Prove that $A^{-1} = \frac{1}{2}(A+I)$

Question

$A$ is $2\times2$, and given eigenvalues of $A$ are $1,-2$, prove $$A^{-1} = \frac{1}{2}(A+I)$$

I've tried expanding matrix $A$ into its components, and solve for components of $A$ with given eigenvalues, but I'm still not getting anywhere. What properties can help here?

Answer 1

Using that $n=2$, since all eigenvalues of $A$ are distinct, we know that it is diagonalizable. So $$ A=S\begin{bmatrix} 1&0\\0&-2\end{bmatrix}S^{-1} $$ for some invertible matrix $S$. Now, since $$ \frac12\,\left(\begin{bmatrix} 1&0\\0&-2\end{bmatrix}+I\right)=\begin{bmatrix} 1&0\\0&-1/2\end{bmatrix}=\begin{bmatrix} 1&0\\0&-2\end{bmatrix}^{-1}, $$ after conjugating with $S$ we get the same equation for $A$.

Edit: How to use the above. \begin{align} A^{-1} &=\left(S\begin{bmatrix} 1&0\\0&-2\end{bmatrix}S^{-1}\right)^{-1}=S\begin{bmatrix} 1&0\\0&-2\end{bmatrix}^{-1}S^{-1}\\ \ \\ &=S\left(\frac12\,\left(\begin{bmatrix} 1&0\\0&-2\end{bmatrix}+I\right)\right)S^{-1}\\ \ \\ &=\frac12\,\left(S\begin{bmatrix} 1&0\\0&-2\end{bmatrix}S^{-1}+SS^{-1}\right)\\ \ \\ &=\frac12\,(A+I) \end{align}

Answer 2

Hint: the characteristic polynomial of $A$ is $$ p_A(x)=(x-1)(x+2)=x^2+x-2$$ and $p_A(A)=0$ by the Cayley-Hamilton theorem.