If AB = BA for any two sqaure matrices, prove by mathematical ind… - Vidyadip

Question

If AB = BA for any two sqaure matrices, prove by mathematical induction that $(AB)^n = A^nB^n.$

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Answer

Let $P(n) : (AB)^n = A^nB^n$
$\therefore$ $P(1) : (AB)^1 = A^1B^1$
$\Rightarrow AB = AB$
So, P(1) is true.
Now, $P(k) : (AB)^k = A^kB^k$
$\text{k}\in\text{N}$
So, P(k) is true, whenever P(k + 1) is true.
$\therefore$ $P(k + 1)$
$\Rightarrow (AB)^{k+1} = A^{k+1}.B^{k+1}$
$\therefore$ $P(k + 1 : AB)^{k+1} = A^{k+1}B^{k+1}$
$\Rightarrow A^k.B^k.AB$
$\Rightarrow A^k.B^kBA$
$\Rightarrow A^kB^{k+1}A$
$\Rightarrow A^kA.B^{k+1}$
$\Rightarrow A^{k+1}B^{k+1}$
$\Rightarrow (A.B)^{k+1} = A^{k+1}B^{k+1}$
So,$ P(k+1)$ is true for all
$\text{n}\in\text{N},$ whenever P(k) is true.
By mathematical induction $(AB) = A^nB^n$ is
true for all $\text{n}\in\text{N}.$

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