00:01
So in the given question we have a system of equations that are given as 2x 2x plus 3 lambda plus 1 y plus 3 times lambda minus 1 z is equal to 0.
00:30
Next we have lambda minus 1 times x plus 4 lambda minus 2 times y plus lambda plus lambda plus 3 times z is equal to 0.
00:51
Lambda minus 1 times x plus 3 lambda plus 1 times y plus 2 lambda plus 2 lambda z and 2 lambda z is equal to 0.
01:09
So this is the system of equations that we are given and we are told to obtain the values of obtain values for lambda so that so that the given linear homogeneous system has non -trivial solutions as non -trivial solutions.
01:58
So this is what we are told in the question, right? so now what we are going to do first is we can form a matrix using the system and using the coefficient of a system of equations and the not a matrix we are teaching the determinant that is with the coefficient of the variables in the matrix in the system of equation and we would have the determinant of 2 3 lambda plus 1 3 times lambda minus 1 lambda minus 1 4 lambda minus 2 lambda plus 3 lambda minus 1 3 lambda minus 1 3 lambda plus 1 2 lambda so this is the determinant and the system would have a trivial solution if the determinant of the coefficient is non -zero.
03:05
So if the determinant is not equal to 0 then if the system would have only trivial solution and for the condition for it the system to have non -trivial solution is that the determinant should be equal to 0 right so let's simplify this determinant first by doing the row operation r3 minus r3 changes to r3 minus r1 so once we do this what we get is we will get the determinant to 3 lambda plus 1 3 times lambda minus 1 lambda minus 1 lambda lambda minus 2 lambda plus 3 0 minus lambda plus 3 lambda minus 3 lambda minus 3 so this is what we get after the row operation next we can take lambda minus 3 as a common factor from the third draw of the determinant right and then we would have we can write it over here as lambda minus 3 times the determinant 2 3 lambda plus 1 3 times lambda minus 1 lambda minus 1 lambda minus 4 lambda minus 2 lambda plus 3 and we have over here 0 minus 1 and 1 right so this is the determinant after we took lambda minus 3 as a common factor next what we are going to do is is we can do the column operation which is c2 changes to c2 plus c3 and then we would have lambda minus 3 times 2 lambda minus 1 0 6 lambda minus 2 5 lambda plus 1 0 0 3 times lambda minus 1 0 3 times lambda minus 1 lambda plus 3 1 and now what we are going to do is to take the determinant over here and we would get lambda minus 3 times 2 times 5 lambda plus 1 2 times 5 lambda plus 1 minus lambda minus 1 times 6 lambda minus 2.
06:05
So this is what we have as the determinant, right? so we can expand this and what we would get is lambda minus 3, lambda minus 3 times minus 6 lambda square here plus 18 lambda and we are having the condition that the system would have a non -trivial solutions only if the determinant is equal to 0 right so we can equate this to 0 which means we can write this as let's first simplify this so we will have lambda minus 3 we can take 6 as a common factor 6 lambda as a common factor from this so 6 times 6 lambda times minus lambda plus 3 which would be equal to 6 times lambda minus 3 times lambda minus 3.
07:12
And this is equal to 0.
07:16
This is equal to 0.
07:19
So this is 3 minus lambda right? this is 3 minus lambda.
07:24
3 minus lambda which we can take we we can multiply minus 1 on both sides of this equation to make it lambda minus 3 so over here we would have lambda minus 3 minus 6 times lambda no we don't have minus 6 over here so 6 times lambda minus 3 times lambda minus 3 equal to 0 or we can write 6 times lambda minus 3 squared is equal to 0 and 6 times lambda minus 3 squared is equal so, what we are having over here is we have two values of lambda from this equation...