Question

If A and B are square matrices of the same order such that \(AB=BA\),then prove by induction that \(AB^n=B^nA\).Further, prove that \((AB)^n=A^nB^n\) for all \(n∈N\)

Answer 1

A and B are square matrices of the same order such that \(AB = BA\). To prove \(P(n):AB^n=B^nA\,\,\, n∈N\) 
For \(n = 1\), we have:
\(P(1):AB^1=B^1A\implies  AB=BA\) 
Therefore, the result is true for \(n = 1\). 
Let the result be true for \(n = k\). 
\(P(k):AB^k=B^kA .....(1)\) 
Now, we prove that the result is true for \(n = k + 1\).
\(AB^k+1=AB^k.B\)
\(=(B^kA)B\,\,\,\,\,\, [by (1)]\)
\(=B^k(AB) [Associative law]\)
\(=B^k(BA) [AB=BA\,\, given]\)
\(=(B^kB)A [Associative\,\, law]\)
\(=B^k+1A\)
Therefore, the result is true for \(n = k + 1.\)
Thus, by the principle of mathematical induction, we have
\(AB^n=B^nA,n∈N\)
Now, we prove that \((AB)^n=A^nB^n\) for all \(n∈N\)
For \(n = 1\), we have:
\((AB)^1=A^1B^1=AB\)
Therefore, the result is true for \(n = 1\).
Let the result be true for \(n = k\).
\((AB)^k=A^kB^k ....(2)\)
Now, we prove that the result is true for \(n = k + 1\).
\((AB)^{k+1}=(AB)^k.(AB)\)
\(=(A^kB^k).(AB) [By(2)]\)
\(= A^k(B^kA)B [Associative\,\, law]\)
\(=A^k(AB^k)B [AB^n=B^nA\,\, for\,\, all\,\, n∈N]\)
\(=(A^kA).(B^kB) [Associative\,\, law]\)
\(=A^{k+1}B^{k+1}\)
Therefore, the result is true for \(n = k + 1\)
Thus, by the principle of mathematical induction, we have \((AB)^n=A^nB^n\) , for all natural numbers.