[BOŞ_SES] Hello. In the previous section we calculate the determinant of a square matrix. These determinants are going to find the inverse of the matrix as an application. Lower similar formula will be used up. Inverse matrix just a matter of paying attention more on tactics. So calculation of the calculation of the inverse of the matrix determinant The parallel secret, a practice that benefited from it. We have found the Laplace expansion formula. That is a determinant of cofactors, as shown by this great, cofactors were being obtained by multiplying the matrix elements. We could do this in two ways. As seen here, I do not do an operation on a constant i. No collection. J collecting on it. The first indicator line dictated. So we keep a fixed line. We stood by this item to proceed in line with the opposing cofactor. So according to this opening line expansion. J holding constant time like this, The second indicator that we're holding steady. The second indicator showed columns. This is determined on the columns along the lines i staying in the same column Or, the opposite stroke coming out, how is selected. This shows the opening of the column. I'yinc the veja the matrix where aij, i'yinc j'yinc leaving the line and column, removing, reducing the resulting N minus one times N minus one determinant of the size of the matrix. We call them the determinant of the sub-matrices. They are also the order of rows and columns selecting as one minus the sum of the forces that are double the total We find değiştirmeyip only by changing the sign of the cofactor. We also can calculate this by multiplying determinants. As you can see here, from the opposite on the same line cofactors or cofactors opposing same column. i and j are coming to the same. We keep constant one, but if we change the other one said, but, in another row with the elements of a line cofactors to hit the'll see it come easily at zero. I can say right now. Cofactor, this fine was calculated by taking the i-th row and throwing j'yinc lines. This k'yınc line. E k'yınc line elements of the i-th row there. Because we find the cofactor of a different line than we were this time i. Therefore aija on the bus, it's in the cofactor. Therefore, there will be zero determinants calculated. Because that contains the same rows It will be determinant. That two lines will be equal determinant of a matrix, which is zero. Something similar to columns. Therefore, immediately thought of these two very simple formula other We get them both. We will show this with an application. Let's see these applications. Matrix 1, 2, 3, 4, are formed. This sub-matrix are as follows. 1, the first row and first column of four staying in one sub-matrix threw it back. We are writing four görütüp here. The first line of the second in column 2 instead of the future cofactor 3pm. Because of this we are taking the first row and the second column. He came here 3. Similarly, this is going 2 be the future instead of the 3 sub-matrices. 3 We take the line and column found in only 2 remain. We are writing here after. Finally, the next sub-matrix instead of 4 4 found that the column and row threw one remains. A number of unions in a matrix of the determinant of this. Now it had this cofactor matrix. Sorry determinant of the matrix of the matrix was lower. Tell me a little bit longer. Here we find the cofactor plus or minus signs of changing. The practical side plus because the first diagonal a line number when we started, a number of columns. Total double number sign does not change. Mark a changing in the next. One line down again after reaching 2 lines as well as one column, the total number only. Here we change the sign. 1 remains the same and the mark. Because the line as 2, 2 rows of columns, double the total number. It does not also change. Now let's write you look after them. The first line (1,2) is made from. The second line (3,4) above occurs. Cofactor A1 we create a vector 1 and from 2 to 4 and minus 3 opponents. Similarly, in other words from the opposite 3 and 4 cofactor line 2 and the opposing second line minus one. See them multiply. I multiply these two vectors, let the domestic product. Minus 1 plus 2 times 3 times 4. 1 2 1 comma 2 to 4 times we stood virgin minus 3 we take domestic product. We found minus 2. Similarly, by opening the second line, This second line is multiplied by the elements of the second row of the cofactor. When we hit them in the vector minus 2 3 4 components, 1 components of the vector gets hit minus 6, minus 2 plus we find 4 again. As you can see from the opposing line matrix cofactors have found hit the determinant. Now let's do the opposite. So the cofactor of the second row to the first row of the matrix multiply. 1 2 The first line again but the second cofactor line, The second line of cofactors minus 1 1. As you can see this when we made the 1.2 vector multiplication minus two, 1 inner product of the vector gives us zero. So different in that matrix, s different in that we find the cofactor gets hit zero. Also see the second row of the matrix 3 4 wherein the second line of this vector, multiply the first row of the cofactor 4, minus 3. As you can see in this hit is multiplied by 4 to 3, 4 to minus 3 are multiplied, the product of these two vectors, this gives zero. We have now seen the brew in the supply of these formulas. Cofactors in the same line of matrix elements we find slammed determinants. We found different sequence of remembering the hit zero. This is a formula of providing the visualization of the we have studied. Let us repeat this formula now. Aija with the