 Hello friends, welcome to the session I am Malka. We are going to discuss matrices. Our given question is let's a equal to matrix 0 1 0 0 show that bracket a i plus b a to the power n equal to a n i plus n a n minus 1 into b a where i is the identity matrix of order 2 and n is the element of natural numbers. Now let me tell you the basic idea behind the question which is mathematical induction method. In this method we first show that the result is true for n equal to 1 then we assume that the result is true for n equal to k and we prove that it is true for n equal to k plus 1. If the result is true for n equal to k plus 1 then we conclude that the result is true for all natural numbers. Let's begin with the solution we are given a equal to matrix 0 1 0 0 and we have to show that a i plus b a to the power n equal to a n i plus n a n minus 1 b a. Now this is our second equation. Now let us say show that the result is true for n equal to 1 on putting n equal to 1 in equation second we have a i plus b a to the power 1 equal to a to the power 1 i plus 1 a 1 minus 1 b a that is a i plus b a equal to a i plus a to the power 0 b a and we all know that a to the power 0 equal to 1. So this can be written as a i plus b a equal to a i plus b a. So the result is true for n equal to 1. Now let us assume that result is true for n equal to k that is a i plus b a to the power k equal to a k i plus k a k minus 1 b a. Let this be our third equation. Now we show that the result is true for n equal to k plus 1 a i plus b a to the power k plus 1 equal to a to the power k plus 1 i plus k plus 1 a to the power k b a. Now we will consider it's LHS first LHS equal to a i plus b a to the power k plus 1. This can also be written as a i plus b a to the power k into a i plus b a. Now we will place the value of a i plus b a to the power k from equation number third. This will give us this is equal to a k i into a i plus a k i into a i plus k a k minus 1 b a into a i plus k a k minus 1 b a into b a. This can be written as a k plus 1 i square plus a k b i a plus k a k minus 1 plus 1 b a i plus k a k minus 1 b square a square. Now since we all know that i square equal to i into i which is equal to i and a i equal to i a equal to a. Therefore we get a k plus 1 i plus a k b i a i a equal to a k a k b then again a i equal to a plus k a k minus 1 b square a square. Now since from equation first we have a equal to 0 1 0 0 matrix this gives a square equal to a into a that is 0 1 0 0 into matrix 0 1 0 0. This will give us a square equal to 0 0 0 0. Now on placing the value of a square in the above equation we get a k plus 1 i plus 1 plus k a to the power k b a plus k a k minus 1 b square into 0 which is equal to a k plus 1 i plus 1 k a k b a and this is 0. So this is our RHS.