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let $A$ be an $n\times n$ matrix with entries $a_{ij}$ such that

$a_{ij}=2$ if $i=j$.

$a_{ij}=1$ if $|i-j|=2$

and $a_{ij}=0$ otherwise.

compute the determinant of $A$.

using the famous formula $\det A=\sum_{i=1}^{n}(-1)^{i+j}a_{ij}\det A^{(ij)}$ where $A{(ij)}$ is the submatrix obtaining from $A$ by omiting it's $i$th row and $j$th colomn, I reached to the formula $\det A=\frac{1}{4}n^2+n+\frac{7}{8}+\frac{1}{8}(-1)^n$. is it correct?

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3 Answers 3

up vote 5 down vote accepted

In Mathematica, the code

f[n_] := Table[
  Which[i == j, 2, Abs[i - j] == 2, 1, True, 0],
  {i, 1, n}, {j, 1, n}
  ];
Table[Det@f@n, {n, 1, 20}]

results in

{2, 4, 6, 9, 12, 16, 20, 25, 30, 36, 42, 49, 56, 64, 72, 81, 90, 100, 110, 121}

Searching for that sequence in the OEIS results in this and your formula is (up to replacing $n$ by $n+2$) the first one given there.

I'd say that the answer is therefore Yes :D

N.B. You should contact the OEIS so that they add this interpretation ofthe sequence of the (pretty impressive!) list they already have.

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so my formula agrees with the real determinant for small vlues of $n$, right? I'm so happy about it!!!! –  Goodarz Mehr Feb 6 '12 at 4:21
1  
To prove it actually works, instead of using the development you used, do row and column operations to obtain a recursion. For example, substract twice the third row to the first row, and use Laplace's formula to develop the determinant along the first column. Now do something similar in the other direction, so as to obtain a smaller version of the matrix. This will give you a recursion. If you have problems doing this, let us know and someone will surely do it in detail. –  Mariano Suárez-Alvarez Feb 6 '12 at 4:29
    
my solution was like this: let $a_n$ be the determinant of this matrix. then using that formula, I got $a_n=2a_{n-1}-2a_{n-3}+a_{n-4}$. solving this recurrence relation gave me that formula. :) –  Goodarz Mehr Feb 6 '12 at 4:34

Cinkir develops in his paper a formula for the determinant of a pentadiagonal Toeplitz matrix. Specializing to your case, let

$$\mathbf P=\begin{pmatrix}a&b&c&&\\b&a&b&c&\\c&b&a&\ddots&\ddots\\&c&\ddots&\ddots&\\&&\ddots&&\end{pmatrix}$$

and consider the associated polynomial $p(x)=cx^4+bx^3+ax^2+bx+c$. The paper gives an expression for the determinant of $\mathbf P$ in terms of the roots of $p(x)$, with limiting cases considered if $p(x)$ has repeated roots.

For your specific case, $p(x)=(x^2+1)^2$; the formula for $\det \mathbf P$ if $p(x)$ takes the form $(x-r)^2(x-s)^2$ goes like

$$\det \mathbf P=\frac{r^{2 n+4}-r^{n+1} s^{n+1} \left((n+2)^2 r^2-2 (n+1) (n+3) r s+(n+2)^2 s^2\right)+s^{2 n+4}}{(r-s)^4}$$

Letting $r=i$ and $s=-i$ yields your formula.

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If $A$ is of this form

$$A_{7\times 7}= \begin{pmatrix} 2 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 2 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 2 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 2 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 2 \\ \end{pmatrix}, $$

I get

$$ \det(A) = \begin{cases} \frac{n^2}{4}+n+1,\quad n \text{ even}\\ \quad\\ \frac{n^2}{4}+n+\frac{3}{4},\quad n \text{ odd} \end{cases} $$

This is just an alternative way of stating the questioner's own succint answer:

$$\det(A) = \frac{n^2}{4}+n+\frac{7}{8} +(-1)^n\frac{1}{8}.$$

This is a nice exercise in symbolic $LU=A$ factorization and would be a good exam question to separate the good from the mediocre students.

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