00:01
For this problem, we have two matrices, a and b, and both are going to be upper triangular or they're both going to be lower triangular.
00:10
I'm going to set this up as if they're upper triangular.
00:13
The same proof applies for lower triangular as well.
00:17
So let me just write down some basic definitions and stuff here, so as we're referencing them, it kind of makes sense.
00:26
So a is a matrix whose elements are a, i, j.
00:30
A is the row, j is the column.
00:33
And b is going to be the same.
00:35
Okay.
00:36
And these are both going to be upper triangular.
00:44
In other words, everything below the diagonal are going to be zeros.
00:49
So the elements in a are going to be zero if the row is greater than the column.
00:58
Okay, those that will give me zeros below the diagonal.
01:01
And the same for b.
01:02
They're both the same type of matrix.
01:06
So again, the element.
01:07
Elements in b will be zero if the row is greater than the column.
01:12
I'm not making anything from the diagonal or above.
01:16
There could be zeros elsewhere.
01:18
We just know for sure that there's zeros below the diagonal.
01:21
Now, if i multiply these together, a times b, and that's going to give us a matrix, and we'll call these elements x that we get when we multiply them together.
01:32
Let's take a look.
01:33
We need to look at the diagonal matrix, the diagonal elements, of matrix a times b.
01:40
We want to prove that that diagonal that's made up of x11, x2, x33, and so on.
01:50
We want to show that those entries are the products of the diagonal entries of a and b.
01:57
So i want to show that x11 is a11111, x2222222, a33333 and b3333.
02:09
And so on.
02:11
Okay.
02:12
So let's see what we have.
02:16
X11, first row, first column in my product, in my solution...