Largest value of determinant












3












$begingroup$



If $alpha,beta,gamma in [-3,10].$ Then largest value of the determinant



$$begin{vmatrix}3alpha^2&beta^2+alphabeta+alpha^2&gamma^2+alphagamma+alpha^2\\
alpha^2+alphabeta+beta^2& 3beta^2&gamma^2+betagamma+beta^2\\
alpha^2+alphagamma+gamma^2& beta^2+betagamma+gamma^2&3gamma^2end{vmatrix}$$




Try: I am trying to break that determinant into product of 2 determinants but not able to break it.



Could someone help me in this question? Thanks.






linear-algebra determinant







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$endgroup$

















    3












    $begingroup$



    If $alpha,beta,gamma in [-3,10].$ Then largest value of the determinant



    $$begin{vmatrix}3alpha^2&beta^2+alphabeta+alpha^2&gamma^2+alphagamma+alpha^2\\
    alpha^2+alphabeta+beta^2& 3beta^2&gamma^2+betagamma+beta^2\\
    alpha^2+alphagamma+gamma^2& beta^2+betagamma+gamma^2&3gamma^2end{vmatrix}$$




    Try: I am trying to break that determinant into product of 2 determinants but not able to break it.



    Could someone help me in this question? Thanks.






    linear-algebra determinant







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$



      If $alpha,beta,gamma in [-3,10].$ Then largest value of the determinant



      $$begin{vmatrix}3alpha^2&beta^2+alphabeta+alpha^2&gamma^2+alphagamma+alpha^2\\
      alpha^2+alphabeta+beta^2& 3beta^2&gamma^2+betagamma+beta^2\\
      alpha^2+alphagamma+gamma^2& beta^2+betagamma+gamma^2&3gamma^2end{vmatrix}$$




      Try: I am trying to break that determinant into product of 2 determinants but not able to break it.



      Could someone help me in this question? Thanks.






      linear-algebra determinant







      share|cite|improve this question











      $endgroup$





      If $alpha,beta,gamma in [-3,10].$ Then largest value of the determinant



      $$begin{vmatrix}3alpha^2&beta^2+alphabeta+alpha^2&gamma^2+alphagamma+alpha^2\\
      alpha^2+alphabeta+beta^2& 3beta^2&gamma^2+betagamma+beta^2\\
      alpha^2+alphagamma+gamma^2& beta^2+betagamma+gamma^2&3gamma^2end{vmatrix}$$




      Try: I am trying to break that determinant into product of 2 determinants but not able to break it.



      Could someone help me in this question? Thanks.






      linear-algebra determinant




      linear-algebra determinant






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      edited 1 hour ago









      StubbornAtom

      6,74931442




      6,74931442










      asked 1 hour ago









      DXTDXT

      5,9812733




      5,9812733






















          2 Answers
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          3












          $begingroup$

          Hint: Using the rules of SARRUS we get after simplifying $$- left( alpha-gamma right) ^{2} left( beta-gamma right) ^{2}
          left( beta-alpha right) ^{2}
          $$






          share|cite|improve this answer









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          • $begingroup$
            I think so! What result do you get?
            $endgroup$
            – Dr. Sonnhard Graubner
            57 mins ago










          • $begingroup$
            This is nice to hear!
            $endgroup$
            – Dr. Sonnhard Graubner
            55 mins ago










          • $begingroup$
            A big typo is only a typo, big or not!
            $endgroup$
            – Dr. Sonnhard Graubner
            55 mins ago



















          1












          $begingroup$

          Hint: The matrix is symmetric. You can use the Cholesky decomposition to rewrite the matrix $A$ as $A=LL^T$, in which $L$ is a lower triangular matrix. Then the determinant is $det A = det L det L^T = det^2 L$. Then $det L$ is given by the product of the diagonal elements of $L$ as it is a lower triangular matrix.






