Proving a matrix is invertible given equation (without identity matrix)

Question

I'm given a square matrix ${A}$ (3×3) and the following equation ${A}^3-2017{A}^2 + {A} = {0}$ and I have to find if ${A}$ is invertible in some cases, no cases, or all cases.

I can find ${A}=0$ as an answer for the non invertible case, but I can't seem to solve the equation. In most other examples I've found, there was an identity matrix, which made it easy to find the invertible of ${A}$ like this: ${A}*invertible=I$ but this is not the case here.

I've tried doing this: $A*(A*(-A+2017*I))=A*I$, but I don't think I can divide both parts by ${A}$ because I haven't proven that ${A}$ is invertible.

Answer 1

Try taking $A$ to be a matrix of the form $\lambda I$, and use the equation to determine what $\lambda$ must be.

Answer 2

Suppose that the $3X3$ matrix is: $$ A=\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix} $$

The characteristic polynomial of A is given by: $$\mu_A[x]=det(A - \lambda I_{3})$$ If we compute this we will find that the constant factor of the characteristic polynomial of A is: $$ \mu_A[x]= ...\lambda^3 + ... \lambda^2 + ... \lambda + (-aei +afh + bdi - bfg-cdh+ceg) $$ As we have given that the constant factor equals to $0$ and we know that $$det(A) = det(A^t) = (-aei +afh + bdi - bfg-cdh+ceg)$$ Which leads us to the fact that $$det(A) = 0$$ Which means that A is never invertible.