If $M=left(begin{smallmatrix} 0 & 1 & -1 \ -1 & 0 & alpha \ 2 & -alpha & 0...
$begingroup$
If $M=begin{pmatrix}
0 & 1 & -1 \
-1 & 0 & alpha \
2 & -alpha & 0
end{pmatrix} $
and $Mx=b$, for some $x$,then find $x^TMx$.
My approach:
If $Mx=b$ has a unique solution $x$ then, $text{Rank(M)}=3$
Thus, there are three linearly independent columns in $M$.
Thus if $alpha neq 0$, $c_1(0,-1,2)+c_2(1, 0 ,-alpha)+c_3(-1, alpha ,0)=(0, 0 ,0)$ has only $c_1,c_2,c_3 =0$ as solutions.
But if $alpha=0$, then the second and third row are linearly dependent, so the $text{rank}(M)=2$ which is not possible.
But then I am not getting any specific value of $alpha$ in $M$.
Am I going wrong somewhere?
linear-algebra matrices quadratic-forms
edited Dec 13 '18 at 14:08
StubbornAtom
5,98811238
asked May 6 '18 at 15:57
Legend KillerLegend Killer
1,6621523
$endgroup$
add a comment |
$begingroup$
If $M=begin{pmatrix}
0 & 1 & -1 \
-1 & 0 & alpha \
2 & -alpha & 0
end{pmatrix} $
and $Mx=b$, for some $x$,then find $x^TMx$.
My approach:
If $Mx=b$ has a unique solution $x$ then, $text{Rank(M)}=3$
Thus, there are three linearly independent columns in $M$.
Thus if $alpha neq 0$, $c_1(0,-1,2)+c_2(1, 0 ,-alpha)+c_3(-1, alpha ,0)=(0, 0 ,0)$ has only $c_1,c_2,c_3 =0$ as solutions.
But if $alpha=0$, then the second and third row are linearly dependent, so the $text{rank}(M)=2$ which is not possible.
But then I am not getting any specific value of $alpha$ in $M$.
Am I going wrong somewhere?
linear-algebra matrices quadratic-forms
edited Dec 13 '18 at 14:08
StubbornAtom
5,98811238
asked May 6 '18 at 15:57
Legend KillerLegend Killer
1,6621523
$endgroup$
$begingroup$
If $alpha = 0$, $det M = 0$
$endgroup$
– GNUSupporter 8964民主女神 地下教會
May 6 '18 at 15:59
1
$begingroup$
Since $M$ is almost antisymmetric, $x^T Mx=x_1 x_3$.
$endgroup$
– J.G.
May 6 '18 at 16:03
1
$begingroup$
It does not matter on the value of $alpha$ actually, because $x^TMx$ is always $x_1x_3$
$endgroup$
– Legend Killer
May 6 '18 at 16:07
$begingroup$
Where in the problem statement does it say that $Mx=b$ has a unique solution? It only says that $x$ satisfies that equation.
$endgroup$
– amd
May 6 '18 at 20:22
add a comment |
$begingroup$
If $M=begin{pmatrix}
0 & 1 & -1 \
-1 & 0 & alpha \
2 & -alpha & 0
end{pmatrix} $
and $Mx=b$, for some $x$,then find $x^TMx$.
My approach:
If $Mx=b$ has a unique solution $x$ then, $text{Rank(M)}=3$
Thus, there are three linearly independent columns in $M$.
Thus if $alpha neq 0$, $c_1(0,-1,2)+c_2(1, 0 ,-alpha)+c_3(-1, alpha ,0)=(0, 0 ,0)$ has only $c_1,c_2,c_3 =0$ as solutions.
But if $alpha=0$, then the second and third row are linearly dependent, so the $text{rank}(M)=2$ which is not possible.
But then I am not getting any specific value of $alpha$ in $M$.
Am I going wrong somewhere?
linear-algebra matrices quadratic-forms
edited Dec 13 '18 at 14:08
StubbornAtom
5,98811238
asked May 6 '18 at 15:57
Legend KillerLegend Killer
1,6621523
$endgroup$
If $M=begin{pmatrix}
0 & 1 & -1 \
-1 & 0 & alpha \
2 & -alpha & 0
end{pmatrix} $
and $Mx=b$, for some $x$,then find $x^TMx$.
My approach:
If $Mx=b$ has a unique solution $x$ then, $text{Rank(M)}=3$
Thus, there are three linearly independent columns in $M$.
Thus if $alpha neq 0$, $c_1(0,-1,2)+c_2(1, 0 ,-alpha)+c_3(-1, alpha ,0)=(0, 0 ,0)$ has only $c_1,c_2,c_3 =0$ as solutions.
