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I was doing a proof from my book and I had a question regarding a step.

Proof:

$I\to E$ (Row switch)

$-\det I=\det E$

$\det E=-1$

$\det(EB)=\det(-1\cdot B) =\mathbf{-1\det B}$

In the last step of the proof(in bold), how was $-1$ taken out of the determinant?

Isn't $\det(n\cdot B)$ not equal to $n\det(B)$ where $n$ is a number?

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I can't really follow the working properly as presented, but you're right to be sceptical of the last step. In general, $\det(kM) = k^n \det(M)$, where $M$ is an $n \times n$ matrix. If $n$ is odd, (e.g. a $3 \times 3$ matrix), then this would be valid, as $(-1)^n = -1$ when $n$ is odd, but my suspicion is that the last step is a typo.

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|EB|=|-1B| =-1|B|

the middle part is not correct. It must be: $$|EB|=|E|\cdot |B|=-1\cdot |B|,$$ because $E$ is an elementary matrix, not a scalar.

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I don't think $\det (EB) = \det (-B)$ is correct, because $E$ only swaps two rows, not the multiplication by $-1$. Specifically the first $=$ on the "last step" is not correct, and the boldface part is not correct either.

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