Value of a such that the system has no solution/infinitely many solutions
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
asked Nov 19 '18 at 2:11
Future Math person
972717
add a comment |
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
asked Nov 19 '18 at 2:11
Future Math person
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
add a comment |
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
asked Nov 19 '18 at 2:11
Future Math person
972717
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
asked Nov 19 '18 at 2:11
Future Math person
972717
asked Nov 19 '18 at 2:11
Future Math person
972717
asked Nov 19 '18 at 2:11
Future Math person
972717
asked Nov 19 '18 at 2:11
Future Math person
972717
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
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 $;aneq4;$ ?
answered Nov 19 '18 at 2:43
DonAntonio
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
add a comment |
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.
answered Nov 19 '18 at 2:44
coffeemath
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
add a comment |
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2 Answers
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active
oldest
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2 Answers
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oldest
votes
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 $;aneq4;$ ?
answered Nov 19 '18 at 2:43
DonAntonio
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
add a comment |
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 $;aneq4;$ ?
answered Nov 19 '18 at 2:43
DonAntonio
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
add a comment |
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 $;aneq4;$ ?
answered Nov 19 '18 at 2:43
DonAntonio
177k1491225
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 $;aneq4;$ ?
answered Nov 19 '18 at 2:43
DonAntonio
177k1491225
answered Nov 19 '18 at 2:43
DonAntonio
177k1491225
answered Nov 19 '18 at 2:43
DonAntonio
177k1491225
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
add a comment |
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
add a comment |
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.
answered Nov 19 '18 at 2:44
coffeemath
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
add a comment |
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.
answered Nov 19 '18 at 2:44
coffeemath
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
add a comment |
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.
answered Nov 19 '18 at 2:44
coffeemath
2,4081413
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.
answered Nov 19 '18 at 2:44
coffeemath
2,4081413
edited Nov 19 '18 at 3:10
answered Nov 19 '18 at 2:44
coffeemath
2,4081413
answered Nov 19 '18 at 2:44
coffeemath
2,4081413
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
add a comment |
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
add a comment |
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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