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If $A$ and $B$ are two square matrices with $\operatorname{det} A=5$ and $\operatorname{det}\left(B^T \cdot A^T\right)=-15$, then $\operatorname{det} B$ is equal to
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-3
Given, $|A|=5 ;\left|B^T A^T\right|=-15$
We know $|A|=\left|A^T\right|$
$\begin{aligned} \therefore \quad\left|B^T\right|\left|A^T\right| & =|B||A|=-15 \\ |B| & =\frac{-15}{|A|}=\frac{-15}{5}=-3\end{aligned}$
We know $|A|=\left|A^T\right|$
$\begin{aligned} \therefore \quad\left|B^T\right|\left|A^T\right| & =|B||A|=-15 \\ |B| & =\frac{-15}{|A|}=\frac{-15}{5}=-3\end{aligned}$
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