[MUSIC] I want to continue with our discussion of Matrix Algebra to define what is a Determinant. To understand determinants. Because to derive Cassini's identity in the course, we need to. We use the concept of Determinants. So let's see what a Determinant is. The easiest way to explain what a Determinant is is to look at a system of first order equations, ax + by = 0, cx + dy = 0. X and y are the variables. A, b, c, d are the known coefficients. And with here the right hand side is identically 0. We can take the system of equations and write it as a matrix equation. So remember our matrix multiplication. This is a, b, c, d times the column vector x y equals 0 as a column vector. So when the question that brings in the Determinant, is when does there exist a nontrivial solution for x and y in these equations? The trivial solution is just x equals 0, y equals 0. because then we know get zero or equal to zero. So when does there exist a nontrivial solution? Well, we can manipulate these two equations. We can eliminate y if we multiply the second equation by b and the first equation by d and then subtract the second equation from the first equation we get rid of y, that will give us (ad- bc) x = 0. Or we can get rid of x, we can multiply the first equation by c and the second equation by a. Subtract the first equation from the second equation and we get (ad- bc) y = 0. So in order for the solution of these two equations to be different than x equals y equals 0, we would require that (ad- bc) equals 0. So there exist a nontrivial solution when ad- bc = 0. That's exactly how we define the determinant of a matrix. So we define the determinant of the matrix (abcd) = ad- bc. How do we view this? We multiply the diagonal elements a, d and then we subtract the multiplication of the off diagonal elements b, c. So that's all there is for the two by two case. If you want me to do higher dimension, you can extend the definition to higher dimension as long as the matrix is square, n by n say, but we don't need that for this class. So we'll just stay with the two by two case. So the theorem we need to prove that Cassini's identity, is that the determinant of the product of two matrices is equal to the product of the determinants. So the determinant of the matrix A times B = det A times det B. We need this only for two by two matrices. So even though this is a general result for A and B n by n matrices, we need to only prove it, for the purpose of this class, for two by two matrices. So I think the easiest way to prove the statement is just to do the calculation. Just to do the calculation. So let’s see how it works. We introduce 2 matrices, matrix A has elements a, b, c, d and matrix B has elements e, f, g, h, right. We can take the product of the matrices so we multiply A times B, so we go across the rows of A and across the columns of B. So here we have (ae + bg af + bh) here. Ce + dg here. Cf + dh here, okay? So we get the product matrix. Now, we're looking at the determinant of AB. So the determinant of AB is the product of the diagonal elements minus the product of the off diagonal elements. We get this expansion. We can do all of the multiplication. So let's do all of the multiplication. And we get this expression. If you want to do the algebra correctly, you have to think how to order the terms. So alphabetical order, I think is good idea. A,b,c,d,e,f,g,h, always keep it in alphabetical order. We can do the expansion. We get four terms from each of these products. We have a minus sign here, so we can look for a cancellation and in fact, acef exactly cancels acef. And the other two don't cancel, but the fourth one bdgh cancels bdgh. So we can do the cancellation. So we're left with four terms, Four terms here. We're trying to find determinant of a times determinant of b, so we can do some nice factoring here. We can factor these terms. So we can factor out an ad. So ad then we get eh- fg, and we can factor out a bc. So then we get. We put a minus sign here. Then get from this bc(eh- fg). And both of these terms then contain an (eh- fg), so we can factor that one out and we get (ad- bc)(eh- fg). So we're getting what we want. So this (ad- bc) is the determinant of A and (eh- fg) is the determinant of B. So we get the determinant of A times the determinant of B. So we've proved then that the determinant of AB, the product ab is equal to the determinant of a times the determinant of b. For the two by two case. And this is exactly what we need to prove Cassini's identity.