Q108 / 160 · TS EAMCET 2010 · Mathematics

If $A$ is a non-zero square matrix of order $n$ with $\operatorname{det}(I+A) \neq 0$ 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

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

Solution

Given, $|I+A| \neq O$ ie, $(A+I)$ is a non-singular matrix. $O \rightarrow$ Null Matrix $I \rightarrow$ Unit Matrix $\because \quad I^3=I$ $\Rightarrow \quad A^3=0$ $\Rightarrow \quad A^3+I=O+I$ $\Rightarrow \quad A^3+I^3=O+I$ $\Rightarrow \quad(A+I)\left(A^2-A+I\right)=(O+I)$ $(A+I)\left(A^2-A+I\right)=I \quad \because(O+I=I)$ Operate $(A+I)^{-1}$ on both sides $\left\{(A+I)^{-1}(A+I)\right\}\left(A^2-A+I\right)$ $=(A+I)^{-1} \cdot I$ $\Rightarrow \quad I \cdot\left(A^2-A+I\right)=(A+I)^{-1}$ $\left(\because I \cdot(A+I)^{-1}=(A+I)^{-1}\right)$ $\Rightarrow \quad(A+I)^{-1}=\left(A^2-A+I\right)$ or $(I+A)^{-1}=\left(I-A+A^2\right)$

Hence, the correct answer is option: 'A'.

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