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

Vedclass · Accepted Answer

$I-A+A^2$

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(A) Given that $\det(I+A) 
eq 0$, 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 $A^3 + I^3 = (A+I)(A^2 - A + I)$.Since $A^3 = O$ and $I^3 = I$, the equation becomes $O + I = (A+I)(A^2 - A + I)$.This simplifies to $I = (A+I)(A^2 - A + I)$.Multiplying both sides by $(A+I)^{-1}$ from the left, we get $(A+I)^{-1} I = (A+I)^{-1} (A+I)(A^2 - A + I)$.Since $(A+I)^{-1}(A+I) = I$, we have $(A+I)^{-1} = I(A^2 - A + I) = A^2 - A + I$.Thus, $(I+A)^{-1} = I - A + A^2$.

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

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