 Hello and welcome to this session. In this session we will discuss a question which says that while the system of equations in A x is equal to B form and the system of equations is given as x plus y is equal to 1, x minus z is equal to 2, y plus z is equal to 4. Now we have to write it in A x is equal to B form that is in matrix form. Now before starting the solution of this question we should know a result. Now a system of equations can be written in matrix form A x is equal to B where A is coefficient matrix, x is variable matrix and B is constant matrix. Now this result will work out as a key idea for solving out the given question. Now let us start with the solution of the given question. The system of equations is given to us and we have to write it in matrix form that is in the form A x is equal to B. Now let this be equation number 1, this be equation number 2 and this be equation number 3. Now we have to write it in matrix form. So first we have to write its coefficient matrix A. Now there are three equations in three variables. So dimension of coefficient matrix will be 3 cross 3. Now in coefficient matrix the elements are the coefficients that is the elements of the matrix are the coefficients of the given variables in the system of equations. First of all let us rewrite these equations then we will write the coefficient matrix. Now here we can see the first equation is x plus y is equal to 1 and here the variable z is not given so we assume coefficient of z as 0. In second equation the variable y is not given so we assume coefficient of y as 0 and in the third equation variable x is not given so we assume coefficient of x as 0. So we rewrite the three equations as plus y plus 0 into z is equal to 1 then x plus 0 into y minus z is equal to 2 and 0 into x plus y plus z is equal to another name these equations as 5 coefficient matrix. Now we know that in coefficient matrix the elements are the coefficients of the given variables in the system of equations. There are three equations in three variations of variables z minus 1 and 1. Now let us write variable matrix nz and they will be written in vector matrix that is in column matrix so variable matrix capital X is equal to the column matrix that is matrix having single columns you can see that we have three constants in terms one having single column with elements one this is the matrix x and this is the matrix can be written in matrix 1 and b is constant matrix that is the matrix with elements in first row as 1 1 0 elements in second row as 1 0 must in third row as 0 1 1 this b that is again a column matrix with elements 1 2 and 4 which is also required matrix forms is equal to b form in matrix form and this completes our session hope you all have enjoyed the session.