 Hello and welcome to the session. In this session, we'll learn about determinant solutions of linear equations by Kramer's rule. First of all, let us discuss that for the system. Now let us consider two equations in two variables, x and y. That is, a1x plus b1y is equal to c1. And a2x plus b2y is equal to c2. Now solving the given equations with multiplication, we are writing the values of equation number 1 and this p equation number 2. b2c1 minus b1c2 whole upon c1 whole upon v2 minus a2b1. 2 minus a2b1 is not equal to 0. Now let us write the solution for the determinant with the elements in the first row as c1b1 and the elements in the second row as c2v2 over the determinant with the elements in the first row as a1v1 and the elements in the second row as a2v2. The determinant with the elements in the first row as and the elements in the second row as the determinant with the elements in the first row as a1v1 and the elements in the second row as a2b2. of the determinant dx over determinant d and y is equal to the determinant dy over determinant d, where d is not equal to 0. Now you can write x is equal to the determinant with elements c, b over the determinant with elements a, b, the determinant with elements a, the determinant with elements, the determinant with element a, b is not equal to 0. Now here you can see that the denominator is really the determinant of the coefficients and in the determinant dx are replaced by the constant terms c1 and c2 in the determinant the coefficients of y that are b1, b2 are replaced by the constant terms c1 and c2 for solving the system of two linear equations with two variables. Now let us discuss the Kramer's rule for the system of three linear equations. Now consider that is a1x plus v1y plus c1z is equal to d1, then a2x plus v2y plus c2z is equal to d2, then a3x plus v3y plus c3z is equal to where d1, d2 and d3 are the constants and they are placed on the right hand side of these equations. Then the determinant with elements in the first row as d1, b1, c1, elements in the second row as d2, b2, c2 and elements in the third row as d3, d3, c3 will be equal to, now from the given equations we will write the determinant as and here d1 will be equal to plus v1y plus c1, then v1 and c1 that is these are the elements in the first row of the determinant and in the second row we will get the elements as now here d2 is equal to a2x plus v2y plus c2z and then c2 and now the elements in the third row will be now d3 will be equal to a3x, c3z, then v3, then c3. Now by the property of determinant this will be equal to the determinant with the elements in the first row as a1x, v1, c1, the elements in the second row as v2, c2, elements in the third row as a3x, v3, c3, the determinant v1, c1, elements in the second row as v2y, v2, c2 and elements in the third row as v3y, v3, the determinant with the elements in first row as c1z, v1, c1, elements in the second row as c2z, v2, c2 and elements in the third row as v3 further this is equal to the determinant with the elements in the first row as a1x, v1, c1, elements in the second row as a2x, v2, c2 and elements in the third row as a3x, v3, c3. Now here this determinant will become into the determinant with the elements in the first row as v1, v1, c1, elements in second row as v2, v2, c2 and elements in the third row as v3, v3, c3 into the determinant with the elements in the first row as c1, v1, c1, elements in the second row as c2, v2, c2 and elements in the third row as c3, v3, c3. Using the property of determinants, the determinant with the elements in the first row as a1, v1, c1, elements in the second row as a2, v2, c2 and elements in the third row as a3, v3, c2, the determinant with the elements in the first row as d1, v1, c1, elements in the second row as d2, v2, c2 and elements in the third row as d3, v3, c3. Now here the determinant with the elements in the first row as a1, v1, c1, elements in the second row as a2, v2, c2 and elements in the third row as a3, v3. But we can write the determinant with elements d1, v2, c3 over the determinant with elements, the determinant with elements d1, v2, c3 is the determinant and with elements a1, v2, c3 of the coefficients d1, d2, d3 respectively. The determinant with elements a1, v2 not equal to 0. Why is equal to the determinant with elements a1, d2, c3 over the determinant with elements a1, b2, c3 the determinant with elements a1, b2, d3 over the determinant with elements the determinant df over determinant d. Then why the determinant dy over determinant d and z is equal to the determinant dz over determinant d where ds not equal to 0. Here you can see that the denominator d x and y and z is the determinant of the coefficients of x, y and z in the given equations. And the determinant dx obtained from the determinant d2 and a3 by the constant terms d1, d2 and d3. The determinant dy is obtained on replacing the y coefficients by the constant terms and the determinant dz is obtained by replacing the z coefficients by the constant terms. Now the Kramer's rule can be used in exactly the same way to solve the system. Now the Kramer's rule in case the determinant d is equal to 0 is not equal to 0 that is the determinant d is not equal to 0. Then the given equations the determinant dx over determinant d then y is equal to the determinant dy over determinant d and z is equal to the determinant dz over determinant d. The given system that is if the determinant d is equal to 0, then there would be 2 plus then the determinant d is equal to the determinant d x, d y and d z is equal to 0, then the given system may be consistent with infinitely many solutions, no solution. If the determinant d is equal to 0, that is in the second term determinant d x, d y, d z is not 0, then the given system of equations is inconsistent. d 2 is equal to d 3 is equal to 0, that is the constant terms is equal to 0, the sum of three linear equations, that is these three constant terms, for example here let the system of two linear equations be 3 y is equal to 6 and minus 6 plus 4 y is equal to 12, then we have to solve these two equations for x and y. First of all, you must note that the constant terms should be on the right hand side of the equations. Now, the determinant d is equal to, now here consider the coefficients of x in the two equations as a 1, terms of y in the two equations as b 1, b 2 and the constant terms as d 1. Now, the determinant d will be equal to the determinant with the elements in the first row as a 1, b 1, that is minus 1 which is equal to 11. So, d is equal to 11 which is not equal to 0, now the given equations exists. Now, the determinant dx can be obtained by replacing the x coefficients in the determinant d by the constant terms. So, it will be equal to the determinant with the elements in the first row as and since the second row as 12 plus we have replaced the coefficients of x constant terms d 1 and d 2. So, this is equal to 24 minus 36 which is equal to minus 12. Now, we will find dy which is equal to, now in the determinant d we will replace the coefficients of y by the constant terms d 1 and d 2 to obtain the determinant dy. So, this is equal to the determinant with elements in first row as 2, 6 and elements in second row as minus 1, 12 which is equal to by the premise rule x is equal to the determinant dx over the determinant d and y is equal to the determinant dy over the determinant d. Now, the determinants d dx and dy are 11 minus 12 and 30 respectively. So, putting these values here this implies x is equal to minus 12 over 11 and y is equal to 30 over 11. So, we have obtained by using Kramer's rule for the given two. So, in this session we have learnt about determinant solutions of linear equations by Kramer's rule and this completes our session. Hope you all have enjoyed this session.