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If $A$ and $B$ are square matrices of the same order such that $\mathbf{A B}=\mathbf{B A}$, then prove by induction that $\mathbf{A B}{ }^n=\mathbf{B}^n A$. Further, prove that $(A B)^n=A^n B^n$ for all $n \in N$.
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Let $\mathrm{P}(\mathrm{n}): \mathrm{AB}^{\mathrm{n}}=\mathrm{B}^{\mathrm{n}} \mathrm{A}$,
But $\mathrm{n}=1, \mathrm{AB}=\mathrm{BA} \quad$ (Given)
$\therefore \mathrm{P}(\mathrm{n})$ is true for $\mathrm{n}=1$ Let $\mathrm{P}(\mathrm{n})$ be true for $\mathrm{n}=\mathrm{K}$
$\mathrm{AB}^{\mathrm{K}}=\mathrm{B}^{\mathrm{K}} \mathrm{A}$
Multiplying both side by $\mathrm{B}$
L.H.S. $=\mathrm{AB}^{\mathrm{K}} \mathrm{B}=\mathrm{A}\left(\mathrm{B}^{\mathrm{K}} \mathrm{B}\right)=\mathrm{AB}^{\mathrm{K}+1}$
R.H.S. $=\left(B^K A\right) B=B^K(A B)$
$=\mathrm{B}^{\mathrm{K}}(\mathrm{BA}) \quad \because \mathrm{AB}=\mathrm{BA}$
$=\left(\mathrm{B}^{\mathrm{K}} \mathrm{B}\right) \mathrm{A}=\mathrm{B}^{\mathrm{K}+1} \mathrm{~A}$ is $\mathrm{P}(\mathrm{n})$ is true for $\mathrm{n}=\mathrm{K}+1$
By principle of mathematical induction. $\mathrm{P}(\mathrm{n})$ is true for all $\mathrm{n} \in \mathrm{N}$.
Choose the correct answer in the following questions (ii) $\mathrm{P}(\mathrm{n}):(\mathrm{AB})^{\mathrm{n}}=\mathrm{A}^{\mathrm{n}} \mathrm{B}^{\mathrm{n}}$
For $n=1$, L.H.S. $=A B$ and R.H.S. $=A B$
$\therefore \mathrm{P}(\mathrm{n})$ is true for $\mathrm{n}=1$ Let $\mathrm{P}(\mathrm{n})$ be true for $\mathrm{n}=\mathrm{k}$
$$
(\mathrm{AB})^{\mathrm{K}}=\mathrm{A}^{\mathrm{K}} \mathrm{B}^{\mathrm{K}}
$$
Multiply both sides by $\mathrm{AB}$
$$
\begin{aligned}
&\text { L.H.S. }=(\mathrm{AB})^{\mathrm{K}}(\mathrm{AB})=(\mathrm{AB})^{\mathrm{K}+1} \\
&\text { R.H.S. }=\mathrm{A}^{\mathrm{K}} \mathrm{B}^{\mathrm{K}}(\mathrm{AB})=\mathrm{A}^{\mathrm{K}} \mathrm{B}^{\mathrm{K}}(\mathrm{BA}) \\
&\because \mathrm{AB}=\mathrm{BA}=\mathrm{A}^{\mathrm{K}}\left(\mathrm{B}^{\mathrm{K}} \cdot \mathrm{B}\right) \mathrm{A}=\mathrm{A}^{\mathrm{K}}\left(\mathrm{B}^{\mathrm{K}+1} \mathrm{~A}\right) \\
&\quad=\mathrm{A}^{\mathrm{K}}\left(\mathrm{AB}{ }^{\mathrm{K}+1}\right)=\mathrm{A}^{\mathrm{K}+1} \mathrm{~B}^{\mathrm{K}+1} \quad\left[\because \mathrm{AB}^{\mathrm{n}}=\mathrm{B}^{\mathrm{n}} \mathrm{A}\right] \\
&\Rightarrow \mathrm{P}(\mathrm{n}) \text { is true for } \mathrm{n}=\mathrm{K}+1
\end{aligned}
$$
By principal of mathematical induction $\mathrm{P}(\mathrm{n})$ is true for all $\mathrm{n} \in \mathrm{N}$.