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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Hint: Using the rules of SARRUS we get after simplifying $$- left( alpha-gamma right) ^{2} left( beta-gamma right) ^{2}
            left( beta-alpha right) ^{2}
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I think so! What result do you get?
              $endgroup$
              – Dr. Sonnhard Graubner
              57 mins ago










            • $begingroup$
              This is nice to hear!
              $endgroup$
              – Dr. Sonnhard Graubner
              55 mins ago










            • $begingroup$
              A big typo is only a typo, big or not!
              $endgroup$
              – Dr. Sonnhard Graubner
              55 mins ago
















            3












            $begingroup$

            Hint: Using the rules of SARRUS we get after simplifying $$- left( alpha-gamma right) ^{2} left( beta-gamma right) ^{2}
            left( beta-alpha right) ^{2}
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I think so! What result do you get?
              $endgroup$
              – Dr. Sonnhard Graubner
              57 mins ago










            • $begingroup$
              This is nice to hear!
              $endgroup$
              – Dr. Sonnhard Graubner
              55 mins ago










            • $begingroup$
              A big typo is only a typo, big or not!
              $endgroup$
              – Dr. Sonnhard Graubner
              55 mins ago














            3












            3








            3





            $begingroup$

            Hint: Using the rules of SARRUS we get after simplifying $$- left( alpha-gamma right) ^{2} left( beta-gamma right) ^{2}
            left( beta-alpha right) ^{2}
            $$






            share|cite|improve this answer









            $endgroup$



            Hint: Using the rules of SARRUS we get after simplifying $$- left( alpha-gamma right) ^{2} left( beta-gamma right) ^{2}
            left( beta-alpha right) ^{2}
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

            79.9k42867




            79.9k42867












            • $begingroup$
              I think so! What result do you get?
              $endgroup$
              – Dr. Sonnhard Graubner
              57 mins ago










            • $begingroup$
              This is nice to hear!
              $endgroup$
              – Dr. Sonnhard Graubner
              55 mins ago










            • $begingroup$
              A big typo is only a typo, big or not!
              $endgroup$
              – Dr. Sonnhard Graubner
              55 mins ago


















            • $begingroup$
              I think so! What result do you get?
              $endgroup$
              – Dr. Sonnhard Graubner
              57 mins ago










            • $begingroup$
              This is nice to hear!
              $endgroup$
              – Dr. Sonnhard Graubner
              55 mins ago










            • $begingroup$
              A big typo is only a typo, big or not!
              $endgroup$
              – Dr. Sonnhard Graubner
              55 mins ago
















            $begingroup$
            I think so! What result do you get?
            $endgroup$
            – Dr. Sonnhard Graubner
            57 mins ago




            $begingroup$
            I think so! What result do you get?
            $endgroup$
            – Dr. Sonnhard Graubner
            57 mins ago












            $begingroup$
            This is nice to hear!
            $endgroup$
            – Dr. Sonnhard Graubner
            55 mins ago




            $begingroup$
            This is nice to hear!
            $endgroup$
            – Dr. Sonnhard Graubner
            55 mins ago












            $begingroup$
            A big typo is only a typo, big or not!
            $endgroup$
            – Dr. Sonnhard Graubner
            55 mins ago




            $begingroup$
            A big typo is only a typo, big or not!
            $endgroup$
            – Dr. Sonnhard Graubner
            55 mins ago











            1












            $begingroup$

            Hint: The matrix is symmetric. You can use the Cholesky decomposition to rewrite the matrix $A$ as $A=LL^T$, in which $L$ is a lower triangular matrix. Then the determinant is $det A = det L det L^T = det^2 L$. Then $det L$ is given by the product of the diagonal elements of $L$ as it is a lower triangular matrix.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Hint: The matrix is symmetric. You can use the Cholesky decomposition to rewrite the matrix $A$ as $A=LL^T$, in which $L$ is a lower triangular matrix. Then the determinant is $det A = det L det L^T = det^2 L$. Then $det L$ is given by the product of the diagonal elements of $L$ as it is a lower triangular matrix.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Hint: The matrix is symmetric. You can use the Cholesky decomposition to rewrite the matrix $A$ as $A=LL^T$, in which $L$ is a lower triangular matrix. Then the determinant is $det A = det L det L^T = det^2 L$. Then $det L$ is given by the product of the diagonal elements of $L$ as it is a lower triangular matrix.






                share|cite|improve this answer









                $endgroup$



                Hint: The matrix is symmetric. You can use the Cholesky decomposition to rewrite the matrix $A$ as $A=LL^T$, in which $L$ is a lower triangular matrix. Then the determinant is $det A = det L det L^T = det^2 L$. Then $det L$ is given by the product of the diagonal elements of $L$ as it is a lower triangular matrix.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                MachineLearnerMachineLearner

                1,906213




                1,906213






























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