If Matrix $A^3 = 0$ then what will be $I+A+A^2$ Matrix $A$ is a square matrix such that $$A^3=0$$ then what would be value of $$I+A+A^2=?$$ such that $I$ is unit matrix of same order as $A$
I firstly supposed sum to be $X$ that is 
$X=I+A+A^2$ then 
$$AX=A+A^2+A^3$$
$$AX=A+A^2$$
and multiplying both sides with $A^{-1}$ we get 
$$K=I+A$$ but my answer was incorrect , what is correct solution to this problem?

If $A^3=0$, then $A^4=0$, $A^5=0$, and so on, so $I+A+A^2=I+A+A^2+A^3+A^4+\cdots$. The infinite sum $I+A+A^2+A^3+A^4+\cdots$ looks a lot like an infinite geometric series. There’s a formula for the sum of an infinite geometric series (of real numbers, if the ratio has absolute value less than $1$), $\frac{a}{1-r}=a+ar+ar^2+\cdots$ so maybe it would be interesting to play with that formula in this case:
$$I+A+A^2+A^3+A^4+\cdots \overset{?}{=}\frac{I}{I-A}.$$
Given that $I+A+A^2+A^3+A^4+\cdots = I+A+A^2$, this becomes
$$I+A+A^2 \overset{?}{=}\frac{I}{I-A}.$$
Even if you don’t believe this makes any sense, it might lead you to look at whether 
$$\left(I+A+A^2\right)\left({I-A}\right) =I.$$
It does, so you’ve discovered that when $A^3=0$, $I+A+A^2=\left(I-A\right)^{-1}$.
