Question

If A is a square matrix such that ${{A}^{2}}=A$, then ${{\left( I+A \right)}^{3}}-7A$ is.
(a) $3I$
(b) $O$
(c) $I$
(d) $2I$

Answer 1

Hint: For solving this question first we will see an important result of matrix algebra,i.e. $A\cdot I=I\cdot A=A$ . Then, we will multiply $\left( I+A \right)$ with itself to get ${{\left( I+A \right)}^{2}}$. After that, we will pre multiply $\left( I+A \right)$ with ${{\left( I+A \right)}^{2}}$ further. And then, we will solve further to find the result of ${{\left( I+A \right)}^{3}}-7A$ .

Complete step-by-step solution -
Given:
It is given that, $A$ is a square matrix such that ${{A}^{2}}=A$ and we have to solve for the result of ${{\left( I+A \right)}^{3}}-7A$.
Now, we have the following equation:
${{A}^{2}}=A............\left( 1 \right)$
Now, we know that $A\cdot {{A}^{-1}}=I$ where $I$ is the identity matrix of the same order. And as we know that when we multiply an identity matrix with any other matrix $A$ , then we will get the resultant matrix as $A$ itself. Then,
$A\cdot I=I\cdot A=A.................\left( 2 \right)$
Now, we will multiply $\left( I+A \right)$ with itself to get ${{\left( I+A \right)}^{2}}$ . Then,
$\begin{align}
  & {{\left( I+A \right)}^{2}}=\left( I+A \right)\times \left( I+A \right) \\
 & \Rightarrow {{\left( I+A \right)}^{2}}=I\cdot I+I\cdot A+A\cdot I+A\cdot A \\
 & \Rightarrow {{\left( I+A \right)}^{2}}={{I}^{2}}+I\cdot A+A\cdot I+{{A}^{2}} \\
\end{align}$
Now, we will write ${{A}^{2}}=A$ from equation (1), and $A\cdot I=I\cdot A=A$ from equation (2) in the above equation and ${{I}^{2}}=I$ . Then,
$\begin{align}
  & {{\left( I+A \right)}^{2}}={{I}^{2}}+I\cdot A+A\cdot I+{{A}^{2}} \\
 & \Rightarrow {{\left( I+A \right)}^{2}}=I+A+A+A \\
 & \Rightarrow {{\left( I+A \right)}^{2}}=I+3A \\
\end{align}$
Now, we will pre multiply by $\left( I+A \right)$ in the above equation to get ${{\left( I+A \right)}^{3}}$. Then,
$\begin{align}
  & {{\left( I+A \right)}^{2}}=I+3A \\
 & \Rightarrow {{\left( I+A \right)}^{3}}=\left( I+A \right)\times \left( I+3A \right) \\
 & \Rightarrow {{\left( I+A \right)}^{3}}={{I}^{2}}+I\cdot \left( 3A \right)+A\cdot I+A\cdot \left( 3A \right) \\
 & \Rightarrow {{\left( I+A \right)}^{3}}={{I}^{2}}+3\left( I\cdot A \right)+A\cdot I+3{{A}^{2}} \\
\end{align}$
Now, we will write ${{A}^{2}}=A$ from equation (1), and $A\cdot I=I\cdot A=A$ from equation (2) in the above equation and ${{I}^{2}}=I$ . Then,
$\begin{align}
  & {{\left( I+A \right)}^{3}}={{I}^{2}}+3\left( I\cdot A \right)+A\cdot I+3{{A}^{2}} \\
 & \Rightarrow {{\left( I+A \right)}^{3}}=I+3A+A+3A \\
 & \Rightarrow {{\left( I+A \right)}^{3}}=I+7A \\
\end{align}$
Now, subtract the $7A$ in the above equation to get ${{\left( I+A \right)}^{3}}-7A$ . Then,
$\begin{align}
  & {{\left( I+A \right)}^{3}}=I+7A \\
 & \Rightarrow {{\left( I+A \right)}^{3}}-7A=I+7A-7A \\
 & \Rightarrow {{\left( I+A \right)}^{3}}-7A=I \\
\end{align}$
Now, from the above result, we conclude that, if ${{A}^{2}}=A$ then, ${{\left( I+A \right)}^{3}}-7A=I$ .
Hence, option (c) will be the correct option.

Note: Here, the student should first understand what is asked in the problem and then proceed in the right direction to get the correct answer quickly. And we should proceed stepwise while solving for smooth calculation. Moreover, though the question is very easy, and then solve further as per the rules of the matrix algebra. And remember $I\cdot A=A\cdot I=A$ as a formula in general for other questions.