 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. Question is, using elementary transformations, find the inverse of each of the matrices. If it exists, the real matrix is 2 0 minus 1 5 1 0 0 1 3. Let us now start with the solution. First of all, let us assume a is equal to matrix given in the question that is matrix 2 0 minus 1 5 1 0 0 1 3. Now, to find inverse by row transformation method, we will write a is equal to i a, where i is the identity matrix. So, we can write 2 0 minus 1 5 1 0 0 1 3 is equal to matrix 1 0 0 0 1 0 0 0 1, multiply by a. Now, we will apply sequence of row operations simultaneously on matrix a on left hand side and the matrix i on the right hand side till we obtain identity matrix on the left hand side. Now, to make this element equal to 1, we will apply on R 1 row operation 1 upon 2 R 1. So, we can write applying R 1 row operation 1 upon 2 R 1, we get matrix 1 0 minus 1 upon 2 5 1 0 0 1 3 is equal to matrix 1 upon 2 0 0 0 1 0 0 1 multiplied by a. To make this element equal to 0, we will apply on R 2 row operation R 2 minus 5 R 1. So, we can write applying R 2 row operation R 2 minus 5 R 1, we get matrix 1 0 minus 1 upon 2 0 1 5 upon 2 0 1 3 is equal to matrix 1 upon 2 0 0 minus 5 upon 2 1 0 0 0 1 multiplied by a. Now, to make this element equal to 0, we will apply on R 3 row operation R 3 minus R 2. So, we can write applying on R 3 row operation R 3 minus R 2 we get matrix 1 0 minus 1 upon 2 0 1 5 upon 2 0 0 1 upon 2 is equal to matrix 1 upon 2 0 0 minus 5 upon 2 1 0 5 upon 2 minus 1 1 multiplied by a. Now, to make this diagonal element equal to 1, we will apply on R 3 row operation 2 R 3. So, we can write applying on R 3 row operation 2 R 3 we get matrix 1 0 minus 1 upon 2 0 1 5 upon 2 0 0 1 is equal to matrix 1 upon 2 0 0 minus 5 upon 2 1 0 5 minus 2 2 multiplied by a. Now, to make this element equal to 0, we will apply on R 1 row operation R 1 plus 1 upon 2 R 3 and to make this element equal to 0, we will apply on R 2 row operation R 2 minus 5 upon 2 R 3. So, we can write applying R 1 row operation R 1 plus 1 upon 2 R 3 and applying on R 2 row operation R 2 minus 5 upon 2 R 3 we get 1 0 0 0 1 0 0 0 1 is equal to matrix 3 minus 1 1 minus 10 6 minus 5 5 minus 2 2 multiplied by a the identity matrix of the order 3 into 3. So, we know identity matrix can be written as a inverse multiplied by a. Now, comparing these two expressions we conclude a inverse is equal to this matrix. So, we can write a inverse is equal to matrix 3 minus 1 1 minus 15 6 minus 5 5 minus 2 2. So, our required inverse is given by the matrix 3 minus 1 1 minus 15 6 minus 5 5 minus 2 2. This completes the session. Hope you enjoyed the session. Goodbye.