yes, no a gives infinitely many solutions. the only possible value 4 of a gives no solutions.
    – coffeemath
    Nov 19 '18 at 8:07














0












0








0






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 '18 at 3:10

























answered Nov 19 '18 at 2:44









coffeemath

2,4081413




2,4081413








  • 1




    In row $3$ column $4$ it must be $$(3-a)[1-a+2]=(3-a)^2=(a-3)^2$$
    – DonAntonio
    Nov 19 '18 at 2:49










  • @DonAntonio Thanks for fix. Put it in, with note.
    – coffeemath
    Nov 19 '18 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 '18 at 4:46










  • Yes, no a gives infinitely many solutions. the only possible value 4 of a gives no solutions.
    – coffeemath
    Nov 19 '18 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 '18 at 2:49










  • @DonAntonio Thanks for fix. Put it in, with note.
    – coffeemath
    Nov 19 '18 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 '18 at 4:46










  • Yes, no a gives infinitely many solutions. the only possible value 4 of a gives no solutions.
    – coffeemath
    Nov 19 '18 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 '18 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 '18 at 2:49












@DonAntonio Thanks for fix. Put it in, with note.
– coffeemath
Nov 19 '18 at 3:11




@DonAntonio Thanks for fix. Put it in, with note.
– coffeemath
Nov 19 '18 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 '18 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 '18 at 4:46












Yes, no a gives infinitely many solutions. the only possible value 4 of a gives no solutions.
– coffeemath
Nov 19 '18 at 8:07




Yes, no a gives infinitely many solutions. the only possible value 4 of a gives no solutions.
– coffeemath
Nov 19 '18 at 8:07


















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