If A=pmatrix2&33&5pmatrix with |A|=1, then the determinant of the matrix A^2025-3A^2024+A^…

Question

If $A=\begin{pmatrix}2&3\3&5\end{pmatrix}$ with $|A|=1$, then the determinant of the matrix $A^{2025}-3A^{2024}+A^{2023}$ is

Accepted Answer

By Cayley--Hamilton for the $2	imes2$ matrix $A$: $A^2-(\operatorname{tr}A)A+|A|I=0$. Here $\operatorname{tr}(A)=2+5=7$ and $|A|=1$, so $A^2=7A-I$.

Then $A^2-3A+I=(7A-I)-3A+I=4A$.

So $A^{2025}-3A^{2024}+A^{2023}=A^{2023}(A^2-3A+I)=A^{2023}(4A)=4A^{2024}$.

$\det(4A^{2024})=4^2\cdot\det(A)^{2024}=16\cdot1=16$.

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