Hello. In fact, we know the determinant calculation, but a kind of definition We have not, give up the definition, we avoid but what we have done until now; three three unknowns in the equation, even the two unknowns, the solution of two equations We found that naturally occur at the determinant, We determine the characteristics and determinants cofactors using the determinant of the features we describe, We have identified sub-matrices, we find the cofactor of their determinants, We calculate the determinant of the matrix by multiplying the number of cofactors. Correct by testing it on various determinants I do not want to say that we have found a proof, but we do not provide. Now the unknown in the equation in two unknowns We brought a recognition by the various generalizations such as checksums. We now have our say in this definition that these propositions, axioms that define a determinant, postulates, propositions, they are equivalent terms, we say that the rows of a matrix If this determinant is multiplied by the mean t t it has been hit. If you think a matrix consisting of two lines to the collection, one single determinant to calculate collection We agree to assume coming. If you think these two vectors of the two planes A similar situation toplanmasının- min. Here is the importance of being well, we assume that if a result of displacement. See here first, then hit a line with a number we calculate the determinant, In the right-hand side before we stood by that number then calculates the determinant. Similarly, we create a line picking up of two vectors, Recognizing individual lines here with this vector We calculate determinants, we collect them. Nevertheless, the only difference between the left and right row are different operations. After picking up the determinant vectors before we get here, We're gathering right before then calculates determinants. This again linearity property. If a line is zero zero determinant, We agree that there should be a determinant of a matrix in the unit We accept that. These do not question anymore. These are our proposition, postulate, axiom if you like what terms. Number calculated with this property because all of them will be released at the end of a number, These features provide a number of determinants that we call the matrix. They also D1, D2, D3, D4, call propositions. This proposition can also be calculated with a single number that thinner Your account kanıtlayabilis but given that the number appears as a single theorem. Now I'm a little generalize these propositions. Why? Because we found a new method of calculation using these propositions, We will achieve a formula. This formula will be a permutation symbol, the matrix multiplication of all the elements will occur. Now we see but what we saw the other until the very non-diverse, even including them, we will reach a formula. We're doing as a generalization; You will recall a line from D1 We stood with t here multiply all rows with a number t1, t2, and TN. We stood in line order, then we get the determinant, We stood at this number right after we receive the determinants ago. This generalization is applied repeatedly to the D1 is easy once we remove it. D1'll get out of this first line's apply only where t1 A1 remains. Before you apply to t2 on the rest, you take it out, a1, a2 remains. In this way a chain zipper opens as t1, t2, tn'll get out determinants change. Here, too, we had a generalization, We thought a line consisting of a total of two in the proposition. Here's a line that is comprised of one per this before though so here in a row After we get the determinant vectors will collect, we will calculate separately with each of the right side first determinant because one has a pin, then we will receive the sum of them. This again proved to be equal because both are here in C1 in this collection reserve the rest to hold together, before C is the C1'Li, others would, then re-take the C2'Li, Take the C triple pth up going here separately p The collection is open to become one of the determinants. If we generalize a little bit more so that only one of any line p Let's also not the collection is the collection of all such line. Here, unfortunately, it uses dual display that can seem confusing, J1, J2, jnr one display. We assume that the first line of the collection consists of many p1, We assume that the second line of the collection consists of many p2. This p1 and p2, the PN is not necessarily equal. Instead of making this collection before, first determinants single after that calculate only one determinant of the We believe that we take the collection. In this way as the left and the right the result of the calculation is different row We find that using the feature to change the equation. This generalization D2. Let's just simple application. If you hit the determinant of a matrix c determinant of this matrix comes equivalent to multiplying the force of n'yinc c. We see this now easily see t1 in this property, t2, TF will all c. Because all lines with a means to multiply matrices c means to multiply by c. E is also t1, t2, tn C at times over for all that c n'yinc forces outside interests so as to c. Here we hit the car all the lines here, one by one out of every c When we received that we can encounter the following error often determinants, if the determinant as to be multiplied by c. Not. det c, matrix means all lines multiplied by c means to multiply by c. Therefore c n'yinc would multiply the force because a car each. Example: Let's start with the matrix, one, two, three, We know this is four minus two earlier, we used the same simple example. Although we have to multiply the Trinity, the three first rows, six would be the second line of nine, would be twelve. As you can see, when we get thirty six minus its determinants nine times six, minus eighteen fifty-four remain. C eighteen minus the second force in ensuring that the formula here, because the two in our example, the second force is the product of the determinants here. Indeed, the second force three nine determinants of minus two were, We provide minus eighteen. Now we move forward a little more, right now we're going to get a formula. Let's take a typical ith line in order to achieve this, the