Express matrix inverse with αA+βI

Question

Hi I have a characteristic polynomial of matrix A: $$p(x)=x^2+x+1$$

and I have to express inverse of a matrix A with $$A^{-1}=αA+βI$$ where α and β are real numbers. I don't know where to go from here.

I know that this is true: $$AA^{-1}=A^{-1}A=I$$

Answer 1

$$P(A) = 0$$ $$A^2 + A + I = 0$$ $$A(A+I) = -I$$ $$A(-A-I) = I$$ so $A^{-1} = -A-I$.

Answer 2

From the Cayley-Hamilton you get that $p(A)=0$; in other words, $A^2+A+\operatorname{Id}=0$. So$$A+\operatorname{Id}+A^{-1}=A^{-1}(A^2+A+\operatorname{Id})=0.$$ In other words, $A^{-1}=-A-\operatorname{Id}$.

Answer 3

Without Cayley Hamilton: The eigenvalues are $\lambda_1=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$ and $\lambda_2=\overline{\lambda_1}$ where $\lambda_2=-\lambda_1-1$ and $\lambda_1=-\lambda_2-1$ and there exists an invertible matrix $T$ such that

$TAT^{-1}=\begin{pmatrix}\lambda_1 & 0\\0&\lambda_2\end{pmatrix}$. Since $|\lambda_1|=|\lambda_2|=1$ one has

$TA^{-1}T^{-1}=\begin{pmatrix}\lambda_2 & 0\\0&\lambda_1\end{pmatrix}=\begin{pmatrix}-\lambda_1-1 & 0\\0&-\lambda_2-1\end{pmatrix}=-TAT^{-1}-I$.

Multiplying by $T^{-1}$ from the left and by $T$ from the right yields the result.