Question

If $B,C$ are $n$ rowed square matrices and if $A=B+C$, $BC=CB$, ${{C}^{2}}=O$, then show that for every $n\in N$, ${{A}^{n+1}}={{B}^{n}}\left( B+\left( n+1 \right)C \right)$.

Answer 1

To prove the required condition, we will consider the given equation i.e. $A=B+C$ and calculates the value of ${{A}^{n+1}}$ by using the above equation. Now we will use the binomial expression ${{\left( a+b \right)}^{n}}={{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...+{}^{n}{{C}_{n-1}}a{{b}^{n-1}}+{{b}^{n}}$ and the remaining given equations that are $BC=CB$, ${{C}^{2}}=0$. After using above equations in the value of ${{A}^{n+1}}$ we will get our final result.

Complete step by step answer:
Given that, $B,C$ are $n$ rowed square matrices and $A=B+C$.
Now the value of ${{A}^{n+1}}$ can be calculated by using the above equation as
${{A}^{n+1}}={{\left( B+C \right)}^{n+1}}$
For this problem we are denoting the matrix $C$ by the letter $D$, then the above equation modified as
${{A}^{n+1}}={{\left( B+D \right)}^{n+1}}$
We have the binomial expansion of ${{\left( a+b \right)}^{n}}$ as ${{\left( a+b \right)}^{n}}={{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...+{}^{n}{{C}_{n-1}}a{{b}^{n-1}}+{{b}^{n}}$ so expanding the term ${{\left( B+D \right)}^{n+1}}$ by using this expansion in the above equation, then we will get
$\begin{aligned} & {{A}^{n+1}}={{\left( B+D \right)}^{n+1}} \\\ & \Rightarrow {{A}^{n+1}}={{B}^{n+1}}+{}^{n+1}{{C}_{1}}{{B}^{n+1-1}}D+{}^{n+1}{{C}_{2}}{{B}^{n+1-2}}{{D}^{2}}+...+{}^{n+1}{{C}_{n+1-1}}{{B}^{n+1-n}}{{D}^{n}}+{{D}^{n+1}} \\\ & \Rightarrow {{A}^{n+1}}={{B}^{n+1}}+{}^{n+1}{{C}_{1}}{{B}^{n}}D+{}^{n+1}{{C}_{2}}{{B}^{n-1}}{{D}^{2}}+...+{}^{n+1}{{C}_{n}}B{{D}^{n}}+{{D}^{n+1}} \\\ \end{aligned}$
In the problem we have given that ${{C}^{2}}=O$, we are denoting matrix $C$ by the letter $D$ $\Rightarrow {{D}^{2}}=O$
If ${{D}^{2}}=O$ then ${{D}^{3}}={{D}^{4}}=...={{D}^{n}}=O$, now the value of ${{A}^{n+1}}$ is modified as
$\begin{aligned} & {{A}^{n+1}}={{B}^{n+1}}+\left( n+1 \right){{B}^{n}}D+O+O+O+...+O \\\ & \Rightarrow {{A}^{n+1}}={{B}^{n+1}}+\left( n+1 \right){{B}^{n}}D \\\ \end{aligned}$
Taking ${{B}^{n}}$ as common in the above equation, then we will have
${{A}^{n+1}}={{B}^{n}}\left[ B+\left( n+1 \right)D \right]$
Here we have denoted matrix $C$ by the letter $D$, so we need to replace the letter $D$ with $C$ to get the result.
$\therefore {{A}^{n+1}}={{B}^{n}}\left[ B+\left( n+1 \right)C \right]$
Hence Proved.

Note: For this type of problems we can also use the substitution method i.e. check the given condition according to the given values for different values of $n$ i.e. $n=1,2,3,..$
For example, for $n=1$
The value of ${{A}^{n+1}}$ is
$\begin{aligned} & {{A}^{n+1}}={{B}^{n}}\left[ B+\left( n+1 \right)C \right] \\\ & \Rightarrow {{A}^{1+1}}={{B}^{1}}\left[ B+\left( 1+1 \right)C \right] \\\ & \Rightarrow {{A}^{2}}=B\left[ B+2C \right] \\\ & \Rightarrow {{A}^{2}}={{B}^{2}}+2BC....\left( \text{i} \right) \\\ \end{aligned}$
Now the value of ${{A}^{2}}$ from the given equation $A=B+C$ is given by
$\begin{aligned} & {{A}^{2}}={{\left( B+C \right)}^{2}} \\\ & \Rightarrow {{A}^{2}}={{B}^{2}}+2BC+{{C}^{2}} \\\ \end{aligned}$
Given that ${{C}^{2}}=O$
$\therefore {{A}^{2}}={{B}^{2}}+2BC...\left( \text{ii} \right)$
From equations $\left( \text{i} \right)$ and $\left( \text{ii} \right)$ we got the same result. So, we can conclude that ${{A}^{n+1}}={{B}^{n}}\left( B+\left( n+1 \right)C \right)$ is valid for every $n\in N$. If it is a multiple-choice question and there is another option with the same result for $n=1$, then it is better to check the condition for $n=2$.