00:01
Okay, for this problem, we have to take the following matrix.
00:09
1, negative 1, 2, 246, 303, 0, negative 5, 10, and put it into reduced, well, put it into row echelon form.
00:28
If you recall, i'll put it right over here.
00:32
Matrix is in row echelon form when there are ones on the main diagonal and there are zeros everywhere else below them and up above here there can be anything you need them to be any non zero number doesn't matter so to start let me get rid of that matrix over there to start our first pivot position here up in the top left our first position there in row one is already a one.
01:10
So we're good there.
01:11
So all we have to do is eliminate the negative one below it.
01:15
So to do that, we'll take row two and make it row two plus row one.
01:24
And we will get one zero, two, two, two, six, six, three, three, zero, negative five.
01:42
Okay.
01:43
After that, we want to eliminate that two below our first one there in the top left.
01:50
So we will take row three and make it negative two, row one, plus row three.
02:03
And we will get one zero zero, two, six two, three, three, negative three, and zero, negative five.
02:24
Okay.
02:27
After that, we'll go down a diagonal...