 Hello and welcome to the session. In this session we will discuss the equation which says that find the inverse of this matrix that is the matrix with elements in first row as 7 minus 5 and elements in second row as 2 minus 1 and solve the following system of equations and the system of equations is given to us as 7x minus 5y is equal to 2, 2x minus y is equal to 1. Now before starting the solution of this question we should know a result and that is if determinant of matrix A is not equal to 0 then the inverse of matrix A exists and x is equal to A inverse into B gives the solution of the given system of equations and here x is the variable matrix B. Now this result will work out as a key idea for solving the given question. Now let us start with the solution of the given question. Now given matrix with elements in first row as 7 minus 5 elements in second row as 2 minus 1 and like this B matrix A and determinant of matrix A is equal to determinant with elements in first row as 7 minus 5 and elements in second row as 2 minus 1 is equal to into minus 1 is equal to minus is equal to A is equal to 3 which is not equal to 0. Now from the key idea we know that if determinant of matrix A is not equal to 0 then the inverse of B gives the solution of the given system of equations A is not equal to inverse of matrix that is determinant of matrix A. Metrics with as D minus B and elements in second row as minus C matrix A that is A is equal to E is equal to minus 1 A inverse which is equal to that is inverse of matrix A is equal to 1 upon determinant of matrix A in first row as D minus B that is minus 1 and minus B will be minus of minus 5 that is 5 is equal to 3 so there inverse of matrix A will be equal to 1 upon 3 into this matrix of this matrix we have inverse of matrix A is equal to matrix with elements in first row as minus 1 upon 3 and 5 upon 3 and elements in second row as minus 2 upon 3 and 7 upon 3. Now we have given the system of equations let this be equation one that is in matrix form. Now here we need a column matrix that is matrix having single with elements x and y D now in these given equations we have two so matrix D is again a column matrix that is matrix having single column with elements the given system matrix A is equal to matrix B the given system of equation now we have found inverse of matrix A into B will give the solution of the given system of equations on matrix x that is the variable matrix is the column matrix having elements inverse of matrix A which is equal to matrix having elements in first row as minus 1 upon 3 5 upon 3 and elements in second row as minus 2 upon 3 which is B that is again a column matrix having elements 2 is 2 matrices now for multiplication of 3 matrices you have to check that the number of columns of first matrix number of rows of second matrix now here you can see first number of columns in first matrix is equal to number of rows in second matrix would be reported 2 plus 1 now for multiplication of the first matrix by the corresponding elements of each column of the second matrix and then add the products matrix elements of the column of resultant matrix to 1 the element in second row multiplying the corresponding element of this column and adding the product minus 2 by 3 plus 5 by 3 and element in second row as minus 4 by 3 plus 7 by 3 which is further equal to matrix with element in first row on solving it will be 3 upon 3 and element in second row will be again 3 upon 3 and this is equal to column matrix thus the given matrix that is the column matrix having elements x and y is equal to column matrix having elements 1 and 1 now using equality of matrices we get 1 and y is equal to and this completes our session hope you all have enjoyed the session