elements of the i-th row AI1, AI2, aija, because it will be a first We satırd sets indicator means that the i-th row, also determines the order of the elements in the second row. Now we have seen previously in this unit vectors, I1 only a first component, i.e. the others zero i, j, k as unit vectors and generalized to the size of the Cartesian vector. This means that a typical vector, wherein the components wherein the vector that typically numbers, these vectors can be written as the sum of the components of the product. Just as a three dimensional vector of the first component second times i Multiply component j, k we say hit the third component. So we can write one line of such a collection. Let this process for all lines. Once A1 A1 indicator for the first line and the second variable and a vector with the collection on the second opening the same here, first let's take a good, Let k1 k is also such a collection. a2 first component of the first indicator 2 matrix elements of the a2, k2 is the first a2 NSAIDs çarpılmış and all the lines in this way I will create from this collection. As you can see, because there's one in every gathering the number one issue n² There k1 are the collection also went up to the one for 1pm. That means our number n² matrix of n rows, We produce n² in the number of columns in this way. Previously we found that generalization, that order of addition and multiplication We get out of this collection are using the feature does not change the result. So determinant for evac. We're getting out collecting coefficients out We get that the number of coefficients in the matrix. Inside, the only unit of the matrix, it's sort of sorry various vectors. How many can? We'll see a concrete example on this as well. n² the sum can be sorted. For example, if we take a case 2 of 2, 1 2 1 1 2 1 2 2 rows and four for coming When one falls to be repeated, but the same vector will be determinant 0. 2 of 2 means that the I1, I1 I2 falls, falls I2, I1, I2 remains, I2, I1 remains Therefore, the observed collection up from just n² number from 0 to n factorial is different. Now let's take this a little more concrete, We say for instance n = 2, the first may be the first. It would be 0. The second may be the second. Yet it would be 0. Look though was 1 1 1 0 1 0 Or, to be more generally will be determinant for a matrix with repeated lines would be 0. 1 2 See we receive the first line, the first vector, the first unit vector, the second unit vector in the second row. But as 2 1 1 determinant in this matter Make a direct determinant of -1, we find accounts. Or if we think in terms of a more general rule, replace the line we come to when we change vectors. Each displacement brought change to a minus sign and place a time As yet we come again to -1. Although we have done for n = 3, See here again, a unit vector, the second vector unit, the third unit vector. These 1, 2, as during 3 1, 1, 1, but could be repeated When the lines is 0 drops Collection. 1, it could be one third. 1, 1, 0 and 2 would be lowered if we back 3 factorial, 1x2x3 remains 6 units. This would also include 6-1,2,3 that sort of reputation. You are ordered as 1,2,3. As 2,3,1 or 3: 2, 1. So write the numbers 1,2,3 trigonometric direction while rotating, 1,2 1,2,3 2,3,1 or 3 of them + that gives value to the trigonometric direction in reverse order If you for example 1 3.2 this time in reverse order Or, it's relocation happens when you return -1 Repeat terms. See the following determinants J2 also a time to take the last line here Let's bring the second line was the first vector, the first unit vector. The third line was third. This is good progress. This time if we changed from 2 to 1 to place the first vector The first line of output, the second vector in the second row. So we were able to come to 1,2,3 well as modifying the two places. Because every change had a one- point -1 a more So we multiply by -1 -1 in will receive the second force. So we get the double force. It also gives +1. Let's change the first line to the second line to see the place again if you look at it. One came up, came to the third row and second row. Sign has changed. But we come to the second line to the third line you change the order of 1,2,3. Minus because it comes at a time but still well because we change twice It was +1. If you look here, if you change once final place If you put in place in the second row to take the line 1,2,3 Welcome to order. This determinant is negative for modifying place once you arrive in the order 1,2,3. We see them remembering where these vectors If the sequence consisting of either 0 turns are repeated, or double room for one turns out +1 or we can have the order changed with 1,2,3,4 The only place we can have modifying minus 1 off. This place is called the permutation change. This is called "permutation Icon" Or, "Transformation Icon" can say. It also ε (epsilon) are showing with. ε n to move in here one indicator k1, k2 ... kN the one going one indication. So we earlier that unit determinant determinants which all possible n of vectors n'nc force so many can see how very easily. 10 10 10th even force will be much larger in number but It does not matter. We are getting here has been reduced to a formula. He tells us the following formula; Get all product possible. Get all of the elements in the matrix multiplication. They either 1 or 0 and -1 with the ε-defined multiplying the framework of rules, taken totals. So it is quite a laborious job, but how immaculate formula I got to see also that in a formula. This theory but make it useful in terms of applications rather To simplify this a laborious wherein n for multiplying the n-1-one Let us assume, for example a1k1 to let out, Let's give him a new name to the remainder of the Ak1j say here j be here if we have what it takes k1, k2 consisting of its To specify the icon. A1j or the opposing Ak1j'y We're going back to that product; a1k1 A1k1 multiplication. These are nothing more than our already our cofactors. We already have such a formulation with a sense front. As a result, this is where our proposition now We arrive at the formula. This is what we call A1k1 A1k1 a1k1 i say get out, taking the total of the rest. These n pieces of this collection