00:01
Okay, this question is about finding a solution of this system of linear equations such that k allows the solution to not be unique.
00:23
Okay, so not unique means that we need something like this kind of form, right? we need a row of all zeros.
00:42
Okay so how do we do that luckily there's this funny little property that uh basically if the determinant of a matrix called the matrix a equals zero um then that's telling you that there is a row or column of all zeros so that's what we want to find you want to find the values of k with the determinant of the matrix equals zero.
01:35
So for this we don't want to use the augmented matrix, right? so we're going to just pretend that doesn't exist, and it's just the ones in the k's.
01:44
So if we take the determinant, should be using the straight line because it's a determinant.
01:52
If we take the determinant of the 3x3 matrix 1 1k1 k1 k1, t111, tongue twister, we should get we want to multiply this by this block of four numbers we get one times k minus one and then we have a subtraction factor here so then we get minus one times then we do this block of numbers which is a big one minus k and then we add k times 1 minus k squared and then we want this all to equal zero so if we just do some algebra here we get negative 2 plus 3k minus k cubed equals 0 okay this is a cubic function which makes it a little bit difficult to factor it and whatnot.
03:21
So there's a different way that you can do this.
03:29
If you would like, you can make your life a lot easier if you do 1 times k minus 1.
03:44
And we just pull out a minus 1 here, so that it's plus 1 times k minus 1 plus, and then just realize that this is k times k minus 1, plus, and then just realize that this is k times.
04:02
Sorry, can you make this minus k? because we had to pull out a subtraction.
04:16
Then that becomes k squared minus 1, which we can factor into k plus 1 times k minus 1.
04:40
So basically we know that k minus 1 is definitely a factor...