If $A \in M(\mathbb{Q},2)$ then it is impossible that $A^3=5I$
I tried to use in Jordan form, but I cannot.
2026-03-27 00:00:16.1774569616
Matrix $A$ cannot satisfy $A^3=5I$
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If it satisfies, $(\det A)^3=25$. And the determinant is not $\mathbb{Q}$-valued. It is a contradiction.