 Hi and welcome to the session I am Shashi and I am going to help you to solve the following question. Question is using elementary transformations find the inverse of each of the matrices if it exists. The given matrices 2 minus 3, 3, 2, 2, 3, 3 minus 2, 2. Let us find the solution now. Let A is equal to matrix 2 minus 3, 3, 2, 2, 3, 3 minus 2, 2. Now to find the inverse by elementary row transformation we will write A is equal to I where I is the probability matrix or we can write matrix 2 minus 3, 3, 3, 3 minus 2, 2 is equal to matrix 1, 0, 0, 0, 1, 0, 0, 0, 1 multiplied by A. Now to make this element equal to 1 we will apply on R1 row operation 1 upon 2 R1. So we can write applying R1 row operation 1 upon 2 R1 we get matrix 1 minus 3 upon 2, 3, 2, 2, 3, 3 minus 2, 2 is equal to matrix 1 upon 2, 0, 0, 0, 1, 0, 0, 0, 1 multiplied by A. Now we know diagonal elements of the identity matrix are 1 and all other elements are equal to 0. So we have to make these elements equal to 1 and all other elements should be equal to 0. Now to make this element equal to 0 we will apply on R2 row operation R2 minus 2 R1 and to make this element equal to 0 we will apply on R3 row operation R3 minus 3 R1. So we can write applying row operation R2 minus 2 R1 and R3 row operation R3 minus 3 R1 we get matrix 1 minus 3 upon 2, 3 upon 2, 0, 5, 0, 0, 5 upon 2 minus 5 upon 2 is equal to matrix 1 upon 2, 0, 0, minus 1, 1, 0, minus 3 upon 2, 0, 1 multiplied by A. Now to make this diagonal element equal to 1 we will apply on R2 row operation 1 upon 5 R2. So we can write applying on R2 row operation 1 upon 5 R2 we get matrix 1 minus 3 upon 2, 0, 1, 0, 0, 5 upon 2 minus 5 upon 2 is equal to matrix 1 upon 2, 0, 0, minus 1 upon 5, 1 upon 2, 0, minus 3 upon 2, 0, 1 multiplied by A. To make this element equal to 0 we will apply on R1 row operation R1 plus 3 upon 2 R2 and to make this element equal to 0 we will apply on R3 row operation R3 minus 5 upon 2 R2. So we can write applying R1 row operation R1 plus 3 upon 2 R2 and R3 row operation R3 minus 5 upon 2 R2 we get matrix 1, 0, 3 upon 2, 0, 1, 0, 0, 0, 0, minus 5 upon 2 is equal to matrix 1 upon 5, 3 upon 10, 0, minus 1 upon 5, 1 upon 5, 0, minus 1, minus 1 upon 2, 1 multiplied by A. Always remember that row operations will take place simultaneously on the matrix A on the left hand side and the matrix I on the right hand side till we obtain the identity matrix on the left hand side. Now, in this diagonal element equal to 1 we will apply on R3 row operation minus 2 upon 5 R3. So we can write applying R3 row operation minus 2 upon 5 R3. So we get matrix 1, 0, 3 upon 2, 0, 1, 0, 0, 0, 1 is equal to matrix 1 upon 5, 3 upon 10, 0, minus 1 upon 5, 1 upon 5, 0, 2 upon 5, 2 upon 5, 1 upon 5, minus 2 upon 5. But we write by A. Now to make this element equal to 0 we will apply on R1 row operation R1 minus 3 upon 2 R3. So we can write time R1 minus 3 upon 2 R3, applying on R1 row operation R1 minus 3 upon 2 R3. We get the matrix 1, 0, 0, 0, 1, 0, 0, 0, 1 is equal to matrix 1 minus 2 upon 5, 0, 3 upon 5, minus 1 upon 5, 1 upon 5, 0, 2 upon 5, 1 upon 5, minus 2 upon 5 multiplied by A. Clearly this is the identity matrix of the order 3 into 3. So we know identity matrix is equal to A inverse multiplied by A. Now comparing the two expressions we get A inverse is equal to this matrix. So we get A inverse is equal to matrix minus 2 upon 5, 0, 3 upon 5, minus 1 upon 5, 1 upon 5, 0, 2 upon 5, 1 upon 5, minus 2 upon 5. So the required inverse is given by the matrix minus 2 upon 5, 0, 3 upon 5, minus 1 upon 5, 1 upon 5, 0, 2 upon 5, 1 upon 5, minus 2 upon 5. This is our required answer. Hope you understood the session. Take care and goodbye.