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I found this problem in A. Kostrikin's algebra book. There is no solution or a hint to it there. Only answer: $\det B=a$.

Let $A = [a_{ij}] \in \mathcal{M}(n,n; K), \ \det A=a, \ \ c \in K, c \neq0$

$B= [b_{ij}] \in \mathcal{M}(n,n; K), \ \ b_{ij}=c^{i-j}a_{ij}, \ \ \ i,j \in \{ 1,2,\dots,n\}$.

I drew (wrote down) both matrices but I don't see how $\det B$ can be equal to $\det A$.

Could you help me?

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What is your definition of $\det$? Are you using permutations? – Sigur Jan 19 '13 at 14:41
It depends. But mostly Laplace's theorem or Gaussian elimination. – Hagrid Jan 19 '13 at 15:06
up vote 3 down vote accepted

This is because $B=\begin{pmatrix}c^1\\&c^2\\&&\ddots\\&&&c^n\end{pmatrix}A\begin{pmatrix}c^1\\&c^2\\&&\ddots\\&&&c^n\end{pmatrix}^{-1}.$

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What is $c_1,\ldots, c_n$? – Sigur Jan 19 '13 at 14:43
@Sigur Typo. Should be $c^1,\ldots,c^n$, i.e. the powers of $c$. – user1551 Jan 19 '13 at 14:44
Thank you. Could you tell me, though, how you came up with this :) ? – Hagrid Jan 19 '13 at 15:10
@Hagrid I simply have seen this kind of matrix similarity many times. – user1551 Jan 19 '13 at 15:59
Ok. Thank you again. – Hagrid Jan 19 '13 at 16:18

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