Determinant of adjoint

Home Forums Math Olympiad, I.S.I., C.M.I. Entrance Determinant of adjoint

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  • #32406
    Aniruddha Bardhan
    Participant

    see the attachement.

    My approach: B=adj(A) and C=adj(B). so det (B)= A^2=2^2 and det (C)=B^2=2^4.

    so det(2AB^TC)= 2^8 . I did not get any option with 2^8

    May be I am wrong kindly check.

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    #34638
    Jatin Kr Dey
    Participant

    I think you should have to apply the following property :

    $$ \displaystyle det(cA) = c^{order(A)} det(A) $$

    • This reply was modified 3 months, 1 week ago by Jatin Kr Dey.
    • This reply was modified 3 months, 1 week ago by Jatin Kr Dey.
    • This reply was modified 3 months, 1 week ago by Jatin Kr Dey.
    • This reply was modified 3 months, 1 week ago by Jatin Kr Dey.
    #34646
    Jatin Kr Dey
    Participant

    Given, $$ cofactors(b_{ij})= c_{ij}  \Rightarrow Adj(B)= C^T $$ \

    $$ cofactors(a_{ij})= b_{ij} \Rightarrow Adj(A)= B^T $$  \

    $$ det(A) = 2 $$

    and the order of the matrices is 3 .

    We have to apply the followings :

    $$ |Adj(A) = (|A|)^{order(A) – 1} $$

    $$ |A| = |A^T| $$

    $$ |ABC| = |A| |B| |C| $$

    $$ |cA| = c^{order(A)}|A| $$

    So now $$ |Adj(A)| = 2^{3-1}=|B^T|=|B| $$ \

    $$ |Adj(B)| = (2^{3-1})^{3-1}=|C^T|=|C| = 2^4$$

    $$ |2A| = 2^{3}|A|  = 2^4 $$

    Therefore , $$ \displaystyle |2AB^TC| = |2A| |B^T| |C| = 2^4 . 2^2 . 2^4 = \sum_{r=1}^{11} {10 \choose r-1} $$

    • This reply was modified 3 months, 1 week ago by Jatin Kr Dey.
    • This reply was modified 3 months, 1 week ago by Jatin Kr Dey.
    • This reply was modified 3 months, 1 week ago by Jatin Kr Dey.
    • This reply was modified 3 months, 1 week ago by Jatin Kr Dey.
    • This reply was modified 3 months, 1 week ago by Jatin Kr Dey.
    • This reply was modified 3 months, 1 week ago by Jatin Kr Dey.
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