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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?

3 Answers 3

5

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.

  • 0
    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. :)2012-02-06
4

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.

1

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.