Answer:
. For n = 1.
L.H.S. = ABn
= AB? = AB
R.H.S. = BbA
= B?A = BA
But AB = BA
L.H.S. =
R.H.S.
Let us suppose
that given statement is true for n = k.
Now consider for
n = k + 1
n = k + 1
[by associatively
of matrix multiplication]
= (Bk
A) B [Using (1)]
= Bk (AB)
= Bk (BA)
ABk+1
= Bk+1 A,
which is true
for n = k + 1.
2nd
Part : To prove
(AB)n
= An Bn
We
shall prove this result also by the principle of Mathematical Induction.
For n = 1.
L.H.S. =
(AB)? = AB
R.H.S. = An
Bn = A?B? = AB
L.H.S. =
R.H.S.
given
statement is true for n = 1
Let us suppose
that given statement is true for n = k.
Consider for n =
k + 1
(AB)k+1
= (AB)k . (AB)1 = AkBk (AB)
[Using
(2)]
[By associatively
of matrix multiplication]
= Ak
(Bk+1A) = AkABk+1
which is true
for n = k + 1.
Hence by PMI,
given statement is true for n = k + 1.
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