# Find a determinant of this matrix.

I've got matrix $$\begin{equation*} A=\left( \begin{array}{ccccccc} a & 0 & 0 & \ldots & 0 & 0 & b \\ 0 & a & 0 & \ldots & 0 & b & 0 \\ 0 & 0 & a & \ldots & b & 0 & 0 \\ \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & b & \ldots & a & 0 & 0 \\ 0 & b & 0 & \ldots & 0 & a & 0 \\ b & 0 & 0 & \ldots & 0 & 0 & a \\ \end{array} \right) \end{equation*}$$

оf $$2n$$ size (number columns and rows). I need to find it's determinant using a traditional formula:

$$\det A = \sum_{\sigma \in S_{2n}}{\text{sgn}(\sigma)\prod_{i=0}^{2n}{A_{i\sigma(i)}}}$$

Where $$\text{sgn}(\sigma)$$ is $$(-1)^{inv(\sigma)}$$, and $$inv(\sigma)$$ is the number of inversions in $$\sigma$$.

It's easy to see that non-null terms in will be if and only if $$\sigma(i) \in \{i, 2n - i + 1\}$$.

So, the answer for me looks like: $$\text{sgn}(\sigma_1)a^{k_1}b^{2n - k_1} + \text{sgn}(\sigma_2)a^{k_2}b^{2n - k_2}\ldots$$. (where each $$sgn(\sigma_i)$$ is $$\pm1$$).

But I have no idea how to expand this and count all $$\sigma$$'s.

• foreach $i \le n$, group rows $i,2n+1-i$ and columns $i,2n+1-i$ together, you get a block-diagonal matrix with $2\times 2$ blocks: $[\begin{smallmatrix}a & b\\ b &a\end{smallmatrix}]$. Commented Dec 14, 2020 at 0:50

We have to focus on permutations $$\sigma \in S_{2n}$$ such that $$\sigma(i) \in\{i, 2n-i+1\}$$ for all $$i \in \{1, \ldots, 2n\}$$.

To uniquely determine such a permutation, we have to pick $$k$$ numbers out of $$1, 2, \ldots, n$$ which will be fixed points of $$\sigma$$. If $$i$$ is one of the remaining $$n-k$$ numbers, then set $$\sigma(i)=2n-i+1$$.

The action of $$\sigma$$ on $$n+1, \ldots, 2n$$ is now uniquely determined like this: $$\sigma(i) = \begin{cases} 2n-i+1, &\quad\text{if }i \in \sigma(\{1,\ldots, n\}),\\ i, &\quad\text{otherwise}.\\\end{cases}$$

To find the sign of $$\sigma$$, notice that $$\sigma$$ is a composition of $$n-k$$ transpositions $$(i, 2n-i+1)$$ each having sign $$-1$$ so $$\operatorname{sgn}(\sigma) = (-1)^{n-k}$$.

Now if $$i$$ is a fixed point of $$\sigma$$ then $$A_{i\sigma(i)} = a$$ and otherwise $$A_{i\sigma(i)} = b$$. Since $$\sigma$$ as above has $$2k$$ fixed points and $$2n-2k$$ other points, it yields the product $$a^{2k}b^{2n-k}$$ in the determinant formula.

All in all, we get

$$\det A = \sum_{\sigma \in S_{2n}} \operatorname{sgn}(\sigma) \prod_{i=1}^{2n} A_{i\sigma(i)} = \sum_{k=0}^{n}{n\choose k}(-1)^{n-k} a^{2k}b^{2n-2k} = (a^2-b^2)^n.$$

This is simply the value $$(a^2-b^2)^n$$

• An answer without any sort of explanation isn't particularly helpful. Commented Dec 14, 2020 at 5:59