How can I prove that for a matrix $B$ with elements $b_{ij} = k^{i-j} a_{ij}$ this is true: $\det(B) = \det(A)$

Question

Let $A$ be a $ n \times n$ real matrix with elements $a_{ij}$ and $k$ a nonzero real number. Prove that for a matrix $B$ with elements $b_{ij} = k^{i-j} a_{ij}$ this is true: $$\det(B) = \det(A)$$

Answer 1

Apply the general formula for a determinant: \begin{align} \det B&=\sum_{\sigma\in\mathfrak S_n}(-1)^{\varepsilon(\sigma)}b_{1\,\sigma(1)}b_{2\,\sigma(2)}\dots b_{n1\,\sigma(n)} \\ &=\sum_{\sigma\in\mathfrak S_n}(-1)^{\varepsilon(\sigma)}k^{1-\sigma(1)}a_{1\,\sigma(1)}k^{2-\sigma(2)}a_{2\,\sigma(2)}\dots k^{n-\sigma(n)} a_{n1\,\sigma(n)}\\ &=k^{1-\sigma(1)+2-\sigma(2)+\dots+n-\sigma(n)}\sum_{\sigma\in\mathfrak S_n}(-1)^{\varepsilon(\sigma)}a_{1\,\sigma(1)}b_{2\,\sigma(2)}\dots a_{n1\,\sigma(n)} \\ & =k^{1+2+\dots+n-(\sigma(1)+\sigma(2)+\dots+\sigma(n))}\det A \end{align} and observe that $\;\sigma(1)+\sigma(2)+\dots+\sigma(n)$ is just $\;1+2+\dots+n$ in another order.

Answer 2

Hint. Write $B=DAD^{-1}$ for some diagonal matrix $D$.

Answer 3

$\det B = \sum (\lambda_i b_{1, i_1}b_{2, i_2}..b_{1, i_n})$ Where $\{i_n\}$ is a permutation of $1,2,..n$ and $|\lambda_i|=1 $ that depends only on choice of $\{i_n\}$.

So:

$\det B = \sum (\lambda_i b_{1, i_1}b_{2, i_2}..b_{1, i_n})= \sum (\lambda_i a_{1, i_1}a_{2, i_2}..a_{1, i_n}k^{\sum j-i_j})=\sum (\lambda_i a_{1, i_1}a_{2, i_2}..a_{1, i_n})=\det A$

Because $\sum i_j=\sum j$ for any choice of permutation