Mixing
Determine the values of a, b and c, for which the systems have (1) exactly one solution, (2) no solutions,...
0
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I will just attach a picture. Can someone help me to solve this? I think I missed some information. 
linear-algebra matrices systems-of-equations determinant
asked Jan 16 at 18:50
Aliaksei KlimovichAliaksei Klimovich
516
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add a comment |
0
$begingroup$
I will just attach a picture. Can someone help me to solve this? I think I missed some information.
linear-algebra matrices systems-of-equations determinant
asked Jan 16 at 18:50
Aliaksei KlimovichAliaksei Klimovich
516
$endgroup$
$begingroup$
Your work looks correct to me. Note that $a-2b+c=0$ is the equation of a plane
$endgroup$
– pwerth
Jan 16 at 18:52
$begingroup$
Thanks, @pwerth , but I need to determine the values for a,b and c. I have no idea what to do with them.
$endgroup$
– Aliaksei Klimovich
Jan 16 at 18:58
$begingroup$
My point is that there is no way to simplify the answer any further than saying "all values of a,b,c such that $a-2b+c=0$", which you could rephrase as "all points on the plane $a-2b+c=0$. There are obviously infinitely many such points so there is no way you can write them all down
$endgroup$
– pwerth
Jan 16 at 19:00
$begingroup$
I'll also point out that since you exhausted both cases, you proved that there are no values of $a,b,c$ for which the system has a unique solution.
$endgroup$
– pwerth
Jan 16 at 19:01
$begingroup$
@pwerth I had this task when my group in college did not start planes. So, there must be a way how to make an answer except yours. I lost my notes, that's why it's hard for me now.
$endgroup$
– Aliaksei Klimovich
Jan 16 at 19:04
add a comment |
0
0
0
$begingroup$
I will just attach a picture. Can someone help me to solve this? I think I missed some information.
linear-algebra matrices systems-of-equations determinant
asked Jan 16 at 18:50
Aliaksei KlimovichAliaksei Klimovich
516
$endgroup$
I will just attach a picture. Can someone help me to solve this? I think I missed some information.
linear-algebra matrices systems-of-equations determinant
linear-algebra matrices systems-of-equations determinant
asked Jan 16 at 18:50
Aliaksei KlimovichAliaksei Klimovich
516
asked Jan 16 at 18:50
Aliaksei KlimovichAliaksei Klimovich
516
asked Jan 16 at 18:50
Aliaksei KlimovichAliaksei Klimovich
516
asked Jan 16 at 18:50
Aliaksei KlimovichAliaksei Klimovich
516
asked Jan 16 at 18:50
Aliaksei KlimovichAliaksei Klimovich
516
516
$begingroup$
Your work looks correct to me. Note that $a-2b+c=0$ is the equation of a plane
$endgroup$
– pwerth
Jan 16 at 18:52
$begingroup$
Thanks, @pwerth , but I need to determine the values for a,b and c. I have no idea what to do with them.
$endgroup$
– Aliaksei Klimovich
Jan 16 at 18:58
$begingroup$
My point is that there is no way to simplify the answer any further than saying "all values of a,b,c such that $a-2b+c=0$", which you could rephrase as "all points on the plane $a-2b+c=0$. There are obviously infinitely many such points so there is no way you can write them all down
$endgroup$
– pwerth
Jan 16 at 19:00
$begingroup$
I'll also point out that since you exhausted both cases, you proved that there are no values of $a,b,c$ for which the system has a unique solution.
$endgroup$
– pwerth
Jan 16 at 19:01
$begingroup$
@pwerth I had this task when my group in college did not start planes. So, there must be a way how to make an answer except yours. I lost my notes, that's why it's hard for me now.
$endgroup$
– Aliaksei Klimovich
Jan 16 at 19:04
add a comment |
$begingroup$
Your work looks correct to me. Note that $a-2b+c=0$ is the equation of a plane
$endgroup$
– pwerth
Jan 16 at 18:52
$begingroup$
Thanks, @pwerth , but I need to determine the values for a,b and c. I have no idea what to do with them.
$endgroup$
– Aliaksei Klimovich
Jan 16 at 18:58
$begingroup$
My point is that there is no way to simplify the answer any further than saying "all values of a,b,c such that $a-2b+c=0$", which you could rephrase as "all points on the plane $a-2b+c=0$. There are obviously infinitely many such points so there is no way you can write them all down
$endgroup$
– pwerth
Jan 16 at 19:00
$begingroup$
I'll also point out that since you exhausted both cases, you proved that there are no values of $a,b,c$ for which the system has a unique solution.
$endgroup$
– pwerth
Jan 16 at 19:01
$begingroup$
@pwerth I had this task when my group in college did not start planes. So, there must be a way how to make an answer except yours. I lost my notes, that's why it's hard for me now.
