 Hello and welcome to the session. In this session we will discuss a question which says that solve the following system of equations by using determinants. x plus 2y plus 3z is equal to 4, 2x plus 4y plus 6z is equal to 5, 6x plus 12y plus 18z is equal to 15. Now before starting the solution of this question we should know some results. First is the Kramer's rule. Now for the system of three linear equations in x, y and z, that is a1x plus b1y plus c1z is equal to d1, a2x plus b2y plus c2z is equal to d2, a3x plus b3y plus c3z is equal to d3. The determinant t is given by the determinant with the elements in the first row as a1, b1, c1, the elements in the second row as a2, b2, c2 and the elements in the third row as a3, b3, c3. And if determinant d is not equal to 0, then x is equal to determinant dx over determinant d, y is equal to determinant dy over determinant d and z is equal to determinant dz over determinant v. Where the determinant dx is obtained on replacing the x coefficients a1, a2, a3 by the constants d1, d2, d3. And the determinant dy is obtained on replacing the y coefficients that is b1, b2, b3 by the constants d1, d2, d3 and the determinant dz is obtained on replacing the z coefficients that is c1, c2, c3 by the constants d1, d2, d3. And secondly, if the determinant g is not equal to 0, then the given system of equations is consistent and has a unique solution. So it will have a unique solution and it is consistent. And if the determinant t is equal to 0, then there will be two cases. That is, if determinant dx is equal to determinant dy is equal to determinant dz is equal to 0, then the system of linear equations will be consistent with infinitely many solutions or it can be inconsistent having no solution. And in the other case, if at least one half determinant dx, determinant dy, determinant dz is not 0, then the system is inconsistent having no solution. Now these results will work out as a key idea for solving out this question. And now we will start with the solution. Now here we have to solve the following system of equations by using determinants. So given the system of equations is x plus 2y plus 3z is equal to 4, 2x plus 4y plus 6z is equal to 5, and 6x plus 12y plus 18z is equal to 15. Let this be equation number 1, this be 2 and this be equation number 3. Now by using this formula we can find out the determinant d. Now the determinant d will be equal to determinant with the elements in the first row as a1, b1, b1, that is 1, 2 and 3. And elements in the second row as a2, b2, c2, that is 2, 4, 6 and elements in the third row as 6, 12, 18, that is a3, b3 and c3. Now this will be equal to 1 into 4 into 18 is 72 minus 12 into 6 is 72 the whole minus 2 into 18 into 2 is 36, minus 6 into 6 is 36 the whole plus 3 into 12 into 2 is 24, minus 6 into 4 is 24 the whole. Further on solving this we will get 1 into 0 that is 0 minus 2 into 0 that is 0 plus 3 into 0 that is 0 which is equal to 0. Therefore determinant d is equal to 0. Now we will find the determinant dx that is we will replace the x coefficients with the constant terms. Now the determinant dx will be equal to determinant with elements in the first row as 4, 2, 3 elements in the second row as 5, 4, 6 and elements in the third row as 15, 12 and 18. Further this is equal to 4 into 18 into 4 is 72 minus 12 into 6 is 72 the whole minus 2 into 18 into 5 is 19 minus 15 into 6 is 19 the whole plus 3 into 5 into 12 is 16 minus 15 into 4 is 60 the whole which is further equal to now 4 into 0 is 0 minus 2 into 0 is 0 plus 3 into 0 is 0. So this is also equal to 0 therefore dx that is the determinant dx is equal to 0. Now we will find the determinant dy that is we will replace the y coefficients with the constant 4, 5 and 15. So the determinant dy will be equal to the determinant with elements in the first row as 1, 4, 3 elements in the second row as 2, 5, 6 and elements in the third row as 6, 15 and 18. So this is equal to 1 into 5 into 18 is 19 minus 6 into 15 is 90 the whole minus 4 into 2 into 18 is 36 minus 6 into 6 is 36 the whole plus 3 into 15 into 2 is 13 minus 6 into 5 is 30 the whole. Which is further equal to now 90 minus 90 is 0 so 1 into 0 will be 0 minus 4 into 0 is 0 plus 3 into 0 is 0. So this is also equal to 0 and now we will find the determinant dz that is we will replace the z coefficients by the constants 4, 5 and 15. Now the determinant dz is equal to the determinant with the elements in the first row as 1, 2, 4 elements in the second row as 2, 4, 5 and elements in the third row as 6, 12 and 15. Further this is equal to 1 into 60 minus 60 the whole minus 2 into 30 minus 30 the whole plus 4 into 24 minus 24 the whole. So this is equal to 1 into 0 which is 0 minus 2 into 0 this is 0 plus 4 into 0 this is 0 so this is equal to 0. So we get determinant dz is equal to 0 and determinant dx is equal to determinant dy is equal to determinant dz is equal to 0. Now using this result which is given in the key idea since the determinant dz is equal to 0 so we cannot apply a parameters rule however we observe that the determinant dx is equal to determinant dy is equal to determinant dz is equal to 0. Therefore the system may have finitely many solutions or no solution. Now this is equation number 1 this is equation number 2 and this is equation number 3. Now to check that the system has infinitely many solutions or no solution consider equation 1. Now let z is equal to any constant say k now putting z is equal to k in equation number 1 and 3 we get x plus 2y is equal to 4 minus 3k and plus 12y is equal to 15 minus 18k. Like this p equation number 4 and this p equation number 5. Now this equation is in two variables and now the determinant d will be equal to the determinant with the elements in the first row as a1 v1 and the elements in the second row as a2 v2. Now a1 here is 1, v1 here is 2, a2 here is 6 and b2 here is 12 and 4 minus 3k and 15 minus 18k are the constants d1 and d2. Further this is equal to 1 into 12 that is 12 minus 6 into 2 that is 12 which is equal to 0. Therefore the t is equal to that is the determinant d is equal to 0. Now in this case we can obtain the determinant dx by replacing the x coefficients with the constants d1 and d2. Now the determinant dx is equal to the determinant with the elements in the first row as 4 minus 3k 2 and the elements in the second row as 15 minus 18k and 12. Now this is equal to 12 into 4 minus 3k the whole will give 40 h minus 36k minus 2 into 15 minus 18k the whole will give minus 13 plus 36k which is equal to now these terms are cancelled with each other so this will be 48 minus 13 which is 18. And this is not equal to 0. Now according to this result which is given in the key idea if at least one of the determinant dx the determinant dy the determinant dz is not 0 then the given system of equations is inconsistent having no solution. Now for the equations 4 and 5 the determinant d is equal to 0 and the determinant dx is not equal to 0. Therefore the system of equations given by equation number 4 at 5 no solution hence the system of given equations will have no solution that is it is inconsistent. So this is the solution of the given question and that's all for this session hope you all have enjoyed the session.