Let $A= begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}$. Find the entry in the first row and second column...












4












$begingroup$


I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.

Where:
$P=begin{pmatrix} 1&1 \ -4&-1 end{pmatrix}$
$D=begin{pmatrix} 0&0 \ 0&6 end{pmatrix}$
$P^{-1}=frac{1}{3}begin{pmatrix} -1&-1 \ 4&1 end{pmatrix}$.



So that $A^{2014}=PD^{2014}P^{-1}$

But i just want to know whether there is an alternate method.






linear-algebra eigenvalues-eigenvectors diagonalization







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is the standard elegant way to do it.I don't know another method other than brute force.
    $endgroup$
    – Ethan Bolker
    Jan 6 at 0:19






  • 1




    $begingroup$
    I think that what you did is very well, other ways will come to the decomposition anyways
    $endgroup$
    – José Alejandro Aburto Araneda
    Jan 6 at 0:19
















4












$begingroup$


I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.

Where:
$P=begin{pmatrix} 1&1 \ -4&-1 end{pmatrix}$
$D=begin{pmatrix} 0&0 \ 0&6 end{pmatrix}$
$P^{-1}=frac{1}{3}begin{pmatrix} -1&-1 \ 4&1 end{pmatrix}$.



So that $A^{2014}=PD^{2014}P^{-1}$

But i just want to know whether there is an alternate method.






linear-algebra eigenvalues-eigenvectors diagonalization







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is the standard elegant way to do it.I don't know another method other than brute force.
    $endgroup$
    – Ethan Bolker
    Jan 6 at 0:19






  • 1




    $begingroup$
    I think that what you did is very well, other ways will come to the decomposition anyways
    $endgroup$
    – José Alejandro Aburto Araneda
    Jan 6 at 0:19














4












4








4


2



$begingroup$


I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.

Where:
$P=begin{pmatrix} 1&1 \ -4&-1 end{pmatrix}$
$D=begin{pmatrix} 0&0 \ 0&6 end{pmatrix}$
$P^{-1}=frac{1}{3}begin{pmatrix} -1&-1 \ 4&1 end{pmatrix}$.



So that $A^{2014}=PD^{2014}P^{-1}$

But i just want to know whether there is an alternate method.






linear-algebra eigenvalues-eigenvectors diagonalization







share|cite|improve this question











$endgroup$




I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.

Where:
$P=begin{pmatrix} 1&1 \ -4&-1 end{pmatrix}$
$D=begin{pmatrix} 0&0 \ 0&6 end{pmatrix}$
$P^{-1}=frac{1}{3}begin{pmatrix} -1&-1 \ 4&1 end{pmatrix}$.



So that $A^{2014}=PD^{2014}P^{-1}$

But i just want to know whether there is an alternate method.






linear-algebra eigenvalues-eigenvectors diagonalization




linear-algebra eigenvalues-eigenvectors diagonalization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 0:18









vadim123

76k897189




76k897189










asked Jan 6 at 0:16









DD90DD90

2648




2648








  • 1




    $begingroup$
    This is the standard elegant way to do it.I don't know another method other than brute force.
    $endgroup$
    – Ethan Bolker
    Jan 6 at 0:19






  • 1




    $begingroup$
    I think that what you did is very well, other ways will come to the decomposition anyways
    $endgroup$
    – José Alejandro Aburto Araneda
    Jan 6 at 0:19














  • 1




    $begingroup$
    This is the standard elegant way to do it.I don't know another method other than brute force.
    $endgroup$
    – Ethan Bolker
    Jan 6 at 0:19






  • 1




    $begingroup$
    I think that what you did is very well, other ways will come to the decomposition anyways
    $endgroup$
    – José Alejandro Aburto Araneda
    Jan 6 at 0:19








1




1




$begingroup$
This is the standard elegant way to do it.I don't know another method other than brute force.
$endgroup$
– Ethan Bolker
Jan 6 at 0:19




$begingroup$
This is the standard elegant way to do it.I don't know another method other than brute force.
$endgroup$
– Ethan Bolker
Jan 6 at 0:19




1




1




$begingroup$
I think that what you did is very well, other ways will come to the decomposition anyways
$endgroup$
– José Alejandro Aburto Araneda
Jan 6 at 0:19




$begingroup$
I think that what you did is very well, other ways will come to the decomposition anyways
$endgroup$
– José Alejandro Aburto Araneda
Jan 6 at 0:19










3 Answers
3






active

oldest

votes


















5












$begingroup$

Hint Show that
$$A^2=6A$$



Note: You can show the stronger statement:
$$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
but this is overkill.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
    $endgroup$
    – Mindlack
    Jan 6 at 0:25










  • $begingroup$
    Very nice indeed!
    $endgroup$
    – Mike
    Jan 6 at 0:33










  • $begingroup$
    Can you please explain it a little further how to proceed with this
    $endgroup$
    – DD90
    Jan 6 at 0:35






  • 1




    $begingroup$
    If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
    $endgroup$
    – Mindlack
    Jan 6 at 0:52










  • $begingroup$
    @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
    $endgroup$
    – N. S.
    Jan 6 at 3:34



















2












$begingroup$

By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.



Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
    $$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$



    Which now immediately gives you
    $$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      Hint Show that
      $$A^2=6A$$



      Note: You can show the stronger statement:
      $$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
      but this is overkill.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
        $endgroup$
        – Mindlack
        Jan 6 at 0:25










      • $begingroup$
        Very nice indeed!
        $endgroup$
        – Mike
        Jan 6 at 0:33










      • $begingroup$
        Can you please explain it a little further how to proceed with this
        $endgroup$
        – DD90
        Jan 6 at 0:35






      • 1




        $begingroup$
        If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
        $endgroup$
        – Mindlack
        Jan 6 at 0:52










      • $begingroup$
        @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
        $endgroup$
        – N. S.
        Jan 6 at 3:34
















      5












      $begingroup$

      Hint Show that
      $$A^2=6A$$



      Note: You can show the stronger statement:
      $$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
      but this is overkill.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
        $endgroup$
        – Mindlack
        Jan 6 at 0:25










      • $begingroup$
        Very nice indeed!
        $endgroup$
        – Mike
        Jan 6 at 0:33










      • $begingroup$
        Can you please explain it a little further how to proceed with this
        $endgroup$
        – DD90
        Jan 6 at 0:35






      • 1




        $begingroup$
        If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
        $endgroup$
        – Mindlack
        Jan 6 at 0:52










      • $begingroup$
        @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
        $endgroup$
        – N. S.
        Jan 6 at 3:34














      5












      5








      5





      $begingroup$

      Hint Show that
      $$A^2=6A$$



      Note: You can show the stronger statement:
      $$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
      but this is overkill.






      share|cite|improve this answer









      $endgroup$



      Hint Show that
      $$A^2=6A$$



      Note: You can show the stronger statement:
      $$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
      but this is overkill.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 6 at 0:23









      N. S.N. S.

      103k6112208




      103k6112208












      • $begingroup$
        This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
        $endgroup$
        – Mindlack
        Jan 6 at 0:25










      • $begingroup$
        Very nice indeed!
        $endgroup$
        – Mike
        Jan 6 at 0:33










      • $begingroup$
        Can you please explain it a little further how to proceed with this
        $endgroup$
        – DD90
        Jan 6 at 0:35






      • 1




        $begingroup$
        If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
        $endgroup$
        – Mindlack
        Jan 6 at 0:52










      • $begingroup$
        @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
        $endgroup$
        – N. S.
        Jan 6 at 3:34


















      • $begingroup$
        This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
        $endgroup$
        – Mindlack
        Jan 6 at 0:25










      • $begingroup$
        Very nice indeed!
        $endgroup$
        – Mike
        Jan 6 at 0:33










      • $begingroup$
        Can you please explain it a little further how to proceed with this
        $endgroup$
        – DD90
        Jan 6 at 0:35






      • 1




        $begingroup$
        If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
        $endgroup$
        – Mindlack
        Jan 6 at 0:52










      • $begingroup$
        @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
        $endgroup$
        – N. S.
        Jan 6 at 3:34
















      $begingroup$
      This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
      $endgroup$
      – Mindlack
      Jan 6 at 0:25




      $begingroup$
      This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
      $endgroup$
      – Mindlack
      Jan 6 at 0:25












      $begingroup$
      Very nice indeed!
      $endgroup$
      – Mike
      Jan 6 at 0:33




      $begingroup$
      Very nice indeed!
      $endgroup$
      – Mike
      Jan 6 at 0:33












      $begingroup$
      Can you please explain it a little further how to proceed with this
      $endgroup$
      – DD90
      Jan 6 at 0:35




      $begingroup$
      Can you please explain it a little further how to proceed with this
      $endgroup$
      – DD90
      Jan 6 at 0:35




      1




      1




      $begingroup$
      If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
      $endgroup$
      – Mindlack
      Jan 6 at 0:52




      $begingroup$
      If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
      $endgroup$
      – Mindlack
      Jan 6 at 0:52












      $begingroup$
      @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
      $endgroup$
      – N. S.
      Jan 6 at 3:34




      $begingroup$
      @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
      $endgroup$
      – N. S.
      Jan 6 at 3:34











      2












      $begingroup$

      By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.



      Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.



        Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.



          Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.






          share|cite|improve this answer











          $endgroup$



          By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.



          Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 6 at 1:08

























          answered Jan 6 at 1:02









          amdamd

          30k21050




          30k21050























              0












              $begingroup$

              Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
              $$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$



              Which now immediately gives you
              $$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
                $$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$



                Which now immediately gives you
                $$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
                  $$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$



                  Which now immediately gives you
                  $$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$






                  share|cite|improve this answer









                  $endgroup$



                  Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
                  $$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$



                  Which now immediately gives you
                  $$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 6 at 3:26









                  obscuransobscurans

                  1,152311




                  1,152311






























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                      Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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                      0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
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                      張江高科駅

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                      張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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