 If our matrix is one of a special type of matrix, then we might have a special way of finding the determinant. And the most common of the special matrices are the triangular matrices. So let's see if we can find the determinant of a triangular matrix. So suppose T is a triangular matrix. Prove or disprove that the determinant of T is equal to... Well, I don't know. Let's find out what it is equal to and see if there's an easy way to find the determinant. Remember, an example is not a proof, but it can be used to guide our proof. So let's find the determinant of an upper triangular matrix. For example, this one. Oh, wait a minute. While this is an upper triangular matrix, we can't find the determinant because it's not a square matrix. So remember, one of the roles of proof in mathematics is it allows us to study mathematics more effectively. In this case, the process of trying to prove something about the determinant of a triangular matrix reminds us of two things. First of all, what a triangular matrix is. But then the other is that it reminds us that we can only take the determinant of a square matrix. So let's get rid of one of these columns. Well, I can't get rid of the first column because I won't have a triangular matrix and the second column likewise will not give me a triangular matrix. So let's get rid of this last column here. And now let's find the determinant. Something that's useful is that we can expand along any row or column. And in this case, it's particularly convenient if we expand along the first column because except for that initial entry, everything else is going to be zero. So if we do that expansion, these last two terms, whatever the determinants of these minors is going to be, they're going to be multiplied by zero so we can drop them out. So my determinant is going to be 1 times 5 times 1 minus 3 times 0. Or let's not do the arithmetic and just write out that product 1 times 5 times 1. What if I have a 4 by 4 triangular matrix? So as before, because my first column has mostly zeros in it, I'll choose to expand along that first column. And so my determinant is going to be 1 times the minor. And now we can try and find the determinant of this matrix. But again, we see that it's probably easiest to expand along the first column because it's mostly a column of zeros. And so I'll expand along that first column. And as before, we won't do the arithmetic, but we'll just write down what we would be doing. And a determinant is going to be 1 times 5 times 1 times 2. And at this point, we might notice some important things. For our triangular matrices, the determinant seems to be the product of the entries along that main diagonal. So let's see if we can generalize it. Again, the value of doing the examples is it gives us some insight into how our proof should proceed. And in our examples, the thing we did is we found the determinant by expanding along the first column. So let's take an upper triangular matrix and we'll repeatedly expand each minor along the first column. So that'll be our t11 times the minor. And all of the other components of this first step are going to be 0. And now when I expand this, because it's an upper triangular matrix, I'll take t22 times the minor. And again, I have another upper triangular matrix. And if I continue this process, what I'm going to end up with is a product of the terms along that first diagonal. And this proves the result as long as t is an upper triangular matrix. So what if t is a lower triangular matrix? So let's take a lower triangular matrix and find its determinant. Here we see that the first row consists of mostly 0, so we'll expand along that first row. And again, we find that our determinant is the product of the entries along that main diagonal. And because an example is not a proof, we need to generalize. But again, having done the example gives us some insight into how we want to proceed with our proof. And so we might proceed as follows. Let t be a lower triangular matrix, and this time we'll expand along the first row. And so our first step will give us t11 times the minor. And again, we have a lower triangular matrix, so we'll expand along the first row again. And if we continue this process, we're going to pick up all of the entries along that main diagonal. And so this proves the result that if t is a triangular matrix, then the determinant of t can be found by multiplying the entries along the main diagonal. What happens if I find the product of two triangular matrices? What is the determinant of the product? Well, that's actually pretty easy. We have this theorem that says the determinant of a product is equal to the product of the determinant. But let's see if we can prove this result without using that theorem. Now you might look at this problem and consider it to be somewhat contrived. We already have a theorem that tells us how to find the determinant of a product. Why wouldn't we use it? And the answer to this comes from sports. Or really, any competitive activity, such as military service. When you train, it's best to train at a disadvantage. That way, when you're in the real situation, it's a lot easier than training is. And we face a similar situation with proofs. The purpose of proof is to reinforce knowledge. So to train harder, start with less. So let's take two upper triangular matrices and multiply them. And then find the determinant of the product matrix, which, because it's a triangular matrix, we do know that the determinant is going to be 2 times 2 times 3. On the other hand, if we find the determinants first, then our determinant of the first matrix will be 1 times 1 times 3. And the second will be 2 times 2 times 1. And we'll rearrange things a little bit and find that the determinants are in fact the same. And again, it's important to remember an example is not a proof. But it can provide insight into how we construct the proof. The key things that we notice here are that when we find the determinant of an upper triangular matrix, then it's going to be the product of the entries along that main diagonal. And so we might begin with asking the question, how do we find those entries? Since we're dealing with triangular matrices, it will be helpful to remember what a triangular matrix is. So we'll pull in our definition. So the entries for the main diagonal of the product A times B are going to come from multiplying the i-th row of A by the i-th column of B. So that entry is going to be given by the sum of the component-wise products. Now, remember, A and B are upper triangular matrices. So Aij and Bij are going to be 0 any time i is larger than j. But now, each of these terms, Aik times Bki, at least one of these factors will be 0 because either i is greater than k or k is greater than i, unless i and k are the same thing. And so that means most of these terms drop out and this entry Cii is just going to be equal to Aii times Bii. Now, the other thing we know is that when we multiply two upper triangular matrices, we get an upper triangular matrix. And since C is an upper triangular matrix, that means that we can write its determinant as the product of these entries along the main diagonal. And so the determinant of C is going to be the product A11B11 times A22B22, and so on, all down the line. Now our example suggests, and we hope to prove, that this is the determinant of A times the determinant of B. So let's write down our ending point. And since we know the determinant of A and the determinant of B, we can fill those in. And if we look, we can go from this line giving us the determinant of C to this line giving us the determinant of A times the determinant of B by rearranging the product, which we can do because our A entries all come from our underlying field.