Determinant of the matrix with $a_{i,j}=2\delta_{i,j}-\delta_{i+1,j}-\delta_{i,j+1}$

Question

How to find the determinant of the matrix of order $n$ with $(i,j)$th entry as $$a_{i,j}=2\delta_{i,j}-\delta_{i+1,j}-\delta_{i,j+1}$$ here $\delta_{i,j}=1$ if $i=j$ and zero otherwise? I tried it for $2\times 2$ and $3\times 3$ matrices and conclude determinant as $n+1.$ But i like directly general method to find determinant. Please help. Thanks.

Answer 1

You can generalise the problem to the computation of determinants: $$D_n(a,b)=\begin{vmatrix} a&b&0&0&\dots&0&0\\ b&a&b&0&\dots&0&0\\ 0&b&a&b&\dots&0&0\\[-1ex] \vdots&&&\!\ddots\!&&&\vdots\\ 0&0&0&0&\dots&a&b\\ 0&0&0&0&\dots&b&a \end{vmatrix}$$ Develop $D_n(a,b)$ along the last row: \begin{align*}D_n(a,b)&=-b\begin{vmatrix} a&b&0&0&\dots&0&0\\ b&a&b&0&\dots&0&0\\ 0&b&a&b&\dots&0&0\\[-1ex] \vdots&&&\!\ddots\!&&&\vdots\\ 0&0&0&0&\dots&a&b \end{vmatrix}+a D_{n-1}(a,b)\\ &=-b^2D_{n-2}(a,b)+a D_{n-1}(a,b) \end{align*}

So one obtains a linear recurrence of order $2$, which is initialised thus: $$D_1(a,b)=a,\quad D_2(a,b)=a^2-b^2.$$

Answer 2

Expansion by cofactors is a general method to find the determinant. For your problem, this involves exactly two terms (for the two nonzero entries in the first row); the two recursive determinants are straightforward:

• the first one is just a smaller instance of the same problem (so you can use your inductive hypothesis to evaluate it)

• the second needs to be evaluated with a further cofactor expansion; one cofactor is evidently zero; the other is an even smaller instance of the same problem.