 Hi, I'm Zor. Welcome to a new Zor education. Continue talking about inverse matrices. Let me just remind you the definition which we actually came up with in the last lecture was that if you have a matrix X, let's say, then the inverse matrix is the one which we symbolically use X to the power of minus one. If it's multiplied by our X on the right, it will give an identity matrix, or on the left will be exactly the same result. We were also talking about matrix X to be a square matrix, dimension n by n, which necessitates the inverse matrix also to be of the same n by n dimension. And so obviously is the dimension of the identity matrix, where all the diagonal elements are ones and all other elements are zeros. Now we have defined this. That's fine. We also investigated certain properties. For instance, certain matrices do not have inverse, like the matrix for instance with one row equals to zero. We already discussed that in the previous lecture, and we have also proven that if the inverse matrix exists, then it must be unique. There are no two different inverse matrix for a given matrix. Now all these discussions left alone one very important aspect. Do inverse matrices exist at all? You see, it's a very important quality of any definition. If you define something, it's important that you define something which really does exist. And then you can investigate the properties, like if it exists then. But this existence is very important thing. So today I'm going to spend relatively short period of time to talk about existence of the matrices, which are inverse of some other matrices. And in this particular case, just for simplicity, I'm restricting myself to a square matrix of two by two dimension. The simplest matrix which can be actually considered. So what I'm going to do is I'm approaching constructively this question of existence. So I will take a square matrix of two by two dimension and I will attempt just to find an inverse to this matrix. If I'm succeeding, it actually means that inverse matrix does exist, at least in the two by two case. And well, if I fail, it means there are no inverse matrices and it doesn't really make any sense to define it anyway, right? But I will succeed. So let's do it and our purpose is just consider any matrix which is a square two by two matrix and I will just take its coefficients as a, b, c and d. And I will try to find another matrix with unknown coefficients, unknown elements, w, x, y and z, such that their product is equal to an identity matrix of two by two size, all right? So I consider this to be like an equation, basically, where this matrix is unknown. This is a known coefficient and this is the known result. Well, if these are not matrices but numbers, let's say you have number a times number x equals number b. I know how to derive the solution to this. If a not equal to zero, this is a condition under which this actually does exist. The solution does exist. Then x is equal to b divided by a, right? So I know that. So now I'm looking for something similar in the world of matrices. But let's just consider we have four different elements, so it's four unknowns. Now, matrix multiplication is actually a certain number of rules which we can derive and really multiply these rows by these columns, etc., to get this. And we will get a certain number of equations, actually four equations because for each element of the result, we have certain linear dependency between the components. So let's just do it. We will have a system of linear equations with a certain number of unknowns and we will just solve it, right? Simply. Okay, how to multiply two matrices? To get an element 1, 1, 1, first row and first column, we have to multiply a scalar product first row, vector row, by first vector column. So it's a b times w y, which is a w plus b y is equal to 1. Next, how to get second row, first column, 2, 1, element 2, 1. Second row, first column. Well, it's second row and first column, scalarly multiplied. And this is 0. Now, this is 0. First row, second column. First row, second column. A x plus b z equals 0. And last one, second row, second column. Second row times second column. C x plus d 0 equals 1. So this is the system of four linear equations with four unknowns. And actually, it's simpler than generalized system because it actually can be considered as two independent systems. You see, there are only w and y here and only x and z here. So I can separately solve this system of two equations with two unknowns and separately from this system of two equations with two unknowns, x and z in this case. So let's try to solve this one. We will multiply, well, I want to get rid of y, so how do I do it? I multiply this by d and this by b, this by d, this by b and subtract. So what will be? It will be a d minus b c w equals d minus 0 d. That would be from this, right? a d w b d y subtract b c w and d by y will be cancelled out and I have a d minus b c. Yeah, that seems to be correct. From which I can derive w equals d divided by a d minus b c. Okay? Now, okay, let me write it here. w equals d divided by a d minus b c. Okay, now, if I will multiply, now to get y, I have to multiply this by c and this by a and subtract, right? So this by c and this by a and subtract. It will be a c w and this will be a c w and subtract. Okay, so I will have b c y minus a d y equals to c, right? Now, mind you that this is, in parenthesis, almost the