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Let $A,B$ be matrices, their entries are real numbers. If $A,B$ are square matrices with the same orders then $$ \det\left(A^{T}BA\right)=\det\left(B\right)\left(\det\left(A\right)^{2}\right). $$ My question is, if their orders are not the same but $B$ is still square, do we have a formula in the form like this $$ \det\left(A^{T}BA\right)=\det\left(B\right)\times\ldots, $$ here $\ldots$ is something that is interseting enough.

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  • $\begingroup$ For example, A is 3x2; B is 3x3. $\endgroup$ – dfghsd Feb 7 '17 at 13:44
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No,See that determinant is always defined for Square matrices only , in your example even though the final outcome for $A^{T}BA$ may be a square mtrix and of course we can calculate its determinant but we cannot write them as individual factors as $A$ is not a square matrix here.

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  • $\begingroup$ I know we cannot have det(A) but may be something else. $\endgroup$ – dfghsd Feb 7 '17 at 13:52
  • $\begingroup$ Ok, since finally determinants are numbers you can decompose the determinant of $A^{T}BA$ into factors and you can write them as determinants of some other matrix [matching of corresponding factors] by trial method ! . $\endgroup$ – BAYMAX Feb 7 '17 at 13:58
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No this formula works only for i × j where i = j.

And both A and B are of same order i × j.

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