Question

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

• A. $I-A+A^2$
• B. $I+A+A^2$
• C. $I+A^{-1}$
• D. I+A

Answer 1

Answer: (A)

Given, $|I+A|\ne O$
e, $(A+I)$ is a non-singular matrix.
$O\to$ Null Matrix
$I\to$ Unit Matrix
$\because I^3=I$
$\Rightarrow A^3=O$
$\Rightarrow A^3+I=O+I$
$\Rightarrow A^3+I^3=O+I$
$\Rightarrow (A+I)\left(A^2-A+I\right)=(O+I)$
$(A+I)\left(A^2-A+I\right)=I\because (O+I=I)$
Operate $(A+I{)}^{-1}$ on both sides
$\left\{(A+I{)}^{-1}(A+I)\right\}\left(A^2-A+I\right)$
$=(A+I{)}^{-1}\cdot I$
$\Rightarrow I\cdot \left(A^2-A+I\right)=(A+I{)}^{-1}$
$\left(\because I\cdot (A+I{)}^{-1}=(A+I{)}^{-1}\right)$
$(A+I{)}^{-1}=\left(A^2-A+I\right)$
or $(I+A{)}^{-1}=\left(I-A+A^2\right)$