Smith normal form of the following matrix












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$begingroup$


Let $$ A = begin{bmatrix} 66 & 30\ 12 & 4 end{bmatrix}$$
I've been trying to find the smith normal form of this matrix, and I keep getting the wrong answer. Here are my workings;



gcd of the entries of the first column is $6$, so we aim to get $6$ in the $(1,1)$ position via row operations. $r_1 : r_1 - 5r_2$ yields $$begin{bmatrix} 6 & 10\ 12 &4 end{bmatrix}$$



Now, the gcd of the first row is $2$, so we aim to get this in the $(1,1)$ position. $c_1 : 2c_1 - c_2$ yields $$begin{bmatrix} 2 & 10\20 &4 end{bmatrix}$$



Now, we aim to get zeroes in the $(1,2)$ and $(2,1)$ position. $c_2 : 5c_1 - c_2$ and $r_2 : r_2 - 10r_1$ yields the matrix begin{bmatrix} 2 & 0\ 0 & 96 end{bmatrix}



But the answer is supposed to be begin{bmatrix} 2 & 0\0 & 48 end{bmatrix}



But I can't see a mistake in my workings anywhere? Am I using the algorithm incorrectly?






linear-algebra abstract-algebra matrices smith-normal-form







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    0












    $begingroup$


    Let $$ A = begin{bmatrix} 66 & 30\ 12 & 4 end{bmatrix}$$
    I've been trying to find the smith normal form of this matrix, and I keep getting the wrong answer. Here are my workings;



    gcd of the entries of the first column is $6$, so we aim to get $6$ in the $(1,1)$ position via row operations. $r_1 : r_1 - 5r_2$ yields $$begin{bmatrix} 6 & 10\ 12 &4 end{bmatrix}$$



    Now, the gcd of the first row is $2$, so we aim to get this in the $(1,1)$ position. $c_1 : 2c_1 - c_2$ yields $$begin{bmatrix} 2 & 10\20 &4 end{bmatrix}$$



    Now, we aim to get zeroes in the $(1,2)$ and $(2,1)$ position. $c_2 : 5c_1 - c_2$ and $r_2 : r_2 - 10r_1$ yields the matrix begin{bmatrix} 2 & 0\ 0 & 96 end{bmatrix}



    But the answer is supposed to be begin{bmatrix} 2 & 0\0 & 48 end{bmatrix}



    But I can't see a mistake in my workings anywhere? Am I using the algorithm incorrectly?






    linear-algebra abstract-algebra matrices smith-normal-form







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $$ A = begin{bmatrix} 66 & 30\ 12 & 4 end{bmatrix}$$
      I've been trying to find the smith normal form of this matrix, and I keep getting the wrong answer. Here are my workings;



      gcd of the entries of the first column is $6$, so we aim to get $6$ in the $(1,1)$ position via row operations. $r_1 : r_1 - 5r_2$ yields $$begin{bmatrix} 6 & 10\ 12 &4 end{bmatrix}$$



      Now, the gcd of the first row is $2$, so we aim to get this in the $(1,1)$ position. $c_1 : 2c_1 - c_2$ yields $$begin{bmatrix} 2 & 10\20 &4 end{bmatrix}$$



      Now, we aim to get zeroes in the $(1,2)$ and $(2,1)$ position. $c_2 : 5c_1 - c_2$ and $r_2 : r_2 - 10r_1$ yields the matrix begin{bmatrix} 2 & 0\ 0 & 96 end{bmatrix}



      But the answer is supposed to be begin{bmatrix} 2 & 0\0 & 48 end{bmatrix}



      But I can't see a mistake in my workings anywhere? Am I using the algorithm incorrectly?






      linear-algebra abstract-algebra matrices smith-normal-form







      share|cite|improve this question









      $endgroup$




      Let $$ A = begin{bmatrix} 66 & 30\ 12 & 4 end{bmatrix}$$
      I've been trying to find the smith normal form of this matrix, and I keep getting the wrong answer. Here are my workings;



      gcd of the entries of the first column is $6$, so we aim to get $6$ in the $(1,1)$ position via row operations. $r_1 : r_1 - 5r_2$ yields $$begin{bmatrix} 6 & 10\ 12 &4 end{bmatrix}$$



      Now, the gcd of the first row is $2$, so we aim to get this in the $(1,1)$ position. $c_1 : 2c_1 - c_2$ yields $$begin{bmatrix} 2 & 10\20 &4 end{bmatrix}$$



      Now, we aim to get zeroes in the $(1,2)$ and $(2,1)$ position. $c_2 : 5c_1 - c_2$ and $r_2 : r_2 - 10r_1$ yields the matrix begin{bmatrix} 2 & 0\ 0 & 96 end{bmatrix}



      But the answer is supposed to be begin{bmatrix} 2 & 0\0 & 48 end{bmatrix}



      But I can't see a mistake in my workings anywhere? Am I using the algorithm incorrectly?






      linear-algebra abstract-algebra matrices smith-normal-form




      linear-algebra abstract-algebra matrices smith-normal-form






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      asked Jan 13 at 22:08









      the manthe man

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      768715






















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          $c_1leftarrow2c_1−c_2$ is not an elementary operation.






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            $begingroup$

            $c_1leftarrow2c_1−c_2$ is not an elementary operation.






            share|cite|improve this answer









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              $begingroup$

              $c_1leftarrow2c_1−c_2$ is not an elementary operation.






              share|cite|improve this answer









              $endgroup$
















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                $begingroup$

                $c_1leftarrow2c_1−c_2$ is not an elementary operation.






                share|cite|improve this answer









                $endgroup$



                $c_1leftarrow2c_1−c_2$ is not an elementary operation.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 13 at 22:23









                user1551user1551

                72.8k566127




                72.8k566127






























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