Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

1 Answer

up vote 3 down vote accepted

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

share|improve this answer
    
What is $c_1,\ldots, c_n$? –  Sigur Jan 19 '13 at 14:43
2  
@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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.