General formula for $f(n)$

Question

Let for $n\geq 3, C_n$ denote the $(2n) \times (2n)$ matrix such that all entries along the diagonal are $2$, all entries along the sub- and super-diagonal are $1$, all entries along the antidiagonal are $1$, all entries along the diagonals directly above and below the antidiagonal are $2$, and all other entries are zero.

Let $f(n) : \mathbb{N}\to\mathbb{N}, f(n) = \det(C_n)$ for $n\geq 3.$ Prove that $$f(n) = \begin{cases}0,&\text{if }n = 3k+2,\ k\in\mathbb{N}\\ 3^n,& \text{otherwise}\end{cases}.$$

I'm not sure how to go about doing this. I tried cofactor expansion along the first column, but I couldn't make much progress. I can't seem to find a recursive relationship. So I just tried converting $C_n$ to an upper triangular matrix using row operations. This results in a matrix satisfying certain patterns, but I can't seem to find a way to prove why reducing the matrix always produces these patterns (I can prove that the $k$th diagonal entry of the resulting upper triangular matrix is $\frac{k+1}k,$ where $1\leq k\leq n$ but I can't deal with the other $n$ diagonal entries well).

Answer 1

$\newenvironment{vsmatrix}{\left|\begin{smallmatrix}}{\end{smallmatrix}\right|}\def\mycolor#1{{\color{blue}#1}}$Denote by $D_n$ the determinant of the $2n × 2n$ matrix to be solved with $2$ replaced by $a$ and $1$ replaced by $b$. In the following calculation, rows to which Laplace's formula for determinants are applied are colored blue. If a determinant has no colored entry, it means the determinant after the next equal sign is derived by applying elementary transformations.$$ D_1 = \begin{vmatrix}a & b \\ b & a\end{vmatrix} = a^2 - b^2,\ D_2 = \begin{vmatrix} a & b & a & b\\ b & a & b & a\\ a & b & a & b\\ b & a & b & a \end{vmatrix} = 0, $$\begin{align*} D_3 &= \begin{vmatrix} a & b &&& a & b\\ b & a & b & a & b & a\\ & b & a & b & a &\\ & a & b & a & b &\\ a & b & a & b & a & b\\ b & a &&& b & a \end{vmatrix} = \begin{vmatrix} a & b &&& a & b\\ b & 0 & 0 & 0 & 0 & a\\ & b & a & b & a &\\ & a & b & a & b &\\ a & 0 & 0 & 0 & 0 & b\\ b & a &&& b & a \end{vmatrix} = \begin{vmatrix} a & b &&& a & b\\ b & 0 & 0 & 0 & 0 & a\\ & \mycolor{0} & \mycolor{a} & \mycolor{b} & \mycolor{0} &\\ & \mycolor{0} & \mycolor{b} & \mycolor{a} & \mycolor{0} &\\ a & 0 & 0 & 0 & 0 & b\\ b & a &&& b & a \end{vmatrix}\\ &= (a^2 - b^2) \begin{vmatrix} a & b & a & b\\ b & 0 & 0 & a\\ a & 0 & 0 & b\\ b & a & b & a \end{vmatrix} = (a^2 - b^2) \begin{vmatrix} 0 & b & a & 0\\ \mycolor{b} & \mycolor{0} & \mycolor{0} & \mycolor{a}\\ \mycolor{a} & \mycolor{0} & \mycolor{0} & \mycolor{b}\\ 0 & a & b & 0 \end{vmatrix}\\ &= -(a^2 - b^2)^2 \begin{vmatrix}b & a \\ a & b\end{vmatrix} = (a^2 - b^2)^3. \end{align*}

