Solution
= Solution
The <matrix trace> and <determinant> of the <matrices> in part (i) are
$$
\operatorname{Tr}Y=2\alpha,\qquad \det Y=\alpha^2+2\beta^2.
$$
Since $\operatorname{Tr}X=0$ and $\det X=2$, the extra conditions force $\alpha=0$ and $\beta^2=1$. Hence
$$
\boxed{Y=X\quad\text{or}\quad Y=-X.}
$$
Both choices are real, commute with $X$, and satisfy the required trace and <determinant> conditions.