Determinant of a particular matrix.












12












$begingroup$


What is the best way to find determinant of the following matrix?



$$A=left(begin{matrix}
1&ax&a^2+x^2\1&ay&a^2+y^2\ 1&az&a^2+z^2
end{matrix}right)$$



I thought it looks like a Vandermonde matrix, but not exactly. I can't use $|A+B|=|A|+|B|$ to form a Vandermonde matrix. Please suggest. Thanks.






linear-algebra matrices determinant







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Did you mean $vert A + B vert = vert A vert + vert B vert$?
    $endgroup$
    – Bladewood
    Feb 15 at 22:20


















12












$begingroup$


What is the best way to find determinant of the following matrix?



$$A=left(begin{matrix}
1&ax&a^2+x^2\1&ay&a^2+y^2\ 1&az&a^2+z^2
end{matrix}right)$$



I thought it looks like a Vandermonde matrix, but not exactly. I can't use $|A+B|=|A|+|B|$ to form a Vandermonde matrix. Please suggest. Thanks.






linear-algebra matrices determinant







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Did you mean $vert A + B vert = vert A vert + vert B vert$?
    $endgroup$
    – Bladewood
    Feb 15 at 22:20
















12












12








12


1



$begingroup$


What is the best way to find determinant of the following matrix?



$$A=left(begin{matrix}
1&ax&a^2+x^2\1&ay&a^2+y^2\ 1&az&a^2+z^2
end{matrix}right)$$



I thought it looks like a Vandermonde matrix, but not exactly. I can't use $|A+B|=|A|+|B|$ to form a Vandermonde matrix. Please suggest. Thanks.






linear-algebra matrices determinant







share|cite|improve this question











$endgroup$




What is the best way to find determinant of the following matrix?



$$A=left(begin{matrix}
1&ax&a^2+x^2\1&ay&a^2+y^2\ 1&az&a^2+z^2
end{matrix}right)$$



I thought it looks like a Vandermonde matrix, but not exactly. I can't use $|A+B|=|A|+|B|$ to form a Vandermonde matrix. Please suggest. Thanks.






linear-algebra matrices determinant




linear-algebra matrices determinant






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 16 at 13:53









Rodrigo de Azevedo

13k41959




13k41959










asked Feb 15 at 16:30









neelkanthneelkanth

2,27621029




2,27621029








  • 1




    $begingroup$
    Did you mean $vert A + B vert = vert A vert + vert B vert$?
    $endgroup$
    – Bladewood
    Feb 15 at 22:20
















  • 1




    $begingroup$
    Did you mean $vert A + B vert = vert A vert + vert B vert$?
    $endgroup$
    – Bladewood
    Feb 15 at 22:20










1




1




$begingroup$
Did you mean $vert A + B vert = vert A vert + vert B vert$?
$endgroup$
– Bladewood
Feb 15 at 22:20






$begingroup$
Did you mean $vert A + B vert = vert A vert + vert B vert$?
$endgroup$
– Bladewood
Feb 15 at 22:20












3 Answers
3






active

oldest

votes


















23












$begingroup$

begin{align}
&|A|\
&=detleft(begin{matrix}
1&ax&a^2+x^2\1&ay&a^2+y^2\ 1&az&a^2+z^2
end{matrix}right) \
&=begin{vmatrix}
1&ax&a^2\1&ay&a^2\ 1&az&a^2
end{vmatrix}+begin{vmatrix}
1&ax&x^2\1&ay&y^2\ 1&az&z^2
end{vmatrix} tag{multilinearity on 3rd column} \
&=0+abegin{vmatrix}
1&x&x^2\1&y&y^2\ 1&z&z^2
end{vmatrix} \
&= a (x-y)(y-z)(z-x)
end{align}






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    i am confused which answer to be accepted ....all are nice....
    $endgroup$
    – neelkanth
    Feb 15 at 16:41






  • 1




    $begingroup$
    @neelkanth I suggest you throw a dice.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 15 at 16:42






  • 1




    $begingroup$
    I rolled I dice three time for better .... got three different choices every time .
    $endgroup$
    – neelkanth
    Feb 15 at 16:47






  • 1




    $begingroup$
    But that requires more resources. :P So, you save time but you have to have multiple dice then. :P
    $endgroup$
    – stressed out
    Feb 15 at 22:58






  • 1




    $begingroup$
    @stressedout You're right. That's why I say if possible. Perhaps OP was busy with a simulation of the Law of Large Number that he couldn't decide which answer to accept. :xd
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 15 at 23:01



















