If ${A^3} = O$, then $I + A + {A^2}$ equals
A.$I - A$
B.$\left( {I + {A^{ - 1}}} \right)$
C.${\left( {I - A} \right)^{ - 1}}$
D.None of these

Hint: As, it is given in the question ${A^3} = O$, and we can use this expression $I + A + {A^2}$, as a starting step, we can assume some value for this expression and multiply both side with A, as this will help us to make use of ${A^3} = O$, then the equation will be more simplified and we can use this equation in the main equation and find the value of the assume variable. Here, we should know one thing that when any matrix is multiplied by an identity matrix it always results in the same matrix.


Complete step by step answer:
Given, ${A^3} = O$, and we need to find the value of $I + A + {A^2}$.
Let, $y = I + A + {A^2}$ …….(1)
Multiply both sides with A.
$Ay = A\left( {I + A + {A^2}} \right)$
Now, simply the above equation,
$Ay = A + {A^2} + {A^3}$,
As the value of ${A^3} = O$, so substitute this value to the above equation.
$Ay = A + {A^2}$
Now, use the above equation in the equation (1).
$
  y = I + Ay \\
  I = y - Ay \\
  I = y\left( {1 - A} \right) \\
  y = I{\left( {1 - A} \right)^{ - 1}} \\
  y = {\left( {I - A} \right)^{ - 1}} \\
 $

Note- When asked to prove in the equation form like this we can directly prove the result by simplifying the equation.This can also be proved by taking a counter example. If any matrix is invertible, then only its inverse can be found out and the inverse can be found out by taking the ratio of adjoint of that matrix and determinant of the matrix.