If $A^3 =0$ such that $A^n≠ I$ for $1 ≤n≤4$, then $(I-A)^{-1}=?$

$A^4$ $A^3$ $I+A$ none of these

none of these

We have, $A^4 (I-A) = A^4-A^5 = A^4-O = A^4 ≠I$ $A^3 (I-A) = A^3-A^4 ≠ I$ and, $(I + A) (I - A) = I-A^2 ≠ I$ Hence, $(I-A)^{-1} ≠A^4, A^3, I+A$.