 Okay friends, so in this session we are going to learn Kramer's rule for solving a pair of linear equations. Now we have learned the method of substitution, elimination and cross multiplication. So this is a new method which is very handy in solving system of linear equations especially when the number of variables is too large. Okay, so what we will start with a simple pair of linear equations in two variables and understand the process and then we can always take it to you know higher number of variables system of linear equations. So let us see how do we solve and it is assumed that you know what determinants are and if you have not gone through the previous session on determinants I would recommend that you please have a basic understanding of determinants in the previous video session. And once you are thorough with determinants and how to find the value of determinants then it becomes easier to understand this rule. So now we have a pair of linear equations here, these are the example equations are there. So before we take up the examples let us first do a generalized study. So let us first write two linear equations in two variables like this. So let us say we have a1x plus b1y is equal to c1 and we have a2x, a2x and b2y and this is equal to c2. Now if you notice the difference between this pair of linear equation and this is that I have written the constant terms on the right hand side. So hence please be careful while solving Kramer's rule you have to have the equations in this format where the variables are on the LHS, variable side variables should be should be in LHS and constant term constant terms should be in RHS. It's not that you can't really solve if the constant terms are on the left hand side but then the particular method which you are adopting in this session would require that variables should be on the left hand side and the constant terms should be on the right hand side. Now let us define a few determinants. Now I am going to define first determinant d and this will be nothing but the determinant of coefficients of x and y. So what are the coefficients of x and y, a1, a2 and b1, b2, isn't it? So the value of this determinant d will be nothing but we know how to find the value of a determinant so you have to multiply first a1, b2 and then subtract a2, b1 from that. This is the value of determinant d. Now I am defining another determinant dx which is nothing but you have to consider the previous determinant d and replace the coefficients of x. So please note this subscript x, so you replace the coefficients of x. So what are the coefficients of x, these are the coefficients of x, isn't it? So I have to replace a1, a2 by what? The constant term. So hence it will be c1, c2 and let the other column be as it is. This is dx. So hence if you see the value of dx is c1, b2 minus c2, b1, this is the value of dx. So here determinant value dx is c1, b2 minus c2, b1. Similarly I have to define another determinant dy and in this case I will take the determinant d and replace the coefficients of y from it and replace it by the constant term. So if you see this is nothing but a1, a2 and then it will be c1, c2 and you know that this determinant value dy will be nothing but a1, c2 minus a2, c1. So these are the three determinants. Now how do I find x? So x is nothing but determinant dx divided by the value of determinant of d. And y is defined as determinant of dy, determinant value dy divided by determinant value d. Clearly if you see these x and y are real, x and y are real only when the denominators must not be 0. So determinant value of d must not be equal to 0. That means what? If I say determinant d it's nothing but a1, b2 minus a2, b1 must not be equal to 0. And if I rearrange it is a1 by a2 should not be equal to b1 by b2 which probably you would have understood or you have come across this particular condition in while studying the consistency of linear equations. So this is what the Gramer's rule is. So hence what did we learn? We defined three determinants. One is the coefficient determinant, another is dx, another was dy. And hence x is nothing but determinant value of dx upon determinant value of d. And y is determinant value of dy divided by determinant value of d. So directly you can find out. Let us now take the two examples which are given above. So here if you see you have to rearrange these equations, these given equations into this form, what form? You have to write 2x plus 3y plus minus 5 equals to 0 can be written as equal to 5. And another equation is 3x minus 2y equals negative 2. Let us now try to solve these equations using determinant method or Gramer's rule. So let us rewrite the equation. So equations were 2x plus 3y equals 5 and 3x minus 2y equals negative 2. Let us define the three determinants, d is equal to coefficient determinant. So it is 2, 3, 3 minus 2. So hence value of d is determinant d is equal to 2 into minus 2, minus 3 into 3, how? Because this is how you have to multiply and subtract. So finally you will get minus 4 minus 9, which is negative 13, which is not equal to 0. That means we will get a solution to this pair of linear equations. Now let us find out dx. dx is replace the coefficient of x that is 2 and 3 here by the constant term. So 5 and minus 2 and 3 and minus 2 like this. So now this implies determinant value of dx will be equal to 5 into minus 2 minus 2 into 3, which will be nothing but minus 10 plus 6, which is equal to negative 4. And what will be dy? Dy is replace the coefficient of b with the constant terms. So 2 and 3 will be as it is and 3 and minus 2 will be replaced by 5 and minus 2. And now we know determinant value of dy is nothing but 2 into minus 2, minus 3 into 5, which is minus 4, minus 15, which is minus 19. So what is x? x is determinant value of dx divided by determinant value of b, which is equal to minus 4 upon negative 13. So it is 4 upon 13. And what is the value of y? y is determinant value of dy divided by determinant value of b, which is minus 19 upon minus 13, which is 19 upon 13. So this is the solution. So x equals 4 by 13 and y equals 19 by 13. So if you deploy these two values back into the system of linear equations, they will be valid. So let us try once. So let us take this equation and say, see whether we have solved it correctly. So 2 into x, x is 4 upon 13 and plus 3 into y, which is 19 by 13. And if you sum them up, you will get 13 and this is 8 plus 57, which is 65 by 13, which is 5. That means our solution x equals to 4 by 13 and y equals to 19 by 13 is correct. So that means this Kramer, the way we found out that is when we deployed Kramer's rule to solve these equations, we got the correct solution. So this is for, this is a solution of linear equation 2 variables using Kramer's rule. In the next session, we would take up solution of linear equations in 3 variables using Kramer's rule. Thank you.