 Hello friends, welcome to the session I am Alka, we are going to discuss matrices. Now our given question is if A equal to matrix 102, 021, 203 then we have to prove that A cube minus 6A squared plus 7A plus 2I equal to 0. Now let's start with the solution. Now we are given that A equal to matrix 102, 021, 203. Now we will calculate the value of A squared, therefore A squared equal to A into A which can be written as, now we will multiply A into A, this will give us 1 into 1 plus 0 into 0 plus 2 into 2 then again 1 into 0 plus 0 into 2 plus 2 into 0 then 1 into 2 plus 0 into 1 plus 2 into 3. Now similarly we will solve for the second row and third row. So this gives A squared equal to 1 plus 0 plus 4, 0 plus 0 plus 0, 2 plus 0 plus 6, 0 plus 0 plus 2, 0 plus 4 plus 0, 0 plus 0 plus 3, 2 plus 0 plus 6, 0 plus 0, 4 plus 0 plus 9, this will give us 5, 0, 8, 2, 4, 5, 8, 0, 13. So this is the value of A squared, let's name this as our first equation. Now we will find A cube where A cube equal to A squared into A, it can be written as A cube equal to A squared that is 5, 0, 8, 2, 4, 5, 8, 0, 13 into matrix 1, 0, 2, 0, 2, 1, 2, 0, 3 which is A. So therefore A cube equal to 5 into 1 plus 0 into 0 plus 8 into 2 then 5 into 0 plus 0 into 2 plus 8 into 0, 5 into 2 plus 0 into 1 plus 8 into 3. Now we will write for the second row and third row. This is equal to 5 plus 0 plus 16, 0 plus 0 plus 0, 10 plus 0 plus 24, 2 plus 0 plus 10, 0 plus 8 plus 0, 4 plus 4 plus 15, 8 plus 0 plus 26, 0 plus 0 plus 0, 16 plus 0 plus 39. This is equal to 21, 0, 34, 12, 8, 23, 34, 0, 55. Now this is the value of A cube, so let's name this as our second equation. Now we will calculate the value of minus 6 A squared that is equal to minus 6 into A squared which is 5, 0, 8, 2, 4, 5, 8, 0, 13 which we have already calculated earlier. So this can be written as this is equal to minus 30, 0, minus 48, minus 12, minus 24, minus 30, minus 48, 0 and minus 78. So this is the value of minus 6 A squared. Let's name it as our third equation. Now we will solve for 7A, 7A equal to 7 into matrix A that is 1, 0, 2, 0, 2, 1, 2, 0, 3 which is equal to therefore 7A equal to 7, 0, 14, 0, 7, 14, 0, 21. This is our fourth equation. Now we will solve the value of 2I which is equal to 2 into identity matrix of third order that is 1, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 2, 0, 0, 0, 2 is the value of 2I. Now let's name it as our fifth equation. Now on adding equation number second, third, fourth and fifth we get this is equal to 21, minus 30, plus 7, plus 2, 0, plus 0, plus 0, 34, minus 48, plus 14, plus 0. Now similarly we will write for the second row and third row. This is equal to 0, 0, 0, 0, 0, 0, 0, 0, 0 equal to 0. Hence A cubed minus 6 A squared plus 7A plus 2I equal to 0. Hence prove. So which is the required solution? Hope you understood it. I'll enjoy the session. Goodbye and take care.