For any $n \geqslant 4$,\begin{align*} D_n &= \begin{vsmatrix} ⋱ & ⋱ &&&&& ⋰ & ⋰\\ ⋱ & a & b &&& a & b & ⋰\\ & b & a & b & a & b & a &\\ && b & a & b & a &&\\ && a & b & a & b &&\\ & a & b & a & b & a & b &\\ ⋰ & b & a &&& b & a & ⋱\\ ⋰ & ⋰ &&&&& ⋱ & ⋱ \end{vsmatrix}_{2n × 2n}\mskip-18mu = \begin{vsmatrix} ⋱ & ⋱ &&&&& ⋰ & ⋰\\ ⋱ & a & b &&& a & b & ⋰\\ & b & 0 & 0 & 0 & 0 & a &\\ && b & a & b & a &&\\ && a & b & a & b &&\\ & a & 0 & 0 & 0 & 0 & b &\\ ⋰ & b & a &&& b & a & ⋱\\ ⋰ & ⋰ &&&&& ⋱ & ⋱ \end{vsmatrix}_{2n × 2n}\mskip-18mu = \begin{vsmatrix} ⋱ & ⋱ &&&&& ⋰ & ⋰\\ ⋱ & a & b &&& a & b & ⋰\\ & b & 0 & 0 & 0 & 0 & a &\\ && \mycolor{0} & \mycolor{a} & \mycolor{b} & \mycolor{0} &&\\ && \mycolor{0} & \mycolor{b} & \mycolor{a} & \mycolor{0} &&\\ & a & 0 & 0 & 0 & 0 & b &\\ ⋰ & b & a &&& b & a & ⋱\\ ⋰ & ⋰ &&&&& ⋱ & ⋱ \end{vsmatrix}_{2n × 2n}\\ &= (a^2 - b^2) \begin{vsmatrix} ⋱ & ⋱ &&&&& ⋰ & ⋰\\ ⋱ & a & b &&& a & b & ⋰\\ & b & a & b & a & b & a &\\ && b & 0 & 0 & a &&\\ && a & 0 & 0 & b &&\\ & a & b & a & b & a & b &\\ ⋰ & b & a &&& b & a & ⋱\\ ⋰ & ⋰ &&&&& ⋱ & ⋱ \end{vsmatrix}_{2(n - 1) × 2(n - 1)}\mskip-54mu = (a^2 - b^2) \begin{vsmatrix} ⋱ & ⋱ &&&&& ⋰ & ⋰\\ ⋱ & a & b &&& a & b & ⋰\\ & b & 0 & b & a & 0 & a &\\ && \mycolor{b} & \mycolor{0} & \mycolor{0} & \mycolor{a} &&\\ && \mycolor{a} & \mycolor{0} & \mycolor{0} & \mycolor{b} &&\\ & a & 0 & a & b & 0 & b &\\ ⋰ & b & a &&& b & a & ⋱\\ ⋰ & ⋰ &&&&& ⋱ & ⋱ \end{vsmatrix}_{2(n - 1) × 2(n - 1)}\\ &= -(a^2 - b^2)^2 \begin{vsmatrix} ⋱ & ⋱ &&&&& ⋰ & ⋰\\ ⋱ & a & b &&& a & b & ⋰\\ & b & a &&& b & a &\\ && b & b & a & a &&\\ && a & a & b & b &&\\ & a & b &&& a & b &\\ ⋰ & b & a &&& b & a & ⋱\\ ⋰ & ⋰ &&&&& ⋱ & ⋱ \end{vsmatrix}_{2(n - 2) × 2(n - 2)}\mskip-54mu = -(a^2 - b^2)^2 \begin{vsmatrix} ⋱ & ⋱ &&&&& ⋰ & ⋰\\ ⋱ & a & b &&& a & b & ⋰\\ & b & a &&& b & a &\\ && \mycolor{0} & \mycolor{b} & \mycolor{a} & \mycolor{0} &&\\ && \mycolor{0} & \mycolor{a} & \mycolor{b} & \mycolor{0} &&\\ & a & b &&& a & b &\\ ⋰ & b & a &&& b & a & ⋱\\ ⋰ & ⋰ &&&&& ⋱ & ⋱ \end{vsmatrix}_{2(n - 2) × 2(n - 2)}\\ &= (a^2 - b^2)^3 \begin{vsmatrix} ⋱ & ⋱ &&&&& ⋰ & ⋰\\ ⋱ & a & b &&& a & b & ⋰\\ & b & a & b & a & b & a &\\ && b & a & b & a &&\\ && a & b & a & b &&\\ & a & b & a & b & a & b &\\ ⋰ & b & a &&& b & a & ⋱\\ ⋰ & ⋰ &&&&& ⋱ & ⋱ \end{vsmatrix}_{2(n - 3) × 2(n - 3)}\mskip-54mu = (a^2 - b^2)^3 D_{n - 3}. \end{align*} Therefore,$$ D_n = \begin{cases} (a^2 - b^2)^n; & n \not\equiv 2 \pmod{3}\\ 0; & n \equiv 2 \pmod{3} \end{cases}. $$

Answer 2

First let us make a simplification: reverse the order of the last $n$ rows and the last $n$ columns. This does not affect the determinant, so we may suppose $$ C_n= \begin{bmatrix} A_n & B_n\\ B_n & A_n \end{bmatrix} $$ where $A_n$ is tridiagonal with parameters $(1,2,1)$ and $B_n$ is tridiagonal with parameters $(2,1,2)$.

Now by block row and column operations we see that $$ \det C_n= \det \begin{bmatrix} A_n & B_n\\ B_n & A_n \end{bmatrix} = \det \begin{bmatrix} A_n + B_n & B_n\\ B_n + A_n & A_n \end{bmatrix} = \det \begin{bmatrix} A_n + B_n & B_n\\ O & A_n - B_n \end{bmatrix}. $$ We therefore have $$ \det C_n= \det 3 G_n\ \det H_n $$ where where $G_n$ is tridiagonal with parameters $(1,1,1)$ and $H_n$ is tridiagonal with parameters $(-1,1,-1)$.

Recall that $f(n)=\det C_n$. Write $g(n)=\det G_n$ and $h(n)=\det H_n$. Let's put $f(0)=g(0)=h(0)=1$ for convenience.

Then with the usual expansion of triadiagonal determinants (expand by first row, then expand second term by first column) we have that both $g(n)$ and $h(n)$ satisfy the recurrence $$ \phi(n+2)=\phi(n+1)-\phi(n) $$ for all $n\geqslant 1$; it's easy to check this is also true for $n=0$.

Moreover $f(0)=g(0)=1$ and $f(1)=g(1)=1$, and with these initial conditions the recurrence has a unique solution: both $f(n)$ and $g(n)$ must cycle through the six values $(1,1,0,-1,-1,0)$.

Then $3^{-n}f(n)=g(n)h(n)$ cycles through the three values $(1,1,0)$. This is exactly what we are asked to prove.

Comment

The question would have been easier to tackle had the $2$s been replaced by $a$ and the $1$s by $b$; the answer is then $(a+b)^{n}(a-b)^{n}$ when $n\not\equiv 2\mod 3$, and $0$ when it is.