Can you check my solution to:
Task
Having two matrices $X,Y\in\mathbb{R}^{n,n}$ where $x,y\in\mathbb{R}$ and matrices are defined as
$ X=\begin{bmatrix} x & 0 &0 & \dots & 0 \\ x & x & 0& \dots & 0\\ x & x & x& \dots & 0\\ \vdots&\vdots&\vdots&\vdots&\vdots \\ x & x & x & \dots & x \\ \end{bmatrix} $ and $ Y=\begin{bmatrix} 0 & \dots & 0 &0 & y \\ 0 & \dots & 0 & y & 0\\ 0 & \dots & y & 0& 0\\ \vdots&\vdots&\vdots&\vdots&\vdots \\ y & \dots & 0 & 0 & 0 \\ \end{bmatrix} $
find the $\det_{2n}\begin{bmatrix} xI_n & Y \\ -Y & X \\ \end{bmatrix} $.
Solution:
We can observe that for matrices $A,B,C,D\in\mathbb{R}^{n,n}$ when matrix $A$ is invertible then
$\begin{bmatrix} A & B \\ C & D \\ \end{bmatrix}=\begin{bmatrix} A & 0 \\ C & I_{n} \\ \end{bmatrix}\cdot\begin{bmatrix} I_n & A^{-1}B \\ 0 & D-CA^{-1}B \\ \end{bmatrix} $
so
$\det_{2n}\begin{bmatrix} A & B \\ C & D \\ \end{bmatrix}=\det_{n}(A) \cdot\det_{n}(D-CA^{-1}B)$
We see that $xI_n$ is invertible matrix when $x\neq 0$ and then we state that
$\det_{2n}\begin{bmatrix} xI_n & Y \\ -Y & X \\ \end{bmatrix}=\det_{n}(xI_n) \cdot\det_{n}(X+Y(xI_n)^{-1}Y) $
and now let's observe that $\det_{n}(xI_n)=x^n$ and
$ X+Y(xI_n)^{-1}Y=X+\frac{1}{x}YI_n^{-1}Y=X+\frac{1}{x}YI_{n}Y =X+\frac{1}{x}Y^2=X+\frac{y^2}{x}I_n $
so $\det_n (X+Y(xI_n)^{-1}Y)=\det_{n}\left(X+\frac{y^2}{x}I_n\right) = (x+\frac{y^2}{x})^n$
and in the end we can write that $\det_{2n}\begin{bmatrix} xI_n & Y \\ -Y & X \\ \end{bmatrix}=x^n (x+\frac{y^2}{x})^n = (x^2+y^2)^n $ when $x\neq 0$.
In the case $x=0$ we see that $\det_{2n}\begin{bmatrix} xI_n & Y \\ -Y & X \\ \end{bmatrix}=\det_{2n}\begin{bmatrix} 0 & Y \\ -Y & 0 \\ \end{bmatrix} =(-1)^n \det_{2n}\begin{bmatrix} Y & 0 \\ 0 & -Y \\ \end{bmatrix} =(-1)^n \cdot y^n \cdot (-y)^n = (-1)^n\cdot (-1)^n \cdot y^{2n} = y^{2n}$
Can you maybe find any simpler solution to this task?
Your derivation of the main result (that the determinant is $(x^2+y^2)^n$) is correct, but your verification is not. You should have $\det\pmatrix{0&Y\\ -Y&0}=\det(Y)^2=y^{2n}$. E.g. if you interchange the first and the last column of $\pmatrix{0&Y\\ -Y&0}$, you get a factor of $-1$. However, the entry $-y$ also contains a minus sign. So, the two minus signs cancel out each other when you compute the determinant. The same holds if you interchange the $j$-th and the $(2n-j)$-th columns of the matrix. Therefore the final determinant does not carry any minus sign.