[MUSIC] Hi everyone this is professor engineer from Kaist. We are now in the week three session known for the Beijing mess for AI beginner part one, which is linear algebra. And this week we are going to study about rural operation more and by using this low operation we try to solve the system of linear on vacation. And for the application we also try to solve the linear combination problem, so this is a big picture of our course. So we start with the view Matrix 1st and then now we are in this in the stage of finding solutions for the system of linear verification which is AX equals B by using real operations. And after that you are going to study about the finding inverse of the square matrix A also using the operation and then you're going to study about determinant of scare matrix A by calculating determinant with cooperation. And in big picture, those three are closely related to each other and also how those three topics are connected with AI application in between. We'll look over the vector vector space and in your independent vector matrix, I can build a problem and die a generalization problem and AI applications such as deep learning and support vector machine. Last session last week we study about how to our system of linear the vacation by using their operation, so I bring also example. So let's solve this system of the near our vacation If the system is inconsistent consistent, so we have for unknown here x, y, z, w and with four linear logic medications. So first occasion is x plus y minus 2z plus 3w equals 2 and second equation is x plus y minus z plus w equals 5. And third one 2x minus y plus four z equals 1, the last one is y plus z plus w equals 8, so followed by last week. So let's use tablet form to solve this system of linear equation by taking the constant which is multiplied with unknowns for unknown. So you can formulate left hand side, the tablet 1, 1, -2, 3 and 1, 1 minus 1, 1 And 2, -1, 4, 0 And 0, 1, 1, 1, did 4 by 4. Numbers are the elements of the 4 by 4 matrix for this system of the algebraic equation and right hand side we have 2, 5, 1, 8. So we can simply like modify or express those four occasions to do this tablet form. Okay now let's use this row operation, so I tried to modify second row and third row which is 1, 1 minus 1 and 5 and 2, -1, 4, 0 and 1. So I want to make the first element of the second row 1, 1 minus 1, 5 become 0. How to do that, the row two become I modify row two minus row one, row two is what 1, -1, 1, 5, the one you've got 1, 1, -2, 3, 2, so I can modify second row R2 are too so how to do that? So 1 minus, so first element for the secondary is 1 and first element for the first lowest 1 so 1 minus 1 becomes 0. And the others 1 -1 becomes 0 and minus 0 minus 1 minus 2 become 1 and 1 minus 3 become- 2, -2 and 5 minus 2 become 3. So new tablet form for the second row become 0, 0, 1, -2, 3, so I can modify this row 2 by using row operation. So to modify third row which is 2, -1, 4, 0, 1 I use R3 minus two times R1. Two times R1, row one is what? Row 1 in the first tabular form 1, 1, -2, 3, 2 so two times there's 2, 2, -4, 6, 4, okay then if I use that operation R3 minus 2R1 it become 0 -3 ache minus six minus three. So now I modify second row and third row by using row operation. And now I change the, exchange the R2 to R4 so it becomes R2 0, 0, 1 minus 2, 3 goes down to the fourth row so 0, 0, 1 minus 2, 3 and fourth row go back to go to the second row. Right this and then the process I try to do is I try to make the opera triangular matrix theater. So I also try to modify third row 0, -3, 8, -6 ,-3 by using R3 plus 3R2, then I can make the minus 3 become 0 and then the other element become 0, 0, 11, -3, 21, here it is. So here only one element is left for making opera triangular matrix What is that Is 0, 0, 1, -2, 3 which is of last row. So I try to make this 0, 0, 1, -2, 3 and this one I want to make it 0 by using row operation so how to do that? 11 times R4 minus R3, then I can make this one becomes 0 and then the other element change like this from minus 2 to minus 19, 3, 2 terms like this. So here I finally get to triangular matrix form like this, okay, now I have the changed form of the system of linear verification. But this miniature application is equivalent to the original occasion because even I use the low operation. The solution of system of linear vacation is same, those two systems of linear justification is equivalent, right? Because I properly use those row operations, first is changing row, second is modified certain no by using other other row. So let's say this, x, y, z like this, so I calculate from the last rule, so -19 from the last row -19 times w of course turf. So in the end w equals -12 over 19 and third row is 11z minus 3w equals 21. Because I already get w from the last row, I just plug in this w to the desiccation and I can get g equals -33 over 19. So I have w and z value here and by using this w and z I can calculate y, so because y in the second row, y plus z plus w equals eight. And first row x plus y minus 2 equals 3w equals 2 so I just used to tell you for the x, y and g I z and w. Then I can find a solution for the system of linear equation which is x, y, z, w and 9 over 19 1/19 nineties. So in this case We found that this system of linear education has unique solution which is this one x, y, z, w and we use the operation to reduce the original matrix AX equals B to the EUX equals C. And we found we can calculate the solution in a very easy way, so after here I just showed that that we can use the operation to find solutions. And then for the next session we are going to study about observing system of linear dedication in a different case. May be inconsistent or the infinitely many solution case. Thank you very much.