$endgroup$
– Aliaksei Klimovich
Jan 16 at 19:04
$begingroup$
Your work looks correct to me. Note that $a-2b+c=0$ is the equation of a plane
$endgroup$
– pwerth
Jan 16 at 18:52
$begingroup$
Your work looks correct to me. Note that $a-2b+c=0$ is the equation of a plane
$endgroup$
– pwerth
Jan 16 at 18:52
$begingroup$
Thanks, @pwerth , but I need to determine the values for a,b and c. I have no idea what to do with them.
$endgroup$
– Aliaksei Klimovich
Jan 16 at 18:58
$begingroup$
Thanks, @pwerth , but I need to determine the values for a,b and c. I have no idea what to do with them.
$endgroup$
– Aliaksei Klimovich
Jan 16 at 18:58
$begingroup$
My point is that there is no way to simplify the answer any further than saying "all values of a,b,c such that $a-2b+c=0$", which you could rephrase as "all points on the plane $a-2b+c=0$. There are obviously infinitely many such points so there is no way you can write them all down
$endgroup$
– pwerth
Jan 16 at 19:00
$begingroup$
My point is that there is no way to simplify the answer any further than saying "all values of a,b,c such that $a-2b+c=0$", which you could rephrase as "all points on the plane $a-2b+c=0$. There are obviously infinitely many such points so there is no way you can write them all down
$endgroup$
– pwerth
Jan 16 at 19:00
$begingroup$
I'll also point out that since you exhausted both cases, you proved that there are no values of $a,b,c$ for which the system has a unique solution.
$endgroup$
– pwerth
Jan 16 at 19:01
$begingroup$
I'll also point out that since you exhausted both cases, you proved that there are no values of $a,b,c$ for which the system has a unique solution.
$endgroup$
– pwerth
Jan 16 at 19:01
$begingroup$
@pwerth I had this task when my group in college did not start planes. So, there must be a way how to make an answer except yours. I lost my notes, that's why it's hard for me now.
$endgroup$
– Aliaksei Klimovich
Jan 16 at 19:04
$begingroup$
@pwerth I had this task when my group in college did not start planes. So, there must be a way how to make an answer except yours. I lost my notes, that's why it's hard for me now.
$endgroup$
– Aliaksei Klimovich
Jan 16 at 19:04
add a comment |
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I have the same result as you: If $$a-2b+c=0$$ then we get infinity many solutions. If $$a-2b+cne 0$$ then we get no solutions.Since the last two equations are $$x_2+2x_3=frac{4a-b}{3}$$ and $$x_2+2x_3=frac{7a-c}{6}$$
answered Jan 16 at 19:16
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
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I have the same result as you: If $$a-2b+c=0$$ then we get infinity many solutions. If $$a-2b+cne 0$$ then we get no solutions.Since the last two equations are $$x_2+2x_3=frac{4a-b}{3}$$ and $$x_2+2x_3=frac{7a-c}{6}$$
answered Jan 16 at 19:16
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
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add a comment |
0
$begingroup$
I have the same result as you: If $$a-2b+c=0$$ then we get infinity many solutions. If $$a-2b+cne 0$$ then we get no solutions.Since the last two equations are $$x_2+2x_3=frac{4a-b}{3}$$ and $$x_2+2x_3=frac{7a-c}{6}$$
answered Jan 16 at 19:16
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
$endgroup$
add a comment |
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$begingroup$
I have the same result as you: If $$a-2b+c=0$$ then we get infinity many solutions. If $$a-2b+cne 0$$ then we get no solutions.Since the last two equations are $$x_2+2x_3=frac{4a-b}{3}$$ and $$x_2+2x_3=frac{7a-c}{6}$$
answered Jan 16 at 19:16
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
$endgroup$
I have the same result as you: If $$a-2b+c=0$$ then we get infinity many solutions. If $$a-2b+cne 0$$ then we get no solutions.Since the last two equations are $$x_2+2x_3=frac{4a-b}{3}$$ and $$x_2+2x_3=frac{7a-c}{6}$$
answered Jan 16 at 19:16
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
answered Jan 16 at 19:16
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
answered Jan 16 at 19:16
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
answered Jan 16 at 19:16
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
76.2k42866
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$begingroup$
Your work looks correct to me. Note that $a-2b+c=0$ is the equation of a plane
$endgroup$
– pwerth
Jan 16 at 18:52
$begingroup$
Thanks, @pwerth , but I need to determine the values for a,b and c. I have no idea what to do with them.
$endgroup$
– Aliaksei Klimovich
Jan 16 at 18:58
$begingroup$
My point is that there is no way to simplify the answer any further than saying "all values of a,b,c such that $a-2b+c=0$", which you could rephrase as "all points on the plane $a-2b+c=0$. There are obviously infinitely many such points so there is no way you can write them all down
$endgroup$
– pwerth
Jan 16 at 19:00
$begingroup$
I'll also point out that since you exhausted both cases, you proved that there are no values of $a,b,c$ for which the system has a unique solution.
$endgroup$
– pwerth
Jan 16 at 19:01
$begingroup$
@pwerth I had this task when my group in college did not start planes. So, there must be a way how to make an answer except yours. I lost my notes, that's why it's hard for me now.
$endgroup$
– Aliaksei Klimovich
Jan 16 at 19:04