 Hello and welcome to the session. In this session we discussed the following question which says, determine the value of lambda so that the following linear equations have no solution. 1 plus 3 lambda whole into x plus 3 y is equal to 2, lambda square plus 1 whole into x plus lambda minus 2 whole into y is equal to 5. So we are given these two equations and we need to find the value of this lambda such that these two equations have no solution. First let's see what is the condition for a given system of equations such that it has no solution. Consider the system of equations A1x plus B1y plus C1 equal to 0, A2x plus B2y plus C2 equal to 0. If this system of equations have no solution then A1 upon A2 is equal to B1 upon B2 is not equal to C1 upon C2. This is the key idea that we use in this question. Now let's move on to the solution. The given system of equations is 1 plus 3 lambda whole into x plus 3y is equal to 2. Let this be equation 1. Then the other equation given is lambda square plus 1 whole into x plus lambda minus 2 whole into y is equal to 5. Let this be equation 2. Now first of all we write the equations 1 and 2 in the form such that all the terms from the right hand side are shifted to the left hand side and we get a 0 on the right hand side. So we rewrite the equation 1 as 1 plus 3 lambda whole into x plus 3y minus 2 equal to 0. We take this as equation 3. Then in the same way we rewrite the equation 2 as lambda square plus 1 whole into x plus lambda minus 2 whole into y minus 5 equal to 0. We take this as equation 4. Let us find out A1, A2, B1, B2 and C1, C2. Let's compare this equation 3. The equation A1 x plus B1 y plus C1 equal to 0. So we get A1 is equal to 1 plus 3 lambda, B1 is equal to 3, C1 is equal to minus 2. Then we compare equation 4 and equation A2 x plus B2 y plus C2 we get A2 is equal to lambda square plus 1, B2 is equal to lambda minus 2 and C2 is equal to minus 5. So we have got A1, B1, C1, A2, B2, C2. So the given system of equations will have no solution. A1 upon A2 is equal to B1 upon B2 is not equal to C1 upon C2. We put the values for A1, A2, B1, B2, C1, C2. We get 1 plus 3 lambda upon lambda square plus 1 is equal to 3 upon lambda minus 2 is not equal to minus 2 upon minus 5. So this means 1 plus 3 lambda upon lambda square plus 1 is equal to 3 upon lambda minus 2 and 3 upon lambda minus 2 is not equal to minus 2 upon minus 5. First let's consider this, that is 1 plus 3 lambda upon lambda square plus 1 is equal to 3 upon lambda minus 2. This gives us 1 plus 3 lambda into lambda minus 2 is equal to 3 into lambda square plus 1. This further gives us 3 lambda square minus 6 lambda plus lambda minus 2 is equal to 3 lambda square plus 3. We have 3 lambda square minus 5 lambda minus 2 is equal to 3 lambda square plus 3. Now from here we get 3 lambda square minus 5 lambda minus 2 minus 3 lambda square is equal to 3. Now 3 lambda square minus 3 lambda square cancels. So we have minus 5 lambda is equal to 3 plus 2 which gives us minus 5 lambda is equal to 5. Further we have lambda is equal to 5 upon minus 5 5 and 5 cancels and this is equal to minus 1. Therefore lambda is equal to minus 1. Now the other condition that we had was 3 upon lambda minus 2 is not equal to minus 2 upon minus 5 that is 3 upon lambda minus 2 is not equal to 2 upon 5 and this condition is true for lambda is equal to minus 1. So as we say the given system of equations will have no solution for lambda is equal to minus 1. So this complete C session I hope you have understood the solution for this question.