>> Here we have a matrix A and a matrix B. The first thing I'd like you to notice is that they're in not the same order. Order is really important in this process of multiplying matrices. Now this is a 3 by 2 matrix and this is a 2 by 4 matrix, and so 3 by 2 because we have 3 rows, 2 columns; 2 by 4 because 2 rows 4 columns. Alright. It turns out that we can only multiply matrices in which the number of columns of the first one corresponds with the number of rows of the second one, alright. And now that therefore takes out the idea of the commutative property holding when we're multiplying matrices that is although we are able to multiply A times B, we cannot multiply B times A because if we swap the rows of these two, we wouldn't have this row-column correspondence idea. Okay, now. Once we establish that we can multiply the two, we can identify the order of the resulting product matrix and the order of that product is gonna be 3 by 4. So A times B is gonna be a 3 by 4 matrix and we can setup the skeleton for that situation like this. Now there are a number of ways of illustrating that and calling attention to the various entries here. One way would be I suppose to do this, we have 3 rows, something like this and then 4 columns, 1, 2, 3, 4, and everywhere where we have an intersection, you see, we have an entry of some kind that we'd have to put in here. Well, that gives a little convoluted so we use another designation here. I'm gonna just put some circles in here to designate the different entry possibilities in this answer matrix. So we'll have entries in these positions and now we have 4 columns. So, this is 1, 2, 3, 4. Okay, now the next trick is to identify the value of each of these--those in each of these circles. Well, let's take this one. Now this item is in the first row, first column of our answer and here's what we do. Now, the first row, first column, we take the first row of our first matrix and the first column of the second matrix and we perform this operation. We take the entries here in this first row and for each entry we're gonna multiply times the entries over there. Here's what we're doing. It is negative 1 times 2. I'm gonna write that out for emphasis. Negative 1 times 2 and then plus, you see, first times first, and then along this row, then 3 times 4. So we have--let's see negative 2 plus 12, this is 10. So the entry that goes right here is 10. Now, let's do, let's take another spot. Let's take a spot, oh, maybe somewhere down here. Now notice that this item is in the--well, let's see, rows and column 1, 2, 3. This is in the third row, second column. This is the A subscript 3, 2 or 3 2 item. Let me write that a little better, third row, second column. Now we use this information to go up here and identify--let's see, third row, boom, second column, second column, these two. So we're taking these two entries and we're multiplying times these two entries in a particular way. It's 5 times 5 is 25 plus 0 times negative 3 is 0. So this entry turns out to be 25. Now--and we have to do that for every single entry in here. Alright. So we're kinda intertwining these two matrices in a very particular and at first complicated kind of way.