00:01
They want us to solve this linear system of equations.
00:05
So let's go ahead and turn this into that upper triangular matrix that we've done so far in this chapter.
00:11
So remember what we are going to do is start over here with a, and then we're going to clear everything below it.
00:18
So what we're going to do is the following.
00:22
So we want to multiply row 1 by negative 3.
00:27
So it would be negative 3, row 1, and then we want to...
00:31
Add this to row 2 and that's going to cancel out 3a and then similarly we want to multiply row 1 by 5 and then add that to row 3 and then that will cancel that 5a so let's go ahead and write that out so our top equation isn't going to change at all so it's still a plus b minus c is equal to negative 1 but but now, when we do negative 3, row 1 plus row 2, so we get 3 times negative 3, so that would be negative 3a, so those cancel.
01:13
Then we'd have b times negative 3 plus negative 1, which would be negative 4b.
01:21
And then we have negative c times negative 3, so that would be positive 3.
01:26
Add that to 2c, so that would give us plus 5c.
01:30
And then lastly, negative 1 times negative 3 is 3 plus 9 is 12.
01:37
Now we do the same thing for the last row.
01:40
So we had negative 5 times a.
01:43
Well, that gives us negative 5.
01:46
Add that to 5a, that gives us 0.
01:48
Now 5 times negative or b times negative 5.
01:54
B plus 3.
01:56
That's going to give us negative 2.
02:00
B, and then for c, so negative 1 times negative 5 is 5 plus 3.
02:10
That would give us 8.
02:11
So we have plus 8c.
02:13
And then negative 1 times negative 5, 5 plus 1 is 6.
02:18
So let me go ahead and get rid of those zeros right there.
02:24
All right.
02:24
So this is the first step.
02:25
So now we have everything below a cleared.
02:29
Now we want to repeat this with the b...