Question

If $A$ and $B$ are square matrices of the same order such that $AB=BA$, then prove by induction that ${AB}^n=B^{n}A$. Further, prove that ${\left(AB\right)}^n=A^n{B}^n$ for all $n\in N$.

Answer 1

$A$ and $B$ are square matrices of the same order such that $AB=BA$

To prove: $P\left(n\right):{AB}^n=B^{n}A,n\in N$

For $n=1$, we have:

$P\left(1\right):AB=BA$ [Given]

$\Rightarrow {AB}^1=B^{1}A$

Therefore, the result is true for $n=1$.

Let the result be true for $n=k$.

$P\left(k\right):{AB}^k=B^{k}A$..........(1)

Now, we prove that the result is rue for $n=k+1$.

${AB}^{k+1}={AB}^k.B$

$=\left(B^{k}A\right)B$ [By(1)]

$=B^k\left(AB\right)$ [Associative law]

$=B^k\left(BA\right)$ [$AB=BA$ (given)]

$=\left(B^{k}B\right)A$ [Associative law]

$=B^{k+1}A$

Therefore, the result is true for $n=k+1$.

Thus, by the principle of mathematical induction , we have ${AB}^n=B^{n}A,n\in N$.

Now, we prove that $\left({AB}^n\right)=A^n{B}^n$ for all $n\in N$

For $n=1$, we have:
${\left(AB\right)}^1=A^1{B}^1=AB$

Therefore, the result is true for $n=1$.

Let the result be true for $n=k$.

${\left(AB\right)}^k=A^k{B}^k$...........(2)

Now we prove that the result is true for $n=k+1$.

${\left(AB\right)}^{k+1}={\left(AB\right)}^k.\left(AB\right)$

$=\left(A^k{B}^k\right).\left(AB\right)$ [By (2) ]

$=A^k\left(B^{k}A\right)B$ [Associative law]

$=A^k\left(A^{k}B\right)B[{AB}^n=B^{n}Aforalln\in N]$

$=\left(A^{k}A\right).\left(B^{k}B\right)$ [Associative law]

$=A^{k+1}{B}^{k+1}$

Therefore, the result is true for $n=k+1$.

Thus, by the principle of mathematical induction , we have ${\left(AB\right)}^n=A^n{B}^n$, for all natural numbers.