[MUSIC] So you reached the final stages of your quantity techniques module. You already know what equations are. You already know what functions are, and you already know what matrices are. So let me give you some content into matrices and why they are important. So I'm taking you back to the first tutorial that we've done together, and I gave you a system of equations that were represented with two functions, x + 2y = 5 and 4x + y = 6. So this was a system of equations that we used in various things. One of the things that we used it's to show what is the equilibrium condition in Cournot duopoly equilibrium. Another interesting thing that we can do with these functions is actually to see how matrices can be used in order to solve the system of equations. So to solve the system equation mean to find equilibrium, the Cournot duopoly situation or Cournot duopoly example. So the way to do that is to first of all, to recognize that there are two, or actually there are two unknowns with three important factors that are involved. The first important factor is the value that are assigned to x, the values that are assigned to y, and the vales that stand as independent value or figures that are represented by 5 and 6. So how do we write these in a matrix form? So because we are talking about system of equation where we have two equations, we have to allocate two rows for those two equations. Because we have two unknowns, we have to allocate two columns to these unknowns. So in the first column, we will put factors that stand in front of x here and x here, which will be 1 and 4. And in front of y and y here, we will assign to the second column values, 2 and 1. So in other words, we can say this will be a two by two matrix. [COUGH] The second thing we have to recognize is that those are just values, and they are assigned to x and y. x and y will stand in one column, x and y. So we have two rows that will represent two unknowns, x and y, as they correspond to the system equations here. And all this system, just to give you some annotation, that this is two by one matrix, or in other words, it's a column vector. And the outcome will be 5 and 6. Again, this will be correspondent to the first equation, and this will be correspondent to the second equation, and again, this is two by one matrix. As you can see, this is another way to write the exactly the same information as decoded in this system equation, but in the way of matrix. As you already see from here, if we have much more than two unknowns, it will be very easy for us to write this down as say, unknown z, unknown b, c, d, and so on and so forth. So rather than writing equations in such way, we will be able to write the equations in the following way. But the idea is the same. The idea of these matrices is to write the same piece of information but in much more condensed, in much more compact and neat way. So let's get rid of that because it's not really important right now. What I want to show you right now is in order to solve this system of equations, we will use some techniques that you studied throughout the lecture on the fourth week of your MOOC tutorials. And you will see that the outcome of that will be exactly the same as the outcome of the equilibrium condition that we found in our first and second tutorials. First of all, [COUGH] what we need to do, we need to find out those bits. In order to find those bits, unlike linear algebra, we need to use the matrix algebra. And in matrix algebra, what we need to recognize is that we want to get rid of this bit, or at least to put this bit, or this matrix, on the other side of the equation, do some manipulation, and to find out the values of x and y here. So how do we transfer this matrix into the other side of the equation? If it was a scalar, what we would have done, so for instance if here we had scalar 2x = 7, we would divide 7 by 2. So in other words, [COUGH] we would, Multiply 7 by 2 in the power of -1. In matrices, such manipulation cannot be done. What can be done is to find an inverse of this big matrix. And in order to find an inverse of a matrix, so if we take matrix A and we want to find an inverse of this matrix, pay attention that the symbols are quite similar, but they do not represent exactly the same thing. So in inverse of the matrix A will be equal to 1 over determinate, not absolute values but determinate which are those type of brackets, times, [COUGH] the Adj(A). So we have to introduce ourselves with two notions, a determinant and an adjugate. So let's deal with them separately. For our purposes, this is matrix A. Matrix A is a two by two matrix, With those values. How do we find a determinant of the matrix? A determinant of the matrix, there are quite a lot of formulas, but for the sake of the quadratic matrix with only two by two dimensions, it's quite straightforward. We'll multiply this value by that value and subtract the product of those two values. So determinant of the matrix is equal to 1 times 1- 2 times 4 or in other words 1- 8 = -7. So this is the first bit that we do. Adjugate of a matrix, Is likely more complex, but in this case, it's not that difficult. So first of all, the dimension of the adjugate matrix. The dimension of adjugate matrix will be exactly the same as the mentioned, or the original matrix. So it's going to be two by two matrix, so we allocate spaces for the four values that we need to plug in. So what will be the value here? In order to find the value here, we need to recognize this is the place that we're interested in, eliminate this row and this column, and we're left with number, that is, will be placed here. And the same exercise we do with all the remaining of the bits here. So for this value, we have to eliminate, This column and this row, and we're left with value 4, but the the only thing we have to remember is to put a negative sign here. Again for this value, we eliminate this column and this row, we're left with 2 with a negative sign. And for this value, we eliminate this column and this row, and we're left with value of 1. If you're more interested in adjugates, I suggest you look at the general expression of adjugate which you would be able to find out in any simple textbook on matrix algebra. One thing about adjugate is that this matrix is not the final stage of the matrix, we'll also need to transpose this matrix. And I will show you exactly how to do that. The transposition of the matrix is, Where, in this example, this will be quite straightforward because the diagonal will remain in the same place, 1 and 1. And the only thing you do is you swap the places of this value and this value, so you have -4 here and -2 here. So this is the adjugate which you need to plug it back, and this is the determinant which we need to plug it back again here. [COUGH] And this will be the outcome of the inverse. So if we plug the inverse back, we see that x y as a column vector is equal to 2, sorry, the inverse of the original matrix, let's call this the original matrix A times 5 and 6. So let's plug back the adjugate and 1 over determinant in its place. So 1 over -7 times 1, -2, -4, 1, and we multiplied by 5 and 6 as a column vector. Let's get rid of that. [COUGH] So the next thing, what we're going to do is we're going to open up this matrices here. In order to multiply matrices, one need to remember that this is two by two matrix, this is two by one matrix. That the inner dimensions of those two matrices need to be the same, and the outer dimensions will represent the dimensions of the outcome matrix. So in this tutorial, I will finish this exercise, and I will try to show you how to solve this system of matrices in order to produce an outcome that will give us the values of x and y. So let's continue from the point that we stopped, so in other words, what we're trying to do is to have a matrix multiplication. In order to have a matrix multiplication, there are two things that we need to keep in mind. First of all, the inner dimensions of the matrix, so this and this, will have to be exactly the same, otherwise you won't be able to multiply the matrices. And the outer dimensions of 2 and 1 will give us the outcome of the matrix. So in other words, what we say that the multiplication of this matrix together will produce us, let's replicate that, 1,- 7. A two by one matrix, or we can say this is just a column vector, and we will see what happens here. So this will be two by one, so two rows and one column. So what are those two rows? In the first row, we will take this row here and multiply it by this column here. So the way to multiply a row and column will be to take the first value here and multiply by the first value here. So 1 x 5 plus, the first value here times the first value here. So -2 times 6 which will be -12. And the same idea will apply to the second row. So first value here times the first value here, 4 times 5, so -20. [COUGH] Second value here, times the second value here, 1 times 6 plus 1 plus 6. What we get now is that it is 1 over 7 with a minus sign times 5 times 1 is 5 minus 12, it is -7. So this is the first column that we get, -20 + 6 = -14. Here, we also have to multiply this vector, this column vector by this scalar which is outside the vector here, and what we get is that x and y is equal to 1 / 7 minus, times 7, it's just 1. -1 over 7 times -14 is just 2. As you can see, with the system of equations that we got before, we transferred it into matrices. We found out how to find an inverse of the matrix, [COUGH] and we also practiced a multiplication of matrices, which resulted in the dimension of the outcome. And at the end what we get, we get exactly the same outcome for x and y. x equal 1, y is equal 2. Which is exactly the same values that we had for equilibria condition of Cournot duopoly. [MUSIC]