showing a matrix has an inverse and how it is constructed from that matrix Can someone explain how you would do these problems, I understand what inverses are, but I really don't understand these problems.
B)Suppose that $A$ is $50\times 50$ and that $A^3-2A^2+9A+7I_{50}=0_{n\times n}$. Show that $A$ must have an inverse and show how it is constructed from $A$.
C) Suppose that $A$ is $50\times 50$ and that $A^2 = A$. Show that if $A$ has an inverse, then $A$ must be a very simple matrix. (What matrix?
For B I was thinking you could expand the equation out so that it becomes 
$AA^2-2AA+9A+7I_{50}=0_{50\times 50}$, but I don't known how that would helped me to show there is an inverse. As for the construction part of the problem, would I just multiple both sides by $A^-$$^1$? And then solve for that?
For C I known  $A^2 = AA$, but again I don't know how that would help me to show the inverse of $A$. 
 A: For part B observe that $$A^3-2A^2+9A+7I_{50}=O_{50\times 50}\quad \implies \quad A^3-2A^2+9A=-7I_{50}$$
Then,
$$A\left(-\frac17A^2+\frac27A-\frac97I_{50}\right)=I_{50}=\left(-\frac17A^2+\frac27A-\frac97I_{50}\right)A$$
A: You are probably overcomplicating the questions. Work by definition:

A matrix $A$ is invertible if there exists a matrix $B$ such that $AB=I$ (although we should also demand that $BA=I$, it is a well known result that $AB=I$ if and only if $BA=I$ for two finite dimensional matrices (and hence $50$-dimensional), so we don't need to specify this)

Now, given that $A^3-2A^2+9A+7I=0$, just transpose the $7I$ to the other side, factorize $A$ out of the left hand side (you can do this because the distributive law for matrices  of multiplication over addition holds), and divide by $-7$:
$$
A^3-2A^2+9A+7I=0 \iff A \left(\frac{-1}{7}(A^2-2A+9)\right) = I
$$
Hence, we found the matrix $B = \left(\frac{-1}{7}(A^2-2A+9)\right)$. Hence, $A$ is invertible, and it is clear how $B$ is constructed from $A$.
As for the second one, use the definition. It is given that $A$ is invertible. Hence, there is a matrix $B$ such that $AB=I$.Now:
$$
A^2=A \implies AA=A \implies (AA)B=AB \implies A(AB) = AB = I \implies A=I
$$
A very simple looking matrix indeed. Hence, we are done. 
