$A=pmatrix1&2&3&4&5\2&3&4&5&6.$ Find $det(A^TA)$. [duplicate]

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  • When A and B are of different order given the $det(AB)$,then calculate $det(BA)$

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Suppose $$A=pmatrix1&2&3&4&5\2&3&4&5&6$$ Find $det(A^TA)$.




I know exactly how to calculate it by writing it as a $5times5$ matrix. But how to calculate it smartly?










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  • When A and B are of different order given the $det(AB)$,then calculate $det(BA)$

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Suppose $$A=pmatrix1&2&3&4&5\2&3&4&5&6$$ Find $det(A^TA)$.




I know exactly how to calculate it by writing it as a $5times5$ matrix. But how to calculate it smartly?










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  • Related : math.stackexchange.com/questions/1782600/…
    – Arnaud D.
    Aug 19 at 15:10












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This question already has an answer here:



  • When A and B are of different order given the $det(AB)$,then calculate $det(BA)$

    3 answers




Suppose $$A=pmatrix1&2&3&4&5\2&3&4&5&6$$ Find $det(A^TA)$.




I know exactly how to calculate it by writing it as a $5times5$ matrix. But how to calculate it smartly?










share|cite|improve this question
















This question already has an answer here:



  • When A and B are of different order given the $det(AB)$,then calculate $det(BA)$

    3 answers




Suppose $$A=pmatrix1&2&3&4&5\2&3&4&5&6$$ Find $det(A^TA)$.




I know exactly how to calculate it by writing it as a $5times5$ matrix. But how to calculate it smartly?





This question already has an answer here:



  • When A and B are of different order given the $det(AB)$,then calculate $det(BA)$

    3 answers







linear-algebra matrices determinant






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edited Aug 19 at 22:54









amWhy

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asked Aug 19 at 14:45









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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • Related : math.stackexchange.com/questions/1782600/…
    – Arnaud D.
    Aug 19 at 15:10
















  • Related : math.stackexchange.com/questions/1782600/…
    – Arnaud D.
    Aug 19 at 15:10















Related : math.stackexchange.com/questions/1782600/…
– Arnaud D.
Aug 19 at 15:10




Related : math.stackexchange.com/questions/1782600/…
– Arnaud D.
Aug 19 at 15:10










1 Answer
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up vote
30
down vote



accepted










  • Rank of $A$ is $2$, hence $A^TA$ cannot be rank $5$ and it must be singular.


  • Hence the determinant must be $0$.






share|cite|improve this answer



























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    30
    down vote



    accepted










    • Rank of $A$ is $2$, hence $A^TA$ cannot be rank $5$ and it must be singular.


    • Hence the determinant must be $0$.






    share|cite|improve this answer
























      up vote
      30
      down vote



      accepted










      • Rank of $A$ is $2$, hence $A^TA$ cannot be rank $5$ and it must be singular.


      • Hence the determinant must be $0$.






      share|cite|improve this answer






















        up vote
        30
        down vote



        accepted







        up vote
        30
        down vote



        accepted






        • Rank of $A$ is $2$, hence $A^TA$ cannot be rank $5$ and it must be singular.


        • Hence the determinant must be $0$.






        share|cite|improve this answer












        • Rank of $A$ is $2$, hence $A^TA$ cannot be rank $5$ and it must be singular.


        • Hence the determinant must be $0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 19 at 14:49









        Siong Thye Goh

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        82.7k1456104












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