But if $alpha=0$, then the second and third row are linearly dependent, so the $text{rank}(M)=2$ which is not possible.
But then I am not getting any specific value of $alpha$ in $M$.
Am I going wrong somewhere?
linear-algebra matrices quadratic-forms
linear-algebra matrices quadratic-forms
edited Dec 13 '18 at 14:08
StubbornAtom
5,98811238
asked May 6 '18 at 15:57
Legend KillerLegend Killer
1,6621523
edited Dec 13 '18 at 14:08
StubbornAtom
5,98811238
asked May 6 '18 at 15:57
Legend KillerLegend Killer
1,6621523
edited Dec 13 '18 at 14:08
StubbornAtom
5,98811238
edited Dec 13 '18 at 14:08
StubbornAtom
5,98811238
edited Dec 13 '18 at 14:08
StubbornAtom
5,98811238
5,98811238
asked May 6 '18 at 15:57
Legend KillerLegend Killer
1,6621523
asked May 6 '18 at 15:57
Legend KillerLegend Killer
1,6621523
asked May 6 '18 at 15:57
Legend KillerLegend Killer
1,6621523
1,6621523
$begingroup$
If $alpha = 0$, $det M = 0$
$endgroup$
– GNUSupporter 8964民主女神 地下教會
May 6 '18 at 15:59
1
$begingroup$
Since $M$ is almost antisymmetric, $x^T Mx=x_1 x_3$.
$endgroup$
– J.G.
May 6 '18 at 16:03
1
$begingroup$
It does not matter on the value of $alpha$ actually, because $x^TMx$ is always $x_1x_3$
$endgroup$
– Legend Killer
May 6 '18 at 16:07
$begingroup$
Where in the problem statement does it say that $Mx=b$ has a unique solution? It only says that $x$ satisfies that equation.
$endgroup$
– amd
May 6 '18 at 20:22
add a comment |
$begingroup$
If $alpha = 0$, $det M = 0$
$endgroup$
– GNUSupporter 8964民主女神 地下教會
May 6 '18 at 15:59
1
$begingroup$
Since $M$ is almost antisymmetric, $x^T Mx=x_1 x_3$.
$endgroup$
– J.G.
May 6 '18 at 16:03
1
$begingroup$
It does not matter on the value of $alpha$ actually, because $x^TMx$ is always $x_1x_3$
$endgroup$
– Legend Killer
May 6 '18 at 16:07
$begingroup$
Where in the problem statement does it say that $Mx=b$ has a unique solution? It only says that $x$ satisfies that equation.
$endgroup$
– amd
May 6 '18 at 20:22
$begingroup$
If $alpha = 0$, $det M = 0$
$endgroup$
– GNUSupporter 8964民主女神 地下教會
May 6 '18 at 15:59
$begingroup$
If $alpha = 0$, $det M = 0$
$endgroup$
– GNUSupporter 8964民主女神 地下教會
May 6 '18 at 15:59
1
1
$begingroup$
Since $M$ is almost antisymmetric, $x^T Mx=x_1 x_3$.
$endgroup$
– J.G.
May 6 '18 at 16:03
$begingroup$
Since $M$ is almost antisymmetric, $x^T Mx=x_1 x_3$.
$endgroup$
– J.G.
May 6 '18 at 16:03
1
1
$begingroup$
It does not matter on the value of $alpha$ actually, because $x^TMx$ is always $x_1x_3$
$endgroup$
– Legend Killer
May 6 '18 at 16:07
$begingroup$
It does not matter on the value of $alpha$ actually, because $x^TMx$ is always $x_1x_3$
$endgroup$
– Legend Killer
May 6 '18 at 16:07
$begingroup$
Where in the problem statement does it say that $Mx=b$ has a unique solution? It only says that $x$ satisfies that equation.
$endgroup$
– amd
May 6 '18 at 20:22
$begingroup$
Where in the problem statement does it say that $Mx=b$ has a unique solution? It only says that $x$ satisfies that equation.
$endgroup$
– amd
May 6 '18 at 20:22
add a comment |
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$begingroup$
If $alpha = 0$, $det M = 0$
$endgroup$
– GNUSupporter 8964民主女神 地下教會
May 6 '18 at 15:59
1
$begingroup$
Since $M$ is almost antisymmetric, $x^T Mx=x_1 x_3$.
$endgroup$
– J.G.
May 6 '18 at 16:03
1
$begingroup$
It does not matter on the value of $alpha$ actually, because $x^TMx$ is always $x_1x_3$
$endgroup$
– Legend Killer
May 6 '18 at 16:07
$begingroup$
Where in the problem statement does it say that $Mx=b$ has a unique solution? It only says that $x$ satisfies that equation.
$endgroup$
– amd
May 6 '18 at 20:22