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**Constatine11** Develop a recursive form for this determinant, for the $\displaystyle n \times n$ case put::

$\displaystyle

K_n=\left[ { \begin{array}{ccccc} k & 1 & 0 & \ldots & 0 \\ 1 & k & 1 & \ldots & 0 \\ 0 & 1 & k & \ldots & 0 \\ \vdots & \, & \ddots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 & k \\ \end{array} } \right]

$

The expand the determinant about the 1 st row:

$\displaystyle \det(K_n) = k \det(K_{n-1}) - 1 \det (L_{n-1})$

where:

$\displaystyle

L_n=\left[ { \begin{array}{ccccc} 1 & 1 & 0 & \ldots & 0 \\ 0 & k & 1 & \ldots & 0 \\ 0 & 1 & k & \ldots & 0 \\ \vdots & \, & \ddots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 & k \\ \end{array} } \right]

$

Now expand the determinant of $\displaystyle L_n$ about the first column:

$\displaystyle \det(L_n) = \det(K_n-1)$

so we have:

$\displaystyle \det(K_n) = k \det(K_{n-1}) - 1 \det (K_{n-2})$

and hopefully you can take it from there.

ZB