same as this one except the sign. So I would like everywhere to have the same. So I will put y equals minus c divided by the same a d minus b c. So I change the sign here and it's here. Okay, so this system is solved. Now this one. Very similarly. To get rid of z, I multiply this by d, this by d and subtract. So what I will get? z will be cancelled. a d minus b c, a d minus b c x equals, now this is by d, this is by b, minus b, from which I conclude that x is equal to minus b divided by the same sinc. And finally, to find z, I multiply this by c, this by a and subtract. And I will get b c minus a d z. Now this is zero, this is a minus a, right? Or again, I reverse the signs and I will get a over a d minus b c. Okay, my system of four equations with four variables is solved. And what's very important is that it is solved with one and only one condition. Because it's the same denominator for every one of them. So if this condition is preserved, then my answer is y w. So it's d minus c minus b a, right? Divided by a d minus b c, which is not equal to zero, right? Since every element of the matrix has this coefficient, I will just use the multiplication of the matrix by constant. That's exactly the same thing. So to multiply the matrix by constant means actually multiplying each element by this constant. So d minus c, okay. So as you see, inverting this particular matrix seems to be very easy. You change the a and d and you reverse the signs of b and c. Well, let's check if we do get the result of this if we multiply this by this particular matrix. You know, checking is always the good thing if you solve the equations, right? So let's just check. Let's forget about this for a while. We will multiply the result by this factor afterwards. But right now we will multiply a b c d by d minus b minus c a, d minus b minus c a. Okay, let's multiply. First row by first column, a d minus b c. Okay. Now, let's go down. Second row, first column, c d minus d c, zero. Now, first row, second column, a b was a minus, so it's minus a b, and b a, zero. And finally, second row, second column, second row, second column. Stop, stop, stop, stop. Something is wrong. Oh, second row. Here, second row, second column, minus b c plus a d. So a d minus b c. So that's my matrix as a result. And with this factor, this will be one, and this will be one, which is exactly the identity matrix. So I have derived with a correct solution of this equation. This is my matrix a b c d minus one. The power of minus one, which means inverse matrix, okay? So the inverse matrix to a b c d is d minus b minus c a with a factor one over a d minus b c under this condition. So this condition is necessary and sufficient to get this solution. Just as a side issue, this is called, this expression is called a determinant of the two by two matrix. And we have a special topic, determinants, matrix determinants, which is already on the website. So any matrix has this kind of an expression called determinant, which is just one real number, which concentrates a lot of power telling something about the matrix. So in a way, it's an equivalent of comparing the numbers, the real numbers with zero. So whatever the properties of the real numbers relative to their being equal to zero are true or not true. For instance, we cannot divide by a real number equal to zero, right? In the matrix world, a similar concept is the determinant of the matrix. If it's equal to zero, it's not a good matrix in some way. Which means, for instance, it does not have an inverse. And there are some other things which I'm actually talking about when I will be talking about determinants in other lectures. But anyway, for a two by two matrix, this is a determinant. And this is the way how the matrix is supposed to be inverted. That's basically it, because something like uniqueness and some other properties of inverse matrices we have discussed already before. Now, if you remember, we were talking about something like linear dependency between rows or columns, which prevent us to have an inverse matrix. Well, guess what? If, let's say, two rows are linearly dependent, then the determinant is equal to zero. Well, just as an example, look at this. If you have two linearly dependent rows, let's say two, four, six, twelve. What's the determinant of this matrix? Well, it's two times twelve, right? A D, this is A, this is D. So the determinant equals two times twelve minus six times four, which is equal to zero. If some more complicated linear dependency exists, the result will be exactly the same. Or among rows. Or if one row is completely equal to zero. For instance, if this is zero and this is zero. Well, this times this is zero minus this times this is zero. So all these separate cases, which I was considering in the previous lecture about inverse matrices, which prevent us to have an inverse matrix, they're all combined together in one inequality. The determinant is not supposed to be equal to zero if we want to find an inverse matrix. Well, that's it for this lecture. I do encourage you to go to Unisor.com, go through the notes for this lecture. And if you are a registered student, you will be able to have certain problems solved, exams. Your supervisor or a parent can enroll or unenroll you in certain courses, etc. So it's a little bit more interesting work with this particular website. Well, other than that, this is it. Thank you very much and good luck.