Question

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

Answer 1

Let $P\left(n\right)$ : If $AB=BA$, then $A{B}^n=B^{n}A$

For $n=1$

L.H.S. $=A{B}^1=AB$

R.H.S.$=B^{1}A=BA=AB$

So, L.H.S.=R.H.S.

$\therefore P\left(n\right)$ is true for $n=1$.

Let us assume $P\left(k\right)$ is true for some $kϵN$

$P\left(k\right)$ : If $AB=BA$, then $A{B}^k=B^{k}A$ …(1)

We will prove that $P(k+1)$ is true.

$P(k+1)$: If $AB=BA$, then $A{B}^{k+1}=B^{k+1}A$

Taking L.H.S.

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

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

$=(B^k.A)B$ (From (1), $A{B}^k=B^{k}A)$

$=B^k\left(AB\right)$

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

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

$=B^k+1A$

=R.H.S.

Hence $P(k+1)$ is true is when $P\left(k\right)$ is true.

$\therefore$ By the principle of mathematical induction,

$P\left(n\right)$ is true for all $nϵN$.

$\therefore$ If $AB=BA$, then $A{B}^n=B^{n}A$ where $nϵN$.

Let $P\left(n\right)$ : If $AB=BA$, then $(AB{)}^n=A^n{B}^n$

For $n=1$

L.H.S. $=(AB{)}^1=AB$

R.H.S. $=A^1{B}^1=AB$

So, L.H.S.=R.H.S.

$\therefore P\left(n\right)$ is true for $n=1$.

We assume that $P\left(k\right)$ is true for some $kϵN$

$P\left(k\right)$ : If $AB=BA$, then $(AB{)}^k=A^k{B}^k$ …(2)

We will prove that $P(k+1)$ is true

$P(k+1)$: If $AB=BA$, then

$(AB{)}^{k+1}=A^{k+1}.B^{k+1}$

Taking L.H.S.

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

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

$=A^k{B}^k\left(BA\right)$ (Given: $AB=BA$)

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

$=A^k\left(B^{k+1}\right)A$

$=A^k\left(B^{k+1}A\right)$

$=A^k\left(A{B}^{k+1}\right)$ (Proved by induction)

$=\left(A^{k}A\right)B^{k+1}$

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

=R.H.S.

Hence $P(k+1)$ is true

$\therefore$ By the principle of mathematical induction, $P\left(n\right)$ is true for all $nϵN$.

$\therefore$ If $AB=BA$, then $(AB{)}^n=A^n{B}^n$ where $nϵN$.