of n-1 a gathered together that consists of n-1 determinant here. So this is not something else from our cofactors. We reduces this dual product. This is because, according to the first line sets the first indicator lines. According to the first line expansion formula. Not a feature of the first row. According could consider opening any line. 1 i Let us instead. k1 is also a little sadeleştirel because now we do not need. The grain is used for the multiplier k1, k2, had to say kN. Let's try it because the jar. According to the opening line we find the formula before you get here. As there is only one fundamental difference, the first of 3 from 3 in this foreboding we We have seen the rules, even if the call is also available, though. However, where a basic mathematical maintenance We propose a robust emerges respectively. Four basic proposition, You can call the axiom postulated instead proposition might say. Accept them as we have had this formula. Here we find the initiative by line. Similarly, the column can be obtained by expansion. In the first, the first indication we keep constant. I gave the first indication for fixed lines, We are moving on to other items by scanning the jar on the i-th row. We keep the j fixed. J columns that define the columns for determining that the second indicator index, indicator, the first indicator variable i.e., mold on mold on the column i of one column j'nc n n'nc line up bringing up the first line, but j'nc column We get this product by staying on. Now, one for them to be able to internalize some of these We have already seen a little more clearly the general formula of the triple application but let's see. Our general formula was like here. Here's a short, short two, including here in all a kN We're getting slammed with typing a two minute and epsilon. See where the order is important, a one, a two, It gives us the chance to repeat the moment you get in the same line item. So here again a one of a calcined, It would be bumped to another with an item in the first row. Then it would give it zero permutation. So this way we have saved a bit of typing. n Instead of taking the shock force until n'nc Minimizing impact until we get to take factorial. For example, we have become the third will be reduced to three the number of Malaria topline. Epsilon's indicators will be three. A bit there will be a two to three. There is now also an indicator to determine the number of couples, get rid of them without defining j, k, m he can describe. Thus, on the one gathering the n j k n m on a one on one to n This is the formula that is becoming more streamlined as much. E we have seen before. Icon on the non-zero permutation, one two three one two three, reverse during the dizilmiş of it with a three-two. They are one, öbürki minus one. Here repeat repeated, though, see if we got an A, a merger, we are becoming one of the two bumped. a a a combined two elements in the first column. When they were going to zero. We did not get them. Others say that the only one here without repeating those two or three using, they're finding the groove six, we can write the determinant. See Epsilon has one two three one two three here. Other j, k, m also comes in a two or three. Here, when we get two or three a second one two three, A one, a two, a three had, is still two to three one coming. Epsilon has three one or two. a still a one two three first indicators, He reiterated his collection in a number of the epsilon others. As you can see they will bring a plus, a minus other three will bring. Determinant, we find the formula for the determinant of a matrix of three three-pointers. If you look at page fifty-three before that, it never works knowing, permutations and so are the determinants of knowing even knowing, three equations with three unknowns that way resolving spontaneously a determinant, we see that such a factor to be. We've had three triple here now using the general rules. We've had a hunch for a general case of three three-pointers in the first place. So we can say that as I have it all in. We tried to go towards a general rule the individual event. Here's terms of logic deduction. Because we have the general rule. We found there a formula. As a general rule we have given. We have four in the sixth with a proposed article. We also provide determinants of these four propositions. We have achieved a formula using them with this permutation icon. Here's three three-pointers in the matrix so by deduction We found DETEMINANTS before, we find the concept of elimination we have achieved with simpler formula. This is going to provide one. If we consider as cofactors that resembles something like, Let's consider an initiative based on the first row. Here we take out a one, a small one. Large air is also a dual collection jar. j fixed, wherein the j fixed, k, and we're picking on me, here we look a one to one, There are two compositions repeat see each other here, one may be two or three, or a third one could not be otherwise. A one though others will fall to zero. So this is just a collection of two triple plus one, three double and two minus one came two two three three three to obtain two multiplied by three. Yet they also before the screening we see that these same cofactor we found. It is also apparent that the work is not easy to formulate. If they have difficulty everyone is forced, them, when he saw the first epsilon work, but all these things here we are able to bring out the essence of such a formula, the permutation icon. Take the hit in the permutations permutations with all multiplications It allows debugging. Either zero or a're doing, once it is doing multiplication, minus doing it once or product. So we're going have found the path calculation. Now it's time to take a break here, I think. After we do the calculations in a practical way how he We'll see. This would return to the Gauss-Jordan elimination. I will say that the reason for the next session. If you want to settle accounts with this huge matrix computer Formula previous epsilonl was not easy to apply the formula. It requires numerous operations. However, we will see the accounts held with the Jordan elimination, We understand that we can find determinant calculate how much less.