But $\mathrm{n}=1, \mathrm{AB}=\mathrm{BA} \quad$ (Given)
$\therefore \mathrm{P}(\mathrm{n})$ is true for $\mathrm{n}=1$ Let $\mathrm{P}(\mathrm{n})$ be true for $\mathrm{n}=\mathrm{K}$
$\mathrm{AB}^{\mathrm{K}}=\mathrm{B}^{\mathrm{K}} \mathrm{A}$
Multiplying both side by $\mathrm{B}$
L.H.S. $=\mathrm{AB}^{\mathrm{K}} \mathrm{B}=\mathrm{A}\left(\mathrm{B}^{\mathrm{K}} \mathrm{B}\right)=\mathrm{AB}^{\mathrm{K}+1}$
R.H.S. $=\left(B^K A\right) B=B^K(A B)$
$=\mathrm{B}^{\mathrm{K}}(\mathrm{BA}) \quad \because \mathrm{AB}=\mathrm{BA}$
$=\left(\mathrm{B}^{\mathrm{K}} \mathrm{B}\right) \mathrm{A}=\mathrm{B}^{\mathrm{K}+1} \mathrm{~A}$ is $\mathrm{P}(\mathrm{n})$ is true for $\mathrm{n}=\mathrm{K}+1$
By principle of mathematical induction. $\mathrm{P}(\mathrm{n})$ is true for all $\mathrm{n} \in \mathrm{N}$.
Choose the correct answer in the following questions (ii) $\mathrm{P}(\mathrm{n}):(\mathrm{AB})^{\mathrm{n}}=\mathrm{A}^{\mathrm{n}} \mathrm{B}^{\mathrm{n}}$
For $n=1$, L.H.S. $=A B$ and R.H.S. $=A B$
$\therefore \mathrm{P}(\mathrm{n})$ is true for $\mathrm{n}=1$ Let $\mathrm{P}(\mathrm{n})$ be true for $\mathrm{n}=\mathrm{k}$
$$
(\mathrm{AB})^{\mathrm{K}}=\mathrm{A}^{\mathrm{K}} \mathrm{B}^{\mathrm{K}}
$$
Multiply both sides by $\mathrm{AB}$
$$
\begin{aligned}
&\text { L.H.S. }=(\mathrm{AB})^{\mathrm{K}}(\mathrm{AB})=(\mathrm{AB})^{\mathrm{K}+1} \\
&\text { R.H.S. }=\mathrm{A}^{\mathrm{K}} \mathrm{B}^{\mathrm{K}}(\mathrm{AB})=\mathrm{A}^{\mathrm{K}} \mathrm{B}^{\mathrm{K}}(\mathrm{BA}) \\
&\because \mathrm{AB}=\mathrm{BA}=\mathrm{A}^{\mathrm{K}}\left(\mathrm{B}^{\mathrm{K}} \cdot \mathrm{B}\right) \mathrm{A}=\mathrm{A}^{\mathrm{K}}\left(\mathrm{B}^{\mathrm{K}+1} \mathrm{~A}\right) \\
&\quad=\mathrm{A}^{\mathrm{K}}\left(\mathrm{AB}{ }^{\mathrm{K}+1}\right)=\mathrm{A}^{\mathrm{K}+1} \mathrm{~B}^{\mathrm{K}+1} \quad\left[\because \mathrm{AB}^{\mathrm{n}}=\mathrm{B}^{\mathrm{n}} \mathrm{A}\right] \\
&\Rightarrow \mathrm{P}(\mathrm{n}) \text { is true for } \mathrm{n}=\mathrm{K}+1
\end{aligned}
$$
By principal of mathematical induction $\mathrm{P}(\mathrm{n})$ is true for all $\mathrm{n} \in \mathrm{N}$.
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