 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. And the first equation is given as x plus y plus z is equal to 6. The second equation is x plus 2 y plus 3 z is equal to 14. And the third equation is x plus 4 y plus 7 z is equal to 30. Now before starting the solution of this question, we should know our result. And that is, whenever we have to solve a system of three simultaneous linear equations involving three unknowns, then in the first step we will find the determinant D where the determinant D is really the determinant of the coefficients. Now after this we can have two cases that is either the determinant D is not equal to 0 or the determinant D is equal to 0. Now when D is not equal to 0 that is the determinant D is not equal to 0, then the given system of equations is consistent and has a unique solution namely, x which is equal to the determinant Dx over D, y which is equal to the determinant Dy over D and z which is equal to the determinant Dz over the determinant D. Where the determinant Dx is obtained by replacing the coefficients of x by the constant terms in the determinant D and the determinant Dy is obtained by replacing the coefficients of y by the constant terms and similarly the determinant Dz is obtained by replacing the coefficients of z by the constant terms. And this result is also called the Cramer's rule. When the determinant D is equal to 0 then further we have two cases that is if D is equal to 0 and the determinants Dx, Dy and Dz are equal to 0 then the given system will be consistent with infinitely many solutions or inconsistent having no solution. The determinant D is equal to 0 and at least one of the determinants Dx, Dy and Dz is not 0 then the given system of equations is inconsistent having no solution. We shall know that the Cramer's rule does not apply if D is equal to 0. Now this result will work out as a key idea for solving out the given question. And now we will start with the solution. Now in the question the following system of equations is given to us so given x plus y plus z is equal to 6, x plus 2 y plus 3 z is equal to 14, 4 y plus 7 z is equal to 13. Now let this be equation number one, this be equation number two and this be equation number three. Now we know that the determinant D is really the determinant of the coefficients so where the determinant D will be equal to the determinant with the elements in the first row as 1, 1 and 1 that is the coefficients of x, y and z respectively in the equation number one and the elements in the second row as 1, 2 and 3 that is the coefficients of x, y and z respectively in equation number two in the third row as 1, 4 and 7 that is the coefficients of x, y and z respectively in equation number further this will be equal to 1 into 14 minus 12 the whole minus 1 into 7 minus 3 the whole plus 1 into 4 minus 2 the whole which is equal to 2 minus 4 plus equal to 0 find the determinant dx which is obtained by replacing the coefficients in the determinant D so the determinant dx will be equal to the determinant with the elements in the first row as 6, 1 and 1 elements in the second row as 14, 2 and 3 and elements in the third row as 2 this will be equal to 14 minus 12 the whole minus 1 into 98 minus 90 the whole plus 1 into 56 minus 6 the whole which is equal to 12 minus 8 minus 4 which is equal to 0 now for obtaining the determinant dy we will replace the coefficients of y by the constant terms in the determinant D so the determinant dy will be equal to the determinant with the elements in the first row as 1 elements in the second row as 1, 14 and 3 elements in the third row as 1, 13 and further this will be equal to 1 into 98 minus 90 the whole minus 6 into 7 minus 3 the whole plus 1 into 13 minus 14 the whole is equal to 8 minus 24 plus 16 which is equal to 0 the determinant dz will be obtained by replacing the coefficients of the constant terms in the determinant dz will be equal to the determinant with the elements in the first row as 1, 1 and 6 elements in the second row as 1, 2 and 14 and elements in the third row as minus 56 the whole minus 1 into 30 minus 14 the whole plus 12 which is equal to 0 now here the determinant d is equal to 0 the determinant dx is equal to 0 determinant dy is equal to 0 which is given in the key idea if the determinant d is equal to 0 and the determinants dx dy and dz are equal to 0 then with infinite instant with no solution equation number 1 therefore putting x is equal to k where k is any real number m equation number 1 and 2 we get y plus z is equal to 6 minus k plus 3 z is equal to 14 minus k equations in two variables y here the determinant d will be equal to the determinant with the elements in the first row as 1, 1 that is the coefficients of y and z respectively the coefficients of y and z respectively in further this will be equal to 3 minus 2 which is equal to 1 so here the determinant d is not equal to 0 which means the premise rule is applicable for finding the solution of these two equations now let us find the determinant dy which will be obtained by replacing the coefficients of y with the constant terms in the determinant d so this is equal to the determinant with the elements in the first row as 6, 1 and elements in the second row as 14 minus k and this is equal to 3 into 6 minus k the whole 14 minus k minus 3 minus 14 which is equal to 4 minus the determinant dz which will be equal to now replacing the coefficients of z by the constant terms in the determinant d so the determinant dz will be equal to the determinant the elements in the first row as 1, 6 in the second row as equal to 14 minus k minus 12 plus 2k which is equal to 2 plus k which is given as the key idea the determinant d is equal to 1 the determinant dy is equal to 4 minus 2k the determinant dz is equal to 2 plus k so y is equal to the determinant dy upon d which is equal to 4 minus 2k 12 upon 1 which is equal to 4 minus 2k and z is equal to the determinant dz over d which is equal to 2 plus k y is equal to 4 minus 2k z is equal to 2 plus k equation number 3 now number 3 we get plus 4 into 4 minus 2k the whole into 30 plus 16 minus i and z in terms of k which is equal to k y is equal to 4 minus 2 this is the equation of the given question and that's all for this session hope you all have enjoyed the session