16












$begingroup$

Note that
$$
detleft(begin{matrix}
1&ax&a^2+x^2\1&ay&a^2+y^2\ 1&az&a^2+z^2
end{matrix}right) =det left(begin{matrix}
1&ax&x^2\1&ay&y^2\ 1&az&z^2
end{matrix}right)=acdotdet left(begin{matrix}
1&x&x^2\1&y&y^2\ 1&z&z^2
end{matrix}right)
$$ which boils down to Vandermonde determinant
$$
a(x-y)(y-z)(z-x).
$$






share|cite|improve this answer









$endgroup$





















    6












    $begingroup$

    I think the best way it's the following:
    $$Delta=sum_{cyc}(ay(a^2+z^2)-ay(a^2+x^2))=asum_{cyc}(x^2z-x^2y)=a(x-y)(y-z)(z-x).$$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      23












      $begingroup$

      begin{align}
      &|A|\
      &=detleft(begin{matrix}
      1&ax&a^2+x^2\1&ay&a^2+y^2\ 1&az&a^2+z^2
      end{matrix}right) \
      &=begin{vmatrix}
      1&ax&a^2\1&ay&a^2\ 1&az&a^2
      end{vmatrix}+begin{vmatrix}
      1&ax&x^2\1&ay&y^2\ 1&az&z^2
      end{vmatrix} tag{multilinearity on 3rd column} \
      &=0+abegin{vmatrix}
      1&x&x^2\1&y&y^2\ 1&z&z^2
      end{vmatrix} \
      &= a (x-y)(y-z)(z-x)
      end{align}






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        i am confused which answer to be accepted ....all are nice....
        $endgroup$
        – neelkanth
        Feb 15 at 16:41






      • 1




        $begingroup$
        @neelkanth I suggest you throw a dice.
        $endgroup$
        – GNUSupporter 8964民主女神 地下教會
        Feb 15 at 16:42






      • 1




        $begingroup$
        I rolled I dice three time for better .... got three different choices every time .
        $endgroup$
        – neelkanth
        Feb 15 at 16:47






      • 1




        $begingroup$
        But that requires more resources. :P So, you save time but you have to have multiple dice then. :P
        $endgroup$
        – stressed out
        Feb 15 at 22:58






      • 1




        $begingroup$
        @stressedout You're right. That's why I say if possible. Perhaps OP was busy with a simulation of the Law of Large Number that he couldn't decide which answer to accept. :xd
        $endgroup$
        – GNUSupporter 8964民主女神 地下教會
        Feb 15 at 23:01
















      23












      $begingroup$

      begin{align}
      &|A|\
      &=detleft(begin{matrix}
      1&ax&a^2+x^2\1&ay&a^2+y^2\ 1&az&a^2+z^2
      end{matrix}right) \
      &=begin{vmatrix}
      1&ax&a^2\1&ay&a^2\ 1&az&a^2
      end{vmatrix}+begin{vmatrix}
      1&ax&x^2\1&ay&y^2\ 1&az&z^2
      end{vmatrix} tag{multilinearity on 3rd column} \
      &=0+abegin{vmatrix}
      1&x&x^2\1&y&y^2\ 1&z&z^2
      end{vmatrix} \
      &= a (x-y)(y-z)(z-x)
      end{align}






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        i am confused which answer to be accepted ....all are nice....
        $endgroup$
        – neelkanth
        Feb 15 at 16:41






      • 1




        $begingroup$
        @neelkanth I suggest you throw a dice.
        $endgroup$
        – GNUSupporter 8964民主女神 地下教會
        Feb 15 at 16:42






      • 1




        $begingroup$
        I rolled I dice three time for better .... got three different choices every time .
        $endgroup$
        – neelkanth
        Feb 15 at 16:47






      • 1




        $begingroup$
        But that requires more resources. :P So, you save time but you have to have multiple dice then. :P
        $endgroup$
        – stressed out
        Feb 15 at 22:58






      • 1




        $begingroup$
        @stressedout You're right. That's why I say if possible. Perhaps OP was busy with a simulation of the Law of Large Number that he couldn't decide which answer to accept. :xd
        $endgroup$
        – GNUSupporter 8964民主女神 地下教會
        Feb 15 at 23:01














