Hi everyone. This is Professor Yongjin Yoon from KAIST. We are now in the first session of the five week class about the basic mathematics for AI beginner Part 1, linear algebra. Up to here we study about the system of linear algebraic equation, AX equals B, how to solve this system of linear algebraic equation by using the row operation and also we study about inverse of a square matrix A and by using the row operation. In this week, we are going to study about determinant over square matrix A, by calculating the determinant, we do a row operation. We already finished to look at a vector space and linearly independent vectors. After the determinant of a square matrix A, we can study about the Matrix Eigenvalue problem and Diagonalization problem, and in the end, we refer to the AI applications. Before we get into the determinant, our determinant is about whether the matrix has an inverse matrix or not. Let me do the ice-breaking with some example. Let's say there is matrix A and B, are two N by N matrix which are invertible. Then show that AB inverse becomes B inverse, A inverse, this is a very famous example of the question then let's works together. Note that, AB inverse denotes the inverse of AB, matrix A times B. What you are asked to do here, is to prove that the inverse of AB is B inverse times A inverse. When we have a prove question like this, just use the definition of inverse matrix is very useful. We need to prove that the inverse of AB is B inverse times A inverse. Then let's use the definition of the inverse of AB. If inverse of AB is B inverse and A inverse, then AB times B inverse, A inverse, should be the what? Identity matrix. Let's say, AB times B inverse, A inverse, then, let's look at the B and B inverse first and B times B inverse is what? From the definition of inverse matrix, B times B inverse become identity matrix, so it becomes A times I times A inverse. It becomes A times A inverse, and in the end, it refers to the identity matrix. AB times B inverse A inverse become matrix I. We proved that, so AB inverse is B inverse, A inverse. Very easy. When you have prove problems, just use the definition of inverse matrix. Now, we look at the idea of the determinant of a square matrix. If you have N-by-N matrix here, A equals like this, the elements A_ij of matrix A can be used to compute the number we called the determinant of A, and if you noticed, det(A) or absolute A, this is determinant of A. For our starting point, our approach is to define the determinant of A in terms of determinants of smaller matrices. The smallest square matrix one can find is the order of 1 by 1 matrix. Let's say A equals a, a small a, 1 by 1 matrix, and we define the determinant of A is a. We calculate the determinant of A using the I-th row of A or either using the j-th column of A. Let's start with the I-th row of A. When you take the I-th row of A, is a_I1, a_I2, a_I3 to the a_IN. This is I-th row of A. Determinant A, we just take the certain a_Ij and then minus 1 to the I plus j, times determinant of A_Ij, which is one order less square matrix, which is N minus 1 by N minus 1 matrix, obtained by deleting the I-th row and the j-th column of A. Then, summation from j equals 1 to the N, the sum over all columns in the row. This, it is not easy to understand. It can be shown that no matter which row we choose, the calculation of the value of the determinant of A is the same. In another way, we can choose J-th column instead of I-th row to calculate the determinant of A. For example, we have the J-th column of A from a_1J, a_2J, and a_iJ and a_NJ. In this case, we took the J-th column of A and the determinant of A is we just took the certain element a_iJ here, and then minus 1 to the i plus J, and times determinant of iJ, and this determinant of iJ is one order less the square matrix, which is N minus 1 by N minus 1 matrix, obtained by deleting the i-th row and J-th column of A. Summation from i equals 1 to the N. It can be shown that no matter which column we choose, to calculate the value of the determinant A is the same. The value is also the same as the one calculated using the row operation. Let's start the study with a simple case, a two-by-two case. A equals a, b, c, d, like this and let's calculate the determinant of A, using the first row. First row in here is what? A and b. Just take a first. Then a times, in the first, minus 1 to the a is what? Row 1 and column 1, so minus 1 to 1 plus 1 and times determinant of d because we exclude ab and ac. Ab is the first row, ac is the first column, including element a, so it remains determinant of d. Then plus-minus 1, we just take the first row to calculate it. When you look at the element b, then minus 1 to the 1 plus 2 and determinant of c because we can exclude the first row, ab, and second column, bd, and only remain the c. Like this, determinant of A, for the a, b, c, d, the two-by-two matrix become ad minus bc. When you take the column to calculate the determinant of A is the same. If we take the second column over the two-by-two matrix, then the determinant of A is minus 1 to the 2 plus 1 because b, element b is the first row and second column, so minus 1 to 2 plus 1 and determinant of c, because we need to exclude row and column which include b so it is only c. Because we took the second column of the matrix a and we need to calculate with d and then d is second row and second column, so minus 1 to the 2 plus 2 and then we exclude the row which includes d, which is cd, second row and also second column, include d, so bd, then it only remains a. When you calculate it, it become ad minus bc again. The determinant of a is ad minus bc by using either taking row or column. Up to here, is the example of the finding and calculate the determinant of A in the simple case for two-by-two case. Therefore, in the next session, we are going to expand to calculate the determinant d, in the higher-order of the square matrix. Thank you very much.
