To my understanding a positive definite matrix is a real symmetric square matrix where all eigenvalues are positive. Therefore for a matrix A and vector v
$Av = {\lambda}v$ where $\lambda$ is an eigenvalue of A
so $det({\lambda}{I_n} - A)=0$.
My next thought would trying to see how the formula changes when A is now A$^2$ and A$^{-1}$, but I'm getting nowhere.
On
Suppose that the eigenvalues of $A$ are $\lambda_1,\ldots,\lambda_n$ with eigenvectors $v_1,\ldots,v_n$. Note that $$A^2 v_i = \lambda_i A v_i$ = \lambda_i^2 v_i$$ and that $$Av_i = \lambda_i v_i \Rightarrow A^{-1}v_i = \lambda^{-1}v_i$$ Therefore the eigenvalues of $A^2$ are $\lambda_1^2,\ldots,\lambda_n^2$ and the eigenvalues of $A^{-1}$ are $\lambda_1^{-1},\ldots,\lambda_n^{-1}$. This should be enough to prove that $A^2$ and $A^{-1}$ are positive definite.
Another definition is that a symmetric matrix $M \in \mathbb R^{n \times n}$ is positive definite if and only if $v^T M v > 0$ for all nonzero vectors $v \in \mathbb R^n$.
Notice that \begin{align} v^T A^2 v &= v^T A^T A v \\ &= u^T u \\ &= \|u\|^2 \end{align} where $u = Av$. If $v \neq 0$ then $u \neq 0$, so $\|u\|^2 > 0$. This shows that $A^2$ is positive definite.
We can show that $A^{-1}$ is positive definite by noting that $$ v^T A^{-1} v = v^T A^{-1} A A^{-1} v = u^T A u $$ where $u = A^{-1}v$. If $v \neq 0$ then $u \neq 0$, so $u^T A u > 0$.