 Hi, this is Meenu and going to help you to solve this question. It says, let A bit is 2 x 2 matrix and B bit, again a 2 x 2 matrix, verify that A bit whole inverse is equal to B inverse A inverse. So, we first find the LHS and then we will find the RHS and we will show that they both are equal. So, let's move on to the solution. Matrix A is 3, 2, 7, 5. Matrix B is 6, 7, 8, 9. We first find the product of the two matrices and then we find the inverse. Now, AB is equal to 3 x 6 plus 7 x 7, 3 x 8 plus 7 x 9, 2 x 6 plus 5 x 7, 2 x 8 plus 5 x 9. And it is equal to 18 plus 49 is 67, 87, 47 and 61. We now find the inverse of the product matrix. For the inverse, we need to find the determinant and the adjoint. So, the determinant of AB is equal to 47, 61 and its value is 61 into 67 minus 47 into 87. And it is equal to 4087 minus 4089 and which is equal to minus 2 which is not equal to 0. So, the determinant AB is minus 2. We now find the adjoint of AB. So, the adjoint AB will be given by the cofactors A11, A21, A12 and A22. A11 will be minus 1 to the power 2. So, we find A21, M11 which is 61, deleting first row and first column will get 61. So minus 1 to the power 2 into 61, which is 61 again. A21 which is minus 1 to the power 3, M21 which is obtained by deleting second row and first column, second row and first column. So it is minus 1 to the power 3 into 87, so it becomes minus 87, then A12 is minus 1 to the power 3, M12 which is obtained by deleting first row and second column. So you will get minus 1 to the power 3 into 47 which is minus 47 and similarly A22 will be equal to minus 1 to the power 4, M22 which is minus 1 to the power 4, M22 is obtained by deleting second row and second column, so you will get 67 which is 67. Now adjoint AB is equal to A11, A21, A12, A22. Now the AB inverse is given by AB inverse is equal to 1 upon determinant AB into adjoint AB, so it will be equal to minus 1 upon 2 because determinant AB is minus 2 and adjoint is 61 minus 87 minus 47, 4767. So this is the LHS that is AB whole inverse. We now find the RHS which is B inverse A inverse. Now B is 6789. So to find B inverse we need to find the determinant of B and adjoint B. So the determinant B is given by 6 into 9 minus 7 into 8 that is minus 2. We now find the adjoint, adjoint B will be given by B11, B21, B12, B22. So we find these cofactors B11 is equal to minus 1 to the power 1 plus 1, M11 where M11 is the determinant obtained by deleting first row and first column. So it is 9 minus 1 to the power 2 into 9 which is 9. Then we find B21 which is minus 1 to the power 3 M21 which is obtained by deleting second row and first column and which is 8. So it becomes minus 8. Then B12 is equal to minus 1 to the power 3 M12 which is obtained by deleting first row and second column. So you will get minus 1 to the power 3 into 7 which is minus 7. Then we find B22 which is minus 1 to the power 4 2 plus 2 that is 4 M22 which is obtained by deleting second row and second column. So the adjoint is equal to 9 minus 8 minus 7 6 and B inverse is given by determinant of B is minus 2. B inverse is equal to 1 upon determinant B into adjoint B which is given by 1 upon minus 2 into 9 minus 7 minus 8 6. We now find A inverse. A is 3 to 7 5 therefore determinant of A is equal to which is obtained by 3 into 5 minus 2 into 7 that is 1. And the adjoint is obtained by obtaining all the cofactors A11 which is minus 1 to the power 2 M11 which is 5 deleting first row and first column. So it is 5 then A21 is equal to minus 1 to the power 3 into second row and first column that is 7 so it becomes minus 7. Then A12 it is minus 1 to the power 3 first row and second column so M12 is 2 and hence it becomes minus 2. Then A22 is equal to minus 1 to the power 4 M22 which is minus 1 to the power 4 obtained by deleting second row and second column which is 3. So the adjoint A is 5 minus 7 minus 2 3 and the A inverse is equal to determinant is 1 so it is 1 upon 1 into 5 minus 2 minus 7 3 which is same as minus 2. We now find B inverse A inverse the product of the two. Now B inverse is 1 upon minus 2 into 9 minus 8 minus 7 6 9 minus 8 minus 7 6 into A inverse which is 5 minus 7 minus 2 3. Now multiplying the two will get 9 into 5 plus minus 8 into minus 2 9 into minus 7 plus minus 8 into 3 then minus 7 into 5 plus 6 into minus 2. Minus 7 into minus 7 plus 6 into 3 6 into 3 and it is equal to minus 1 by 2 61 minus 87 minus 47 67. So this is the RHS and we can see that RHS is same as the LHS. It is also minus 1 by 2 61 minus 87 minus 47 67 and RHS is also the same. LHS is same as the RHS. LHS is equal to RHS. So this completes the question. Bye for now. Take care. Hope you enjoyed the session.