 Hello friends, welcome to the session I am Malka. We are going to discuss determinants are given question is find the inverse of each of the matrices if it exists given in exercises 5 to 11. Our 7th exercise is matrix 1, 2, 3, 0, 2, 4, 0, 0, 5. Now let's start with the solution. We are given A equal to matrix 1, 2, 3, 0, 2, 4, 0, 0, 5. Now we will first find the determinant of A which is now we will express it by along the third row. So it will give us 0 minus 0 plus 5. Now we will leave this. This will give us 1, 2, 0, 2. So this implies determinant of A equal to 5 into 2 minus 0 which is equal to 10. So this implies determinant of A is not equal to 0 therefore A inverse exists. Now we will find the joint of A which is transpose of the matrix A. So for finding this we will find the cofactors of the matrix A. Cofactor of 1 equal to minus 1 to the power 1 plus 4 into 2, 4, 0, 4 which is equal to minus 1 square 10 minus 0 that is equal to 10. Now we will find the cofactor of other elements. Cofactor of 2 equal to 5, cofactor of 4 equal to 0, cofactor of 0 equal to 2, cofactor of 0 equal to 2, cofactor of 0 equal to minus 4, cofactor of 5 equal to 2. Therefore the matrix formed by the cofactors is matrix 10, 0, 0, minus 10, 5, 0, 2, minus 4, 2. Now we will find the joint of A which is transpose of matrix 10, 0, 0, minus 10, 5, 0, 2, minus 4, 2. So this is equal to 10, 0, 0, minus 10, 5, 0, 2, minus 4, 2. Now we will find the A inverse. Therefore A inverse equal to 1 upon determinant of A into a joint of A which is equal to 1 upon 10 into matrix 10 minus 10, 2, 0, 5, minus 4, 0, 0, 2. A inverse equal to 1 upon 10 into matrix 10 minus 10, 2, 0, 5, minus 4, 0, 0, 2 which is the required answer. Hope you understood the solution and enjoyed the session. Goodbye and take care.