Question

If $B={{B}^{2}}$ and $I-B=A$ then ( $A$ , $B$ are square matrix of same order)
(i) ${{A}^{2}}=B$
(ii) ${{A}^{2}}=A$
(iii) ${{A}^{2}}=I$
(iv) ${{A}^{2}}=-A$

Answer 1

Hint: In this problem we need to calculate the value of ${{A}^{2}}$ according to the given data. We have given that $B={{B}^{2}}$ and $I-B=A$ where $A$ , $B$ are square matrices of the same order. We will first consider the equation $I-B=A$ and square the equation on both sides. We will factorize the term ${{\left( I-B \right)}^{2}}$ as $\left( I-B \right)\left( I-B \right)$ and simplify the equation by using the distribution law of multiplication. Now we will use some matrix rules like the square of the identity matrix is always equal to the identity matrix and the product of a matrix with identity matrix is always equal to the multiplied matrix. After applying these two rules in the equation we will use the given value which is $B={{B}^{2}}$ and simplify the equation to get the required result.

Complete step by step answer:
Given that, $B={{B}^{2}}$ and $I-B=A$ where $A$ , $B$ are square matrices of the same order.
Considering the equation $I-B=A$. Applying square on both sides of the above equation to calculate the value of ${{A}^{2}}$, then we will get
${{A}^{2}}={{\left( I-B \right)}^{2}}$
Factoring the term ${{\left( I-B \right)}^{2}}$ as $\left( I-B \right)\left( I-B \right)$ in the above equation, then we will have
${{A}^{2}}=\left( I-B \right)\left( I-B \right)$
Applying distribution law of multiplication in the above equation, then we will get
$\begin{align}
  & {{A}^{2}}={{I}^{2}}-BI-BI-B\left( -B \right) \\
 & \Rightarrow {{A}^{2}}={{I}^{2}}-2BI+{{B}^{2}} \\
\end{align}$
We know that the square of the identity matrix is always an identity matrix, so the above equation is modified as
${{A}^{2}}=I-2BI+{{B}^{2}}$
We have the matrix rule that the product of a matrix with identity matrix will give the multiplied matrix as the result. Applying this rule in the above equation, then we will have
${{A}^{2}}=I-2B+{{B}^{2}}$
We have given the value $B={{B}^{2}}$. Substituting this value in the above equation, then we will get
$\begin{align}
  & {{A}^{2}}=I-2B+B \\
 & \Rightarrow {{A}^{2}}=I-B \\
\end{align}$
In the problem they have mentioned that $I-B=A$. Substituting this value in the above equation, then we will have
${{A}^{2}}=A$

So, the correct answer is “Option (ii)”.

Note: We can observe that all of the matrix problems similar to this problem are solved by assuming any matrix value. But in this problem we have given that $B={{B}^{2}}$. So we need to consider the value of matrix $B$ such that it satisfies the equation $B={{B}^{2}}$.