A is a 3x3 matrix, and A^3 + A^2 + A + I = 0. How do I show that A has an inverse? Is it enough to show that A(-A^2 - A - 1) = I ?
Here is one way that I might do it.
$\displaystyle A^3 + A^2 + A + I = 0\quad \Rightarrow \quad A^4 + A^3 + A^2 + A = 0$
$\displaystyle A^4 + A^3 + A^2 + A = 0\quad \Rightarrow \quad A^4 + A^3 + A^2 + A + I = I$
$\displaystyle A^4 + A^3 + A^2 + A + I = I\quad \Rightarrow \quad A^4 + 0 = I$
$\displaystyle A\left( {A^3 } \right) = \left( {A^3 } \right)A = I$
You are assuming:
$\displaystyle
A^3 + A^2 + A + I = 0
$
So multiply both sides by the matrix A which gives you:
$\displaystyle
A^4 + A^3 + A^2 + A = 0
$
Now add the identity matrix to both sides of the equation:
$\displaystyle
A^4 + (A^3 + A^2 + A + I) = I
$
Observe that the bit in parantheses is the expression we assumed is exactly equal to the matrix of all 0's. Sustitute using that fact to arrive at:
$\displaystyle
A^4 + 0 = I
$
The matrix $\displaystyle A^4 + 0$ is just $\displaystyle A^4$ which is really $\displaystyle A^3(A)$ or $\displaystyle (A)A^3$.
So not only have you shown that the inverse exists but also that the inverse of $\displaystyle A$ is in fact $\displaystyle A^3$ since $\displaystyle (A)A^3 = A^3(A) = I$ which is the definition of an inverse matrix.