$A$ is reflection matrix $2\times2$. $$B=A^4-2A^3-A-5I.$$ Find numbers $k$, $t$ in $\mathbb R$ so that $B^{-1}=kB+tI$.

I know that reflection matrix have eigenvalues of $1$, $-1$ ($A^2=I$) I got this: $$B=A^4-2A^3-A-5I=(A^2)^2-2(A^2)A-A-5I$$ and then:


But what I can do from here?


  • $\begingroup$ Could you expand $BB^{-1}=I$ and then solve for k and t? $\endgroup$ – JB King Feb 22 '13 at 17:21

One quick and dirty way to do this: As you said, $A^2=I$ gives $B= -3A - 4I$. Now, orthogonally diagonalize $A$ as $A=QDQ^T$ where $D=\begin{pmatrix}1\\&-1\end{pmatrix}$. So $\operatorname{trace}(B)=-8$ and $\det B=\det\left(-3\begin{pmatrix}1\\&-1\end{pmatrix}-4I\right)=7.$ Therefore the characteristic equation of $B$ is $B^2+8B+7I=0$. Hence $B(B+8I)=-7I$ and $B^{-1}=-(B + 8I)/7=(3A-4I)/7$.

Alternatively, by Cayley-Hamilton theorem, if $B$ is invertible, its inverse would be a degree-1 polynomial in $B$ and hence a degree-1 polynomial in $A$. So, you just need to find $p$ and $q$ such that $(pA+qI)(-3A-4I)=I$.


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