If $A$ is a $n\times n$ matrix that satisfies ${A^3} - 15{A^2} + 41A - c{I_n} = {0_{n,n}}$. I want to prove that $A$ has an inverse. ($c \in \mathbb{R}$)
I'm thinking you rearrange so that you get the identity matrix on its own on the right and then change the left hand side somehow so that it is a product of $A$ and something else which then has to be the inverse. Is this the correct approach?
$$A(A^2-15A+41)=(A^2-15A+41)A=cI$$ so for $c\neq 0$ the inverse is $A^{-1}=\frac{1}{c}(A^2-15A+41)$.
If $c=0$ then $A$ is not necessarily invertible (for example $A=0$ satisfies the equation), but if it is then the inverse is given by $A^{-1}=-\frac{1}{41}(A-15I)$.
Another approach is to use the Cayley–Hamilton theorem which states that given the characteristic polynomial of $A$: $$p(t)=\det(A-t I_n)$$ which can be defined for any scalar $t$ and the roots of which is the eigenvalues of $A$, but polynomials can also be defined for matrices and so the theorem says that $p(A)=0$ in other words the characteristic polynomial of a matrix $A$ evaluated at $A$ itself gives the $0$ matrix.
This approach has a few advantages: When you see a homogeneous matrix polynomial equation of degree $n$ in a matrix $A$, then you can immediately deduce some properties of $A$:
The charateristic polynomial of an $n\times n$ matrix is of degree $n$ and has the general form $$p(\lambda)=(-1)^n\lambda^n+(-1)^{n-1}\mathrm{tr}(A)\lambda^{n-1}+\dots+\det(A),$$ when it is defined as $p(\lambda)=\det(A-\lambda I)$, where $\mathrm{tr}(A)=\sum_i a_{ii}$ is the trace of $A$ - the sum along the diagonal.
So the degree of the polynomial tells you the dimensionality of $A$. Also the identities $$\mathrm{tr}(A)=\sum_{i=1}^n \lambda_i\qquad\text{and}\qquad\det(A)=\prod_{i=1}^n\lambda_i,$$ where $\lambda_i$ is the $i$'th eigenvalue of $A$, tells us something about the eigenvalues.
In your case though it seams that the characteristic polynomial has been defined as $p(\lambda)=\det(\lambda I -A)$ which means that for an $n\times n$ matrix $A$: $$p(\lambda)=\lambda^n-\mathrm{tr}(A)\lambda^{n-1}+\dots+(-1)^n\det(A).$$ So just looking at the equation tells you the characteristic polynomial in other words that $$p(\lambda)=\lambda^3-15\lambda^2+41\lambda-c,$$ which in turns tells you that $$\mathrm{tr}(A)=a_{11}+a_{22}+a_{33}=15$$ $$a_{12}a_{21}+a_{13}a_{31}+a_{32}a_{23}-a_{11}a_{22}-a_{11}a_{33}-a_{22}a_{33}=41$$ $$\det(A)=c.$$ This has now provided us with the necessary and sufficient condition for the inconvertibility of $A$, since $c=\det(A)$ we have: $$A^{-1}\text{ exits}\quad\iff\quad c\neq0.$$
