 Hello friends welcome to the session we are going to discuss matrices if a equal to matrix 1 1 1 1 1 1 1 1 1 we have to prove that a to the power n equal to 3 to the power n minus 1, 3 to the power n minus 1, 3 to the power n minus 1, 3 to the power n minus 1, 3 to the power n minus 1, 3 to the power n minus 1, 3 to the power n minus 1 and 3 to the power n-1 matrix where n is the element of natural numbers. Now let's start with the solution. In this question you are also going to follow the mathematical induction method. Let me tell you the basic idea. It is 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 start with the solution we are given a equal to matrix 1 1 1 1 1 1 1 1 and we have to show that a n equal to 3 to the power n minus 1 3 to the power n minus 1 3 to the power n minus 1 3 to the power n minus 1 3 to the power n minus 1 3 to the power n minus 1 3 to the power n minus 1 3 to the power n minus 1 and 3 to the power n minus 1 matrix for all n being the element of n. Let's name this as our second equation. Now we'll prove the result for n equal to 1. We get a 1 equal to 3 to the power 1 minus 1 3 to the power 1 minus 1 3 to the power 1 minus 1 similarly second row and third row. This implies a equal to 3 to the power 0 3 to the power 0 3 to the power 0. Similarly for the second row and third row this implies a equal to 1 1 1 1 1 1 1 1 1 1. As we all know that any number to the power 0 equal to 1 so the result is true for n equal to 1. Now for n equal to k we get a to the power k equal to 3 to the power k minus 1 3 to the power k minus 1 3 to the power k minus 1 similarly second row and third row and we name it as our third equation. Now we'll prove the result for n equal to k plus 1 a to the power k plus 1 can be written as a to the power k into a. Now we'll substitute the value of a to the power k from equation number third and value of a from equation number first. This gives us this can be written as 3 into 3 to the power k minus 1 3 into 3 to the power k minus 1 3 into 3 to the power k minus 1 similarly we'll write for the second row and third row this is equal to 3 k minus 1 plus 1 3 to the power k minus 1 plus 1 3 to the power k minus 1 plus 1 similarly for the second row and this is equal to 3 to the power k 3 to the 3 to the power k, 3 to the power k, 3 to the power k, 3 to the power k, 3 to the power k and 3 to the power k matrix. Now we see that the result is true for n equal to k plus 1. Hence the result is true for n equal to k plus 1 and by the principle of mathematical induction the result is true for any natural number in. So hope you understood the solution and enjoyed the session, goodbye and take care.