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Let $$ A= \begin{bmatrix} 0 & a^2 & b^2 & c^2\\ a^2 & 0 & z^2 & y^2\\ b^2 & z^2 & 0 & x^2\\ c^2 & y^2 & x^2 & 0\\ \end{bmatrix} $$ and $$ B= \begin{bmatrix} 0 & ax & by & cz\\ ax & 0 & cz & by\\ by & cz & 0 & ax\\ cz & by & ax & 0\\ \end{bmatrix} $$

Show that $\det(A)=\det(B)$.

I have tried by multiplying and dividing $xyz$ and $abc$ to symmetric rows and columns; however, I was unable to take out the common. So please help.

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  • $\begingroup$ I guess it's det(A) = det(B)? $\endgroup$
    – Greg P.
    Commented Aug 5, 2014 at 15:28
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    $\begingroup$ @GregP. OP is using the notation $|M| = \det M$ $\endgroup$ Commented Aug 5, 2014 at 15:29

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Instead of doing something clever, you could observe that each of your determinants has only nine non-zero terms. So you could just compute both determinants (as polynomials) and compare them.

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  • $\begingroup$ yes but he does not want to expand the determinants... so I guess he does not want to compute them either. $\endgroup$
    – anderstood
    Commented Aug 5, 2014 at 18:29
  • $\begingroup$ @anderstood My impression was that what he didn't want (because he had tried it and failed) was to pull out common factors from some rows or columns, to simplify the determinant. $\endgroup$ Commented Aug 5, 2014 at 18:49
  • $\begingroup$ @anderstood Now I think you're right; I was concentrating on the question and ignored the title. $\endgroup$ Commented Aug 5, 2014 at 19:02
  • $\begingroup$ Anyway this seems to be homework, so he'll probably get the solution from his teacher... :) $\endgroup$
    – anderstood
    Commented Aug 5, 2014 at 19:28
  • $\begingroup$ I want to solve the problem without expanding the determinant, as it would be a lengthy method to expand any row or column of A and B and then compare and equate them.Further these type of questions may come in our exams where time is very less,so the shortest and simplest method would be appreciated $\endgroup$
    – Sourajit
    Commented Aug 6, 2014 at 9:30

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