 Hello and welcome to this session. Let us understand the following problem today. If A inverse is equal to 3 minus 1, 1 minus 15, 6, minus 5, 5, minus 2, 2 and B is equal to 1, 2, minus 2, minus 1, 3, 0, 0, minus 2, 1, find A v the whole inverse. Now let us write the solution. We know that A v the whole inverse is equal to B inverse A inverse. A inverse is already given to us so we need to find B inverse given B is equal to 1, 2, minus 2, minus 1, 3, 0, 0, minus 2, 1. Now let us first find the determinant which is equal to 1 multiplied by 3 minus 0, minus 2, minus 1, minus 0 and minus 2, 2, minus 0, which is equal to 3 plus 2, minus 4 which is equal to 5 minus 4 which is equal to 1. Therefore B is invertible. Now let us find the cofactors. Now first A11 which is equal to minus 1 to the power 1 plus 1, 3 plus 0 which is equal to 3. Now A12 which is equal to minus 1 to the power 1 plus 2, minus 1 minus 0 which is equal to 1. Now A13 is equal to minus 1 to the power 1 plus 3, 2 minus 0 is equal to 2. A21 is equal to minus 1 to the power 2 plus 1, 2 minus 4 is equal to 2. A22 is equal to minus 1 to the power 2 plus 2, 1 plus 0 is equal to 1. A23 is equal to minus 1 to the power 2 plus 3 multiplied by minus 2 minus 0 is equal to 2. Now A31 is equal to minus 1 to the power 3 plus 1, 0 plus 6 is equal to 6. A32 is equal to minus 1 to the power 3 plus 2, 0 minus 2 is equal to 2. A33 is equal to minus 1 to the power 3 plus 3, 3 plus 2 is equal to 5. Now the matrix formed by cofactors are it is equal to 312212625. Now let us find a joint V. Therefore a joint V is equal to taking transpose of this matrix so we get 326 by interchanging rows and columns 326, 112, 225. Therefore V inverse is equal to 1 by determinant B multiplied by a joint of B which is equal to 1 multiplied by determinant 3, 2, 6, 112, 225 which is equal to 326, 112, 225. Therefore this is our V inverse. Now we have to find V inverse A inverse which is equal to 326, 112, 225 multiplied by another matrix 3 minus 1, 1 minus 15, 6 minus 5, 5 minus 2, 2. This is our A inverse. This is our B inverse and this is our A inverse. Now we solve it further. We will use multiplication of matrices now which is equal to multiplying this row by this column we get first 9 minus 30 plus 30. Now this row by this column we get minus 3 plus 12 minus 12. Now this row by this column we get 3 minus 10 plus 12. Now this row by this column we get 3 minus 15 plus 10. Now this row by this column we get minus 1 plus 6 minus 4. Now by this column we get 1 minus 5 plus 4. Now this row by first this column we get 6 minus 30 plus 25. Now this row with this column we get minus 2 plus 12 minus 10. Now this with this we get 2 minus 10 plus 10. Now solving this this and this gets cancelled so the first element of the matrix will be 9. This and this get cancelled so minus 3. Now 3 plus 12 is 15 minus 10 is 5. Now here 10 plus 3 is 30 minus 5 is minus 2. Now here 6 minus 15 minus 4 is 1. 1 plus 4 is 5 minus 5 is 0. 6 plus 25 is 31. 31 minus 30 is 1. 12 minus 10 is plus 2 plus 2 minus 2 so this all gets cancelled so we are left with 0. Now 2 minus 10 plus 10 here this gets cancelled so we are left with 2. Hence required matrix is 9 minus 3 5 minus 2 1 0 1 0 2. I hope you understood the problem. Bye and have a nice day.