Show that $\det(A_n) =(a-1)^{n-1}(a+n-1)$

Asked Jul 09 '20 at 16:07

Active Jul 09 '20 at 16:24

Viewed 42 times

Show that $\det(A_n) =(a-1)^{n-1}(a+n-1)$ with K as a field and $ n \in N$

$ A_n = \begin{pmatrix} a & 1 & 1& 1 & ... & 1 \\ 1 & a & 1 & 1 & ... & 1 \\ 1 & 1 & a & 1 & ... & 1 \\ ...& ... & ... & ... & ... & ... \\ 1 & ... & ... & ... & ... & 1 \\ 1 & 1 & ... & ... & 1 & a \end{pmatrix} \in K^{nxn}$

So i subtract everything with the first column like this

$ A_n = \begin{pmatrix} a & 1 & 1& 1 & ... & 1 \\ 1-a & a-1 & 0 & 0 & ... & 0 \\ 1-a & 0 & a-1 & 0 & ... & 0 \\ ...& ... & ... & ... & ... & ... \\ 1-a & ... & ... & ... & ... & 0 \\ 1-a & 0 & ... & ... & 0 & a-1 \end{pmatrix} \in K^{nxn}$

And then i use Laplace on this

$\det(A_n)= a \cdot (a-1)^{n-1}+ \det\begin{pmatrix} 1-a & 0 & 0 & ... & 0 \\ 1-a & a-1 & 0 & ... & 0 \\ ...& ... & ... & ... & ... \\ 1-a & ... & ... & ... & 0 \\ 1-a & ... & ... & 0 & a-1 \end{pmatrix} + 0$

So $\det\begin{pmatrix} 1-a & 0 & 0 & ... & 0 \\ 1-a & a-1 & 0 & ... & 0 \\ ...& ... & ... & ... & ... \\ 1-a & ... & ... & ... & 0 \\ 1-a & ... & ... & 0 & a-1 \end{pmatrix} $ got to be $(n-1)(a-1)^{n-1}$ if my calculation is correct. But i can't find a way to 'show' it.

edited Jul 09 '20 at 16:24

Ben Grossmann

225,327

asked Jul 09 '20 at 16:07

Payvand

See this post. – Ben Grossmann Jul 09 '20 at 16:22
@Omnomnomnom which one exactly? – Payvand Jul 09 '20 at 16:26
They are all essentially the same, but this one (fifth on the list) is probably the easiest to follow – Ben Grossmann Jul 09 '20 at 16:28
thanks! @Omnomnomnom – Payvand Jul 09 '20 at 16:50

Show that $\det(A_n) =(a-1)^{n-1}(a+n-1)$

0 Answers0