 Hi, I'm Meenu and I'm going to help you to solve this question. It says for the matrix A, show that A cubed minus 6A squared plus 5A plus 11I is equal to 0 and hence find A inverse. So let's move on to the solution. To prove this, we need to find A cubed A squared for the given matrix. Now A is 1, 1, 2, 1, 2, minus 1, 1, minus 3, 3. We now find A squared 1 minus 3, 3, 1, 1, 2, 1, 2, minus 1, 1, minus 3, 3. Let's now multiply the two matrices 1 into 1 plus 1 into 1 plus 1 into 2, 1 into 1, plus 1 into 2, plus 1 into minus 1, then 1 into 1, plus 1 into minus 3, plus 1 into 3, then 1 into 1, plus 2 into 1, plus minus 3 into 2, then 1 into 1, plus 2 into 2, plus minus 3 into minus 1, then 1 into 1, 2 into minus 3, minus 3 into 3, then 2 into 1, plus minus 1 into 1 plus 3 into 2 2 into 1 plus minus 1 into 2 plus 3 into minus 1 2 into 1 plus minus 1 into minus 3 plus 3 into 3 and it is again equal to 4 to 1 minus 3 8 minus 14 3 7 minus 3 and 14 3 plus 9 is 12 12 plus 2 is 14 we now find a cube so a cube is a square into a which is a square is 4 to 1 minus 3 8 minus 14 7 minus 3 14 into a a is 111 1 2 minus 3 2 minus 1 3 we now multiply the two matrices so we get 4 into 1 plus 2 into 1 plus 1 into 2 4 into 1 plus 2 into 2 plus 1 into minus 1 4 into 1 2 into minus 3 plus 1 into 3 then minus 3 into 1 8 into 1 plus 8 into 1 plus minus 14 into 2 then minus 3 into 1 plus 8 into 2 plus minus 14 into minus 1 then minus 3 into 1 8 into minus 3 plus minus 14 into 3 and again it becomes 7 into 1 plus minus 3 into 1 plus 14 into 2 7 into 1 plus minus 3 into 2 plus 14 into minus 1 then 7 into 1 minus 3 into 3 minus 3 into minus 3 14 into 3 n equal to 4 plus 2 plus 2 is 8 4 plus 4 is 8 minus 1 is 7 4 minus 6 is minus 2 minus 2 plus 3 is 1 minus 3 plus 8 is 5 5 plus 5 minus 28 is minus 23 minus 3 plus 16 is 13 13 plus 14 is 27 minus 3 minus 24 is minus 27 minus 27 minus 42 is minus 69 7 minus 3 is 4 plus 28 is 32 7 minus 4 is 7 minus 3 is 4 14 to the is 28 which is 32 7 minus 6 is 1 1 minus 14 is minus 13 7 plus 9 is 16 16 plus 42 is 58 this is a cube now we need to find 6 a square we have found out a square so we multiply each entry by 6 and we will get 6 a square now a square is equal to 4 to 1 minus 3 8 minus 14 that implies 6 a square is equal to 6 into 4 is 24 12 6 minus 18 6 8 is 48 minus 84 7 6 is 42 minus 18 84 now we have to prove that a cube minus 6 a square plus 5 a plus 11 i is equal to 0 so we will solve this and will prove that this is equal to 0 now a cube is 8 minus 23 32 7 27 minus 13 1 minus 69 58 minus 6 a square is 24 minus 18 42 12 48 minus 18 6 minus 84 84 plus 5 a 5 a is multiplying the matrix a with 5 5 5 with 5 so it will give 5 5 5 5 10 minus 15 then 10 minus 5 15 plus 11 i where i is the identity matrix of order 3 cross 3 so it is 11 0 0 0 11 0 0 0 11 so this becomes equal to 8 minus 24 plus 5 plus 11 7 minus 12 plus 5 plus 0 1 minus 6 plus 5 plus 0 minus 23 plus 18 minus minus is plus 18 plus 5 plus 5 plus 0 then we have 27 plus 4 minus 48 plus 10 plus 11 then we have minus 69 minus minus plus 84 minus 15 plus 0 then 32 minus 42 plus 10 plus 0 then minus 13 minus minus plus 18 minus 5 plus 0 then 58 minus 84 plus 15 plus 11 and then we will check this will be a 0 matrix so hence it is proved that a cube minus 6 a square plus 5 a plus 11 i is equal to 0 now we have to find the inverse of a using this equation which is a cube minus 6 a square plus 5 a plus 11 i is equal to 0 which can be written as a square into a minus 6 a square plus 5 a is equal to minus 11 i now we need to find the inverse and for that we post multiply by a inverse so we will get a square into a a inverse minus 6 inverse plus 5 into a a inverse which is equal to minus 11 i into a inverse now we know that any matrix multiplied with its inverse gives us the identity matrix and if we multiply the identity matrix with a matrix it gives us the same matrix keeping using this formula will get a square into i minus 6a into i plus 5i is equal to minus 11 i into a inverse is a inverse and again it is equal to a square minus 6a a multiplied with identity matrix gives us the matrix a itself plus 5i is equal to minus 11 a inverse and again it implies a inverse is equal to 1 upon minus 11 into a square minus 6a plus 5i. Now we will find the a inverse using this equation. Now a square is 4 to 1 minus 3 8 minus 14 7 minus 3 14 minus 6a 6a is 6 6 6 1 2 6 we are multiplying a with 6 to get 6a so it becomes 6 12 minus 18 12 minus 6 18 plus 5i where i is the identity matrix of orders 3 cross 3 so 5i becomes 5 0 0 0 5 0 0 0 5 and it is again equal to minus 1 upon 11 into 4 minus 6 plus 5 2 minus 6 plus 0 1 minus 6 plus 0 minus 3 minus 6 plus 0 8 minus 12 plus 5 minus 14 minus minus plus 18 plus 0 7 minus 12 plus 0 minus 3 plus 6 plus 0 14 minus 18 18 plus 5 and it is equal to minus 1 upon 11 minus 3 4 5 9 this minus 3 4 plus 5 is 9 9 minus 6 is 3 2 minus 6 is minus 4 1 minus 6 is minus 5 then we have minus 9 1 4 minus 5 3 1 now multiplying minus 1 with each term we will get 1 upon 11 into minus 3 4 5 9 minus 1 minus 4 5 minus 3 minus 1 hence the inverse is 1 upon 11 minus 3 9 5 4 minus 1 minus 3 5 minus 4 minus 1 so this completes the question bye for now take care hope you enjoyed the session