Finding a matrix inverse when an equation involving it is a multiple of the identity matrix

Question

Say you had a matrix $A$, and you did an equation like $A^2 - A$, and proved that it was a multiple of $I$. How could you find $A^{-1}$ in the form $rA + sI$ after proving that?

I want to do it myself, which is why I didn't provide the matrix, but I don't know where to go. I know that for a matrix to be invertible, the determinant can't be 0, but I don't think that connects to it in any way.

Answer 1

Suppose $A^2-A=aI$, with $a\ne0$; then (assuming $A^{-1}$ exists) $$ A^{-1}(aI)=A^{-1}(A^2-A) $$ or $$ aA^{-1}=A-I $$ Then it's easy to close the circle.

Answer 2

Hint: Take the equation $A^2 - A = kI$ and multiply both sides by $A^{-1}$. Then, solve for $A^{-1}$.