Short proof for the determinant of a $4$ by $4$ matrix

Hint:

$$ \begin{pmatrix} x & y & z & t \\ -y & x & -t & z \\ -z & t & x & -y \\ -t & -z & y & x \end{pmatrix} \begin{pmatrix} x & -y & -z & -t \\ y & x & t & -z \\ z & -t & x & y \\ t & z & -y & x \end{pmatrix} = (x^2+y^2+z^2+t^2) \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$

The matrix has the two-by-two block form $$M=\begin{pmatrix} xI+yJ & zI+t J \\ -zI+t J & xI-yJ \end{pmatrix}$$ where $I$ is the 2-by-2 identity matrix and $J=\begin{pmatrix} 0&1\\-1&0 \end{pmatrix}$. Since these four matrices commute, the determinant of $M$ by treating the elements as scalars and then taking the determinant of the result: \begin{align}\det{M}&=\det[(xI+y J)(xI-yJ)-(-zI+tJ)(-zI-tJ)]\\&=\det[(x^2+y^2+z^2+t^2)I]\\&=(x^2+y^2+z^2+t^2)^2.\end{align}