So now, some of you may point out that this process would be faster if we preallocated memory. You may come up with something slightly different that still works. Now, I just want to mention that the solution that I'm providing is non-unique. So if we run this, we should get the matrix we expect. So I'll write B of colon comma I equals 2 times B of colon comma I minus 1 and then end. Next up is writing the pattern in MATLAB code. So when we set up our loop, we'll say, for I equals 2:4. And this time, we're appending three more columns to B. So just like before, let's create our initial vector, B. In this case, each column's values are double the values in the column before it. So the first step is to figure out the pattern. And we want to produce the following matrix. This time, we'll start with this column vector.

And since we left the statement unsuppressed, we can see each iteration as well. As we can see, we get the matrix we were expecting. Now, let's see what happens when we hit Run. And we'll continue to iterate through the loop until we've gone through all the index variables. The result will be that A is a 3-by-5 matrix. We index into all of the columns in the third row and set those values equal to 1 plus the row above it. So after one iteration, A is now a 2 by 5 matrix. And the second part, equals A of I minus 1 comma colon plus 1 means that we are setting the second row equal to 1 plus the values in the row before it, in this case row 1, and then end. A of I comma colon means we are indexing into all of the columns in row I, which in this case is row 2. Then inside the loop, I'll write the following command: A of I comma colon equals A of I minus 1 comma colon plus 1, and then end. I'll go ahead and create an index variable for i equals 2 through 5. So this means we're going to need four iterations in the loop. We know that we need to append four rows to our current vector. So now that we have our vector, we need to think about creating our matrix in a loop. And I'll show you why at the end of the video. I'm not going to suppress the outputs in this example. So now that we found the pattern, the question is, how do we create this matrix in a loop? Well, the first step is going to be to create the initial vector. And the second column reads 3, 4, 5, 6, 7, et cetera, et cetera. See, this first column reads 1, 2, 3, 4, 5. The values in each row are equal to 1 plus the values above it. If we look closely at this matrix, we can see a pattern. And I want to make the following matrix from it, this one right here. We're going to do this by answering a few questions. Today, we're going to talk about creating a matrix in a loop. Hello, and welcome back to another MATLAB video.

Matlab vector times matrix