Matrices are your friends.

It is always possible to do this same process to find the inverse of a certain linear combination of a matrix and the identity matrix given that the matrix is 2x2, invertible, and has two eigenvalues that do not differ only by sign.

Were $A$ noninvertible, then ${A}^{2}$ would also be noninvertiable just from a single determinant argument that $det({A}^{2}) = {det(A)}^{2}$ with the determinant of $A$ being zero as it is noninvertible. The basis for finding this inverse of the linear combination of $A$ and $I$ in question is that the square of the matrix is invertible. Thus, this inverse of this specific linear combination doesn't not exist when $A$ is noninvertible.

Let us suppose that $A$ is a 2x2 matrix whose solutions to $det(A - \lambda I) = 0$ are elements of the field that $A$ is defined over. Then $A$ has two eigenvalues $\lambda_{1}$ and $\lambda_{2}$ .

Now this specific linear combination would be:

$tr(A)A - det(A)I = (\lambda_{1} + \lambda_{2})A - \lambda_{1} \lambda_{2} I$ , where $tr(A)$ is the trace of A.

Were $A$ to be invertible, then its characteristic equation would look like

$p(\lambda) = (\lambda - \lambda_{1})(\lambda - \lambda_{2}) = 0$ with choosing to keep it monic

$\Longrightarrow p(\lambda) = {\lambda}^{2} - (\lambda_{1} + \lambda_{2})\lambda + \lambda_{1} \lambda_{2} = {\lambda}^{2} - tr(A)\lambda + det(A) = 0$ .

By Cayley-Hamilton Theorem, $A$ must satisfy this equation such that

${A}^{2} - tr(A)A + det(A)I = 0 \Longrightarrow {A}^{2} = tr(A)A - det(A)I$

Following the same steps as in the solution, ${A}^{2}$ can be found to be a factor of

${A}^{2}{(A - tr(A)I)}^{2} = {det(A)}^{2}I$

$\Longrightarrow \frac{1}{{det(A)}^{2}}{(A - tr(A)I)}^{2} = {(tr(A)A - det(A)I)}^{-1}$ .

Note here that $\frac{1}{{det(A)}^{2}}$ would be undefined were ${A}^{2}$ noninvertible. Also note that were the eigenvalues to only differ by a sign, then trace of $A$ would be zero making it impossible to after out an A in order to produce another ${A}^{2}$ .

Note that when $tr(A) = 0$ that any even power of any invertible 2x2 matrix $A$ or any 2x2 nilpotent matrix of order 2 as well as the zero matrix will just be the negative determinant of $A$ to an even power times the identity matrix. Will elaborate further later.

This process can be extended to nxn matrices with n > 2. Just as the 2x2 case considered ${A}^{2}$ , the nxn case would consider ${A}^{n}$ , but the inverse of ${A}^{n}$ would not necessarily be a linear combination that is taken to a power. Will elaborate later.

I have modified the question to the core of what question I am actually posing.