 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 11th exercises Matrix 1 0 0 0 cos alpha sin alpha 0 sin alpha minus cos alpha Now let's start with a solution. We are given a equal to matrix 1 0 0 0 cos alpha sin alpha 0 sin alpha minus cos alpha Now we'll find the determinant of a 1 into minus cos square alpha then minus sin square alpha This is equal to minus 1. This implies Determinant of a not equal to 0 therefore a inverse exists Now we'll find the co-factor of the different elements co-factor of 1 equal to Minus 1 to the power 1 plus 1 into minus cos square alpha minus sin square alpha, which is equal to minus 1 co-factor of 0 Equal to minus 1 to the power 1 plus 2 into 0 minus 0 equal to 0 Similarly, we'll find the co-factor of other elements Co-factor of 0 is 0. Co-factor of 0 is 0. Co-factor of cos alpha is minus cos alpha Co-factor of sin alpha is minus sin alpha Co-factor of 0 is 0. Co-factor of sin alpha is minus sin alpha and co-factor of minus cos alpha is cos alpha Therefore matrix formed by the co-factor equal to matrix minus 1 0 0 0 minus cos alpha minus sin alpha 0 minus sin alpha and cos alpha Now we'll find the adjoint of a which is the transpose of the matrix formed by the cofector. This is equal to minus 1, 0, 0, then 0 minus cos alpha minus sin alpha, then 0 minus sin alpha and cos alpha. This is the value of adjoint of A. Therefore, A inverse equal to 1 upon determinant of A into adjoint of A. This is equal to minus 1 upon 1 into matrix minus 1, 0, 0, 0, minus cos alpha minus sin alpha, 0 minus sin alpha and cos alpha. Hence A inverse equal to 1, 0, 0, 0, cos alpha, sin alpha, 0, sin alpha minus cos alpha, which is the required answer. Hope you understood the solution and enjoyed the session. Goodbye and take care.