      23












      23








      23





      $begingroup$

      begin{align}
      &|A|\
      &=detleft(begin{matrix}
      1&ax&a^2+x^2\1&ay&a^2+y^2\ 1&az&a^2+z^2
      end{matrix}right) \
      &=begin{vmatrix}
      1&ax&a^2\1&ay&a^2\ 1&az&a^2
      end{vmatrix}+begin{vmatrix}
      1&ax&x^2\1&ay&y^2\ 1&az&z^2
      end{vmatrix} tag{multilinearity on 3rd column} \
      &=0+abegin{vmatrix}
      1&x&x^2\1&y&y^2\ 1&z&z^2
      end{vmatrix} \
      &= a (x-y)(y-z)(z-x)
      end{align}






      share|cite|improve this answer









      $endgroup$



      begin{align}
      &|A|\
      &=detleft(begin{matrix}
      1&ax&a^2+x^2\1&ay&a^2+y^2\ 1&az&a^2+z^2
      end{matrix}right) \
      &=begin{vmatrix}
      1&ax&a^2\1&ay&a^2\ 1&az&a^2
      end{vmatrix}+begin{vmatrix}
      1&ax&x^2\1&ay&y^2\ 1&az&z^2
      end{vmatrix} tag{multilinearity on 3rd column} \
      &=0+abegin{vmatrix}
      1&x&x^2\1&y&y^2\ 1&z&z^2
      end{vmatrix} \
      &= a (x-y)(y-z)(z-x)
      end{align}







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Feb 15 at 16:35









      GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會

      13.3k72549




      13.3k72549








      • 1




        $begingroup$
        i am confused which answer to be accepted ....all are nice....
        $endgroup$
        – neelkanth
        Feb 15 at 16:41






      • 1




        $begingroup$
        @neelkanth I suggest you throw a dice.
        $endgroup$
        – GNUSupporter 8964民主女神 地下教會
        Feb 15 at 16:42






      • 1




        $begingroup$
        I rolled I dice three time for better .... got three different choices every time .
        $endgroup$
        – neelkanth
        Feb 15 at 16:47






      • 1




        $begingroup$
        But that requires more resources. :P So, you save time but you have to have multiple dice then. :P
        $endgroup$
        – stressed out
        Feb 15 at 22:58






      • 1




        $begingroup$
        @stressedout You're right. That's why I say if possible. Perhaps OP was busy with a simulation of the Law of Large Number that he couldn't decide which answer to accept. :xd
        $endgroup$
        – GNUSupporter 8964民主女神 地下教會
        Feb 15 at 23:01














      • 1




        $begingroup$
        i am confused which answer to be accepted ....all are nice....
        $endgroup$
        – neelkanth
        Feb 15 at 16:41






      • 1




        $begingroup$
        @neelkanth I suggest you throw a dice.
        $endgroup$
        – GNUSupporter 8964民主女神 地下教會
        Feb 15 at 16:42






      • 1




        $begingroup$
        I rolled I dice three time for better .... got three different choices every time .
        $endgroup$
        – neelkanth
        Feb 15 at 16:47






      • 1




        $begingroup$
        But that requires more resources. :P So, you save time but you have to have multiple dice then. :P
        $endgroup$
        – stressed out
        Feb 15 at 22:58






      • 1




        $begingroup$
        @stressedout You're right. That's why I say if possible. Perhaps OP was busy with a simulation of the Law of Large Number that he couldn't decide which answer to accept. :xd
        $endgroup$
        – GNUSupporter 8964民主女神 地下教會
        Feb 15 at 23:01








      1




      1




      $begingroup$
      i am confused which answer to be accepted ....all are nice....
      $endgroup$
      – neelkanth
      Feb 15 at 16:41




      $begingroup$
      i am confused which answer to be accepted ....all are nice....
      $endgroup$
      – neelkanth
      Feb 15 at 16:41




      1




      1




      $begingroup$
      @neelkanth I suggest you throw a dice.
      $endgroup$
      – GNUSupporter 8964民主女神 地下教會
      Feb 15 at 16:42




      $begingroup$
      @neelkanth I suggest you throw a dice.
      $endgroup$
      – GNUSupporter 8964民主女神 地下教會
      Feb 15 at 16:42




      1




      1




      $begingroup$
      I rolled I dice three time for better .... got three different choices every time .
      $endgroup$
      – neelkanth
      Feb 15 at 16:47




      $begingroup$
      I rolled I dice three time for better .... got three different choices every time .
      $endgroup$
      – neelkanth
      Feb 15 at 16:47




      1




      1




      $begingroup$
      But that requires more resources. :P So, you save time but you have to have multiple dice then. :P
      $endgroup$
      – stressed out
      Feb 15 at 22:58




      $begingroup$
      But that requires more resources. :P So, you save time but you have to have multiple dice then. :P
      $endgroup$
      – stressed out
      Feb 15 at 22:58




