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(A) Given that $\operatorname{det}(I+A) 
eq 0$,which implies that $(I+A)$ is an invertible matrix.We are given $A^3 = O$.We know the algebraic identi

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$I-A+A^2$

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(A) Given that $\operatorname{det}(I+A) 
eq 0$,which implies that $(I+A)$ is an invertible matrix.We are given $A^3 = O$.We know the algebraic identity $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$.Substituting $x=A$ and $y=I$,we get:$A^3 + I^3 = (A+I)(A^2 - AI + I^2)$.Since $I^3 = I$ and $A^3 = O$,the equation becomes:$O + I = (A+I)(A^2 - A + I)$.$I = (A+I)(A^2 - A + I)$.Since $(A+I)$ is invertible,we can multiply both sides by $(A+I)^{-1}$ from the left:$(A+I)^{-1} I = (A+I)^{-1} (A+I)(A^2 - A + I)$.$(A+I)^{-1} = I(A^2 - A + I)$.$(A+I)^{-1} = I - A + A^2$.

If $A$ is a non-zero square matrix of order $n$ with $\operatorname{det}(I+A) \neq 0$ and $A^3=O$,where $I$ and $O$ are the unit and null matrices of order $n \times n$ respectively,then $(I+A)^{-1}$ is equal to

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