Find this determinant

mathemagics · **Joined:** Sun, 23 Mar 2008 10:55:47 UTC **Posts:** 27

Find the determinant of the matrix

robbwrr · **Posted:** Thu, 13 Nov 2008 20:45:05 UTC

Using a for alpha, if $a=1, |A| =(-1)^{n+1};$ if a=0, |A|=1

,otherwise:

From each row of A, subtract the row directly below to get:

$B=\begin{pmatrix} 1-a & 1& 1&\cdots & 1\\ 0&1-a& 1&\cdots&1\\ \vdots &\vdots&\ddots&\cdots&1\\ 0&0&\cdots&1-a&1\\a&a&\cdots&a&1\end{pmatrix}$

Now starting with row 1 to row (n-1) successively replace Row n with:

$(\frac{-a}{1-a})^k \cdot R_k+ R_n \rightarrow R_n$ to get:

$C=\begin{pmatrix} 1-a & 1& 1&\cdots & 1\\ 0&1-a& 1&\cdots&1\\ \vdots &\vdots&\ddots&\cdots&1\\ 0&0&\cdots&1-a&1\\0&0&\cdots&0&s(a)\end{pmatrix}$

$\displaystyle \\ s(a) = 1 + (\frac{-a}{1-a}) +(\frac{-a}{1-a})^2 + \cdots + (\frac{-a}{1-a})^{n-1} \\ \\= (1-a)\left(1 - \left(\frac{-a}{1-a}\right)^n\right) \\ \\= \frac{(1-a)^n - (-a)^n}{(1-a)^{n-1}}$

Now det(A)=det(B)=det(C)=product of diagonal elements of C :

$det(C) =(1-a)^{n-1} \cdot s(a) = (1-a)^n - (-a)^n$

robbwrr · **Posted:** Thu, 13 Nov 2008 22:35:51 UTC

An Inductive derivation might be as follows:

robbwrr wrote:

Using a for alpha, if $a=1, |A| =(-1)^{n+1};$ if a=0, |A|=1

,otherwise:

From each row of A, subtract the row directly below to get:

$B=\begin{pmatrix} 1-a & 1& 1&\cdots & 1\\ 0&1-a&1&\cdots&1\\ \vdots &\vdots&\ddots&\cdots&1\\ 0&0&\cdots&1-a&1\\a&a&\cdots&a&1\end{pmatrix}$

Spoiler:

Or a faster direct way, finally noticed this morning:

Spoiler:

S.O.S. Mathematics CyberBoard

Find this determinant

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