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$A$ and $B$ are matrices and I found the determinants of

$$A + B,\, A - B,\, AB,\, A^{-1},\, B^T.$$

If we know the determinants of $A$ and $B$ but don't remember the matrices $A$ and $B$, which of the determinants above could we have found?

I think it's only $B^T$ and $A^{-1}$. Can someone tell me if that is correct?

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    $\begingroup$ $\det(AB)=\det A\det B$ $\endgroup$ – user126154 Feb 20 '15 at 9:29
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    $\begingroup$ Also, $\;A\;$ is regular (invertible, non-singular) iff $\;\det A\neq 0\implies \det A^{-1}=\left(\det A \right)^{-1}\;$ . Also, it is always true that $\;\det A=\det A^t\;$ $\endgroup$ – Timbuc Feb 20 '15 at 9:32
  • $\begingroup$ See also: math.stackexchange.com/questions/466043/deta-b-deta-detb $\endgroup$ – Martin Sleziak Feb 20 '15 at 11:30
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From some basic facts about determinants we know that:

So if we know $|A|$ and $|B|$, we also know the determinants of $AB$, $B^T$ and $A^{-1}$.

To show that we cannot say anything about the value of $|A\pm B|$ from the values of $|A|$ and $|B|$, we can simply try some examples:

  • For $A=B=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ we have $|A|=|B|=1$ and $|A+B|=4$ and $|A-B|=0$.
  • For $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $B=\begin{pmatrix}1&0\\1&1\end{pmatrix}$ we again have $|A|=|B|=1$. But in this case $|A+B|=\begin{vmatrix}2&1\\1&2\end{vmatrix}=3$ and $|A-B|=\begin{vmatrix}0&1\\-1&0\end{vmatrix}=1$.

So in both cases $|A|=|B|=1$, but determinants of $A\pm B$ have different values. Which means that they are not determined by the values of $|A|$ and $|B|$.

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