 Hi, and welcome to the session. I'm Shashi and I'm going to help you to solve the following question Question is if a and b are square matrices of the same order such that a b is equal to b a then prove by induction that a multiplied by b raised to the power n is equal to b raised to the power n multiplied by a Further prove that a b whole raised to the power n is equal to a raised to the power n multiplied by b raised to the power n for all n belonging to set of natural numbers First of all, let us understand That in mathematical induction method we first prove that the result is true for n is equal to 1 Then we assume that the result is true for n is equal to k and then prove that it is true for n is equal to k plus 1 If so, then we conclude that the result is true for every n belonging to set of natural numbers n belongs to the set of natural numbers in the given question So we have taken n in the set of natural numbers Now this is the key idea to solve the given question Now let us start with the solution We are given that a b is equal to b a And we have to prove that a b raised to the power n is equal to b raised to the power n multiplied by a For every n belonging to set of natural numbers This is given to us and We have to prove a multiplied by b raised to the power n is equal to b raised to the power n multiplied by a First of all by key idea, we know we will first prove that the result is true for n is equal to 1 so putting n is equal to 1 in a b raised to the power n is equal to b raised to the power n multiplied by We get a multiplied by b raised to the power 1 is equal to b raised to the power 1 multiplied by a This implies a b is equal to b a Which is already given to us So It is true. Let us name this equation as 1. So we can write which is true by Equation 1 Now let us assume that the result is true for n is equal to k. So we can write a multiplied by b raised to the power k is equal to b raised to the power k multiplied by a Let us name this equation as 2 Now we will prove that the result holds for n is equal to k plus 1 now, let us consider a Multiplied by b raised to the power k plus 1 It can be written as a multiplied by b raised to the power k multiplied by b Or we can write it as b raised to the power k Multiplied by a b Now from equation 1, we know a b is equal to b a so we can write b raised to the power k Multiplied by b a now You know bases are same. So we can write b raised to the power k plus 1 multiplied by a Is equal to a multiplied by b raised to the power k plus 1 So the result is true for n is equal to k plus 1 Thus by principle of mathematical induction the result is true for every natural number n So we can write by principle of mathematical induction a multiplied by b raised to the power n is equal to b raised to the power n Multiplied by a for every n belonging to set of natural numbers So this completes the first part of the question Now we have to prove that a b whole raised to the power n is equal to a raised to the power n Multiplied by b raised to the power n for every n belonging to set of natural numbers Now let us name this expression as 3 now again we will use the principle of Mathematical induction We will first prove that the result represented by the equation 1 is true for n is equal to 1 So putting n is equal to 1 in equation 3 We get a b whole raised to the power 1 is equal to a raised to the power 1 Multiplied by b raised to the power 1 this implies a b is equal to a b Now since left-hand side is equal to right-hand side, so obviously It is true For n is equal to 1 Now let us assume that result is true for n is equal to a that is a b whole raised to the power k is Equal to a raised to the power k multiplied by b raised to the power k now We will prove that the result holds for n is equal to k plus 1 Now let us consider a b whole raised to the power k plus 1 which is equal to a b whole raised to the power k Multiplied by a b Now let us name this equation as 4 now using equation 4 we get a raised to the power k b raised to the power k Multiplied by a b now this is further equal to a raised to the power k multiplied by a Multiplied by b raised to the power k multiplied by b We know from equation 2 a Multiplied by b raised to the power k is equal to b raised to the power k multiplied by a so Using equation 2 we can write b raised to the power k multiplied by a is equal to a Multiplied by b raised to the power k Now, this is further equal to a raised to the power k plus 1 multiplied by b raised to the power k plus 1, so a b whole raised to the power k plus 1 is equal to a raised to the power k plus 1 multiplied by b raised to the power k plus 1. So, we can write the result a b whole raised to the power n is equal to a raised to the power n multiplied by b raised to the power n is true for n is equal to k plus 1. By principle of mathematical induction, the result a b whole raised to the power n is equal to a raised to the power n multiplied by b raised to the power n is true for every n belonging to set of natural numbers. Hence, we have proved both the parts of the question. So, this completes the session. Hope you understood the session. Take care and goodbye.