      1




      1




      $begingroup$
      @stressedout You're right. That's why I say if possible. Perhaps OP was busy with a simulation of the Law of Large Number that he couldn't decide which answer to accept. :xd
      $endgroup$
      – GNUSupporter 8964民主女神 地下教會
      Feb 15 at 23:01




      $begingroup$
      @stressedout You're right. That's why I say if possible. Perhaps OP was busy with a simulation of the Law of Large Number that he couldn't decide which answer to accept. :xd
      $endgroup$
      – GNUSupporter 8964民主女神 地下教會
      Feb 15 at 23:01











      16












      $begingroup$

      Note that
      $$
      detleft(begin{matrix}
      1&ax&a^2+x^2\1&ay&a^2+y^2\ 1&az&a^2+z^2
      end{matrix}right) =det left(begin{matrix}
      1&ax&x^2\1&ay&y^2\ 1&az&z^2
      end{matrix}right)=acdotdet left(begin{matrix}
      1&x&x^2\1&y&y^2\ 1&z&z^2
      end{matrix}right)
      $$       which boils down to Vandermonde determinant
      $$
      a(x-y)(y-z)(z-x).
      $$






      share|cite|improve this answer









      $endgroup$


















        16












        $begingroup$

        Note that
        $$
        detleft(begin{matrix}
        1&ax&a^2+x^2\1&ay&a^2+y^2\ 1&az&a^2+z^2
        end{matrix}right) =det left(begin{matrix}
        1&ax&x^2\1&ay&y^2\ 1&az&z^2
        end{matrix}right)=acdotdet left(begin{matrix}
        1&x&x^2\1&y&y^2\ 1&z&z^2
        end{matrix}right)
        $$         which boils down to Vandermonde determinant
        $$
        a(x-y)(y-z)(z-x).
        $$






        share|cite|improve this answer









        $endgroup$
















          16












          16








          16





          $begingroup$

          Note that
          $$
          detleft(begin{matrix}
          1&ax&a^2+x^2\1&ay&a^2+y^2\ 1&az&a^2+z^2
          end{matrix}right) =det left(begin{matrix}
          1&ax&x^2\1&ay&y^2\ 1&az&z^2
          end{matrix}right)=acdotdet left(begin{matrix}
          1&x&x^2\1&y&y^2\ 1&z&z^2
          end{matrix}right)
          $$           which boils down to Vandermonde determinant
          $$
          a(x-y)(y-z)(z-x).
          $$






          share|cite|improve this answer









          $endgroup$



          Note that
          $$
          detleft(begin{matrix}
          1&ax&a^2+x^2\1&ay&a^2+y^2\ 1&az&a^2+z^2
          end{matrix}right) =det left(begin{matrix}
          1&ax&x^2\1&ay&y^2\ 1&az&z^2
          end{matrix}right)=acdotdet left(begin{matrix}
          1&x&x^2\1&y&y^2\ 1&z&z^2
          end{matrix}right)
          $$           which boils down to Vandermonde determinant
          $$
          a(x-y)(y-z)(z-x).
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 15 at 16:36









          SongSong

          14.4k1635




          14.4k1635























              6












              $begingroup$

              I think the best way it's the following:
              $$Delta=sum_{cyc}(ay(a^2+z^2)-ay(a^2+x^2))=asum_{cyc}(x^2z-x^2y)=a(x-y)(y-z)(z-x).$$






              share|cite|improve this answer









              $endgroup$


















                6












                $begingroup$

                I think the best way it's the following:
                $$Delta=sum_{cyc}(ay(a^2+z^2)-ay(a^2+x^2))=asum_{cyc}(x^2z-x^2y)=a(x-y)(y-z)(z-x).$$






                share|cite|improve this answer









                $endgroup$
















                  6












                  6








                  6





                  $begingroup$

                  I think the best way it's the following:
                  $$Delta=sum_{cyc}(ay(a^2+z^2)-ay(a^2+x^2))=asum_{cyc}(x^2z-x^2y)=a(x-y)(y-z)(z-x).$$






                  share|cite|improve this answer









                  $endgroup$



                  I think the best way it's the following:
                  $$Delta=sum_{cyc}(ay(a^2+z^2)-ay(a^2+x^2))=asum_{cyc}(x^2z-x^2y)=a(x-y)(y-z)(z-x).$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 15 at 16:36









                  Michael RozenbergMichael Rozenberg

                  104k1892197




                  104k1892197






























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                      Is it possible to collect Nectar points via Trainline?

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                      2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
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