 Hi, and welcome to the session. I'm Shashi and I'm going to help you to solve the following question. The question is, using elementary transformations, find the inverse of each of the matrices if it exists. The given matrix is 1 minus 1, 2, 3. First of all, let us understand the PRD to solve the given question. We can use either row transformation or column transformation to find the inverse of a matrix. To find a inverse using row transformation, write a is equal to ia and apply row operations on a is equal to ia. Till we get i is equal to ba, then b is the inverse of a. Now, let us discuss when does a inverse does not exist. If we obtain all zeros in one or more rows of the matrix a on left hand side, then a inverse does not exist. Let us start with the solution now. Let a is equal to matrix 1 minus 1, 2, 3. To find inverse of this matrix by row transformation method, we will write a is equal to ia or we can write matrix 1 minus 1, 2, 3 is equal to matrix 1, 0, 0, 1 multiplied by a. Now, we will apply sequence of row operations simultaneously on matrix a on the left hand side and matrix i on the right hand side. We will apply row operations till we obtain identity matrix on the left hand side. Or we can say we have to change this matrix into identity matrix. Now, we know the identity matrix has diagonal elements equal to 1 and all other elements are equal to 0. Now, first of all, we will change this element to 0. So, we will apply on R2 row operation R2 minus 2 other. So, we can write applying on R2 row operation R2 minus 2 R1 we get matrix 1 minus 1, 0, 5 is equal to matrix 1, 0, minus 2, 1 multiplied by a. To change this diagonal element 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 1, 0, 1 is equal to matrix 1, 0, minus 2 upon 5, 1 upon 5, multiplied by a. Now, to make this element equal to 0, we will apply on R1 row operation R1 plus R2. So, we can write applying on R1 row operation R1 plus R2 we get matrix 1, 0, 0, 1 is equal to matrix 3 upon 5, 1 upon 5, minus 2, 1 upon 5, minus 2, 1 upon 5, minus 2 upon 5, 1 upon 5 multiplied by a. Now, clearly this is the identity matrix of the order 2 into 2. So, we know identity matrix is equal to a inverse multiplied by a. Now, comparing the two equations we get a inverse is equal to this matrix. So, we get a inverse is equal to matrix 3 upon 5, 1 upon 5, minus 2 upon 5, 1 upon 5. So, the required inverse is given by the matrix 3 upon 5, 1 upon 5, 2 upon 5 and 1 upon 5. This completes our session. Hope you understood the session. Take care and goodbye.