For some ,R, let A=pmatrix & 2 1 & 2pmatrix and B=pmatrix1 & 1 1 & pmatrix be such that A^…

Question

For some $\alpha,\beta\in\mathbb{R}$, let $A=\begin{pmatrix}\alpha & 2\ 1 & 2\end{pmatrix}$ and $B=\begin{pmatrix}1 & 1\ 1 & \beta\end{pmatrix}$ be such that $A^2-4A+2I = B^2-3B+I = O$. Then $\big(\det(\operatorname{adj}(A^3-B^3))\big)^2$ is equal to \_\_\_\_.

Answer: 225

Accepted Answer

The characteristic equation of $A$ is $(\alpha-\lambda)(2-\lambda)-2=0$, i.e. $\lambda^2-(\alpha+2)\lambda+(2\alpha-2)=0$, so by Cayley--Hamilton $A^2-(\alpha+2)A+(2\alpha-2)I=0$. Comparing with $A^2-4A+2I=0$ gives $\alpha+2=4\Rightarrow \alpha=2$, and indeed $2\alpha-2=2$ checks out.

Similarly for $B$: $(1-\lambda)(\beta-\lambda)-1=0\Rightarrow \lambda^2-(1+\beta)\lambda+(\beta-1)=0$, so $B^2-(1+\beta)B+(\beta-1)I=0$. Comparing with $B^2-3B+I=0$ gives $1+\beta=3\Rightarrow\beta=2$ (and $\beta-1=1$ checks out).

So $A=\begin{pmatrix}2&2\1&2\end{pmatrix}$, $B=\begin{pmatrix}1&1\1&2\end{pmatrix}$. Computing directly, $A^3-B^3=\begin{pmatrix}15&20\6&7\end{pmatrix}=C$, so $\det(C)=105-120=-15$.

For a $2	imes2$ matrix, $\det(\operatorname{adj}(C))=\det(C)^{2-1}=\det(C)=-15$. Hence $\big(\det(\operatorname{adj}(C))\big)^2=(-15)^2=225$.

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