Question

A matrix ‘A’ is given and it satisfies the given condition ${{A}^{3}}=0$. Find the value of $I+A+{{A}^{2}}$.
(a) I + A.
(b) ${{\left( I+A \right)}^{-1}}$.
(c) I – A.
(d) ${{\left( I-A \right)}^{-1}}$.

Answer 1

Hint: We start solving the problem by multiplying with ‘–1’ on both sides of ${{A}^{3}}=0$. We then add both sides with the Identity matrix. Now we take the product $\left( I-A \right).\left( I+A+{{A}^{2}} \right)$ and calculate it. We use the fact that if $A.B=I$, then $B={{A}^{-1}}$ and make subsequent arrangements to get the required result.

Complete step by step answer:
Given that we have matrix ‘A’ and it satisfies the condition ${{A}^{3}}=0$. We need to find the value of $I+A+{{A}^{2}}$.
$\Rightarrow $ ${{A}^{3}}=0$.
Let us multiply each side with ‘–1’.
$\Rightarrow $ $-1\times {{A}^{3}}=-1\times 0$.
$\Rightarrow $ $-{{A}^{3}}=0$.
We add the Identity matrix ‘I’ on both sides.
$\Rightarrow $ $I-{{A}^{3}}=I+0$.
$\Rightarrow $ $I-{{A}^{3}}=I$ ---(1).
Let us multiply $\left( I-A \right)$ and $\left( I+A+{{A}^{2}} \right)$ with each other.
$\Rightarrow $ $\left( I-A \right).\left( I+A+{{A}^{2}} \right)=I.\left( I+A+{{A}^{2}} \right)-A.\left( I+A+{{A}^{2}} \right)$.
$\Rightarrow $ $\left( I-A \right).\left( I+A+{{A}^{2}} \right)=I+A+{{A}^{2}}-\left( A+{{A}^{2}}+{{A}^{3}} \right)$.
$\Rightarrow $ $\left( I-A \right).\left( I+A+{{A}^{2}} \right)=I+A+{{A}^{2}}-A-{{A}^{2}}-{{A}^{3}}$.
$\Rightarrow $ $\left( I-A \right).\left( I+A+{{A}^{2}} \right)=I-{{A}^{3}}$ ---(2).
We substitute the result obtained from equation (2) in equation (1).
$\Rightarrow $ $\left( I-A \right).\left( I+A+{{A}^{2}} \right)=I$ ---(3).
We know that if $A.B=I$, then $B={{A}^{-1}}$. We use this result in equation (3).
$\Rightarrow $ $\left( I+A+{{A}^{2}} \right)={{\left( I-A \right)}^{-1}}$.
We got the value of $I+A+{{A}^{2}}$ as ${{\left( I-A \right)}^{-1}}$.
∴ If ${{A}^{3}}=0$ then the value of $I+A+{{A}^{2}}$ is equal to ${{\left( I-A \right)}^{-1}}$.

So, the correct answer is “Option D”.

Note: Here the matrix $\left( I-A \right)$ is considered as invertible while solving the problem. To say that the matrix $\left( I-A \right)$ invertible, we use the fact that for a nilpotent matrix of order ‘n’ we always have an matrix $\left( I-A \right)$ which is invertible. Here matrix ‘A’ is a nilpotent matrix of order 3 in the given problem.