Determinant of matrix of submatrices 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.