EU, when we hit big aija are Want to gather over my lines, i collection jar on the stationary, You want to gather on columns, I fixed on collection jar. Gather we find zero or remembering the different rows in them. Now this We can reduce the formulas more simple. See the opening on the line we take this aijkj'y When either data or data determinant zero. Determinant to define it that we have to hit it with a number matrix i Let a time k is equal to zero when there are not different. This unit is going matrix. As it's reached the equivalent of a symbol: a time where i and k are equal, when they are different is zero. Similarly, in the opening according to the column you have the same situation. See, I have collection on here. These are the same but different j columns so different. When we hit zero different columns, receiving, we find, If that is the count of the column when we hit the same column We find the same in the kofaktörl has hit determinants. We find different queues gets hit zero. Again, therefore, this jm'y wherein when an equal, This same situation is happening in Ik. This means that we are actually four formulas have been reduced to one. If we bring them to see the structure of matrix multiplication here are the product of two matrices. Matrix multiplication case were as follows; Meanwhile the rest were repeated. With a circulation of something with j j and got kbps. j and this had taken kbps to bring in the large matrix transpose, then we get the transpose. This gives us the times the unit matrix determinant. This tidy display of, we show the components of a holistic matrix separately, A matrisiyle kofaktörün transpozesi, the product of the unit matrix transpose gives us determinants times. Here is the reverse order. • See the j I'd like to bring out that i have, here again as the great A'la Should we change the order we need to bring this to transpose. It did little with the times to transpose times IM See by hit middle has remained the same. Therefore, once this is still a determinant or a zero matrix so that the matrix unit revenue. We can write it in the same way, but you look at them in these two equations transpose of one another. For example taken transpose the first equation. During the inverse of the transpose it sounded. Transpose the transpose of a matrix, so the lines will turn columns, It turns down once again come to the same matrix. Is the transpose of this means that cofactors Or, if you change it comes to the same thing here as well. Unit matrix transpose it anyway because it's symmetrical it is equal to itself. Both of these formulas are valid and therefore the two It gives the same result by row or column. This summer it tidy, once we have achieved this to work cofactor is obtained. Now we're getting out of here immediately following; Once transposed inasmuch as the cofactor Once the unit was determinant matrix, that after the merger cons I'm so in reverse with the left multiply multiply, because it comes out right in a number of these determinants. To multiply the inverse matrix of the unit, the unit matrix if you hit what hit a matrix, a matrix of neutral neutral. It does not change. Unit times the inverse matrix, the inverse is staying. A number of determinants. This side consists of the unit matrix. The opposite occurs with the hit of the unit matrix. Just staying here cofactors. When we divide it determinants, as well as see if a changing cofactor determinant divided by here böldük decisive. In times cofactor cofactor already stayed here a minus. If you read it backwards; that is to say the inverse of a matrix It is obtained by dividing cofactor matrix determinant. And here comes one more thing immediately added. Both way, we find the determinant calculation, as well as when we find to be a reverse. If determinant is zero if there is no minus 1. Because you will divide the cofactor matrix to zero. E indicates that the inverse of this matrix. If you're still here, if this is zero, the more you look ahead, You will not find that time is missing because there will always contend with zero length. Here you will also bring division by zero. Reset cofactor transposed A minus here means that if you hit a matrix if ther It is the determinant of the inverse of the matrix We understand that one size determines that it is not. Again, characterized in that a determinant. To the already determined the origin of the word, for it is the means to identify, and only in the inverse matrix. If this process were the determinant is nonzero faint found only one inverse matrix. If there is no inverse matrix determinant is zero. This variety is also called singular matrix matrix. Now let's review briefly this method of calculation. We're doing that; This is our formula, We calculate the cofactor matrix, we get the transpose, We take a controlling stake in the determinant of the transpose. As görselleştirel. Matrix given. Sub-matrix of this matrix will account. M ij are. Do it as follows; Me and j'nc to say I find ij'y ith line item where the item in the column opposing j'nc n minus one, The minus unity we find the determinant of a sub-matrix. So this column will j'nc column and discard the i-th row. When we see him at minus one here j j j plus one are jumping. That is a no 1j. J'li are not all. Similarly, the i i minus one plus one is omitted. So no i-th row. Here are a minus n minus the wheel of a matrix. It gives us the determinant of M ij'y. This M ij of the individual elements of those marks We obtain the matrix by changing j. We call cofactor matrix. We take this cofactor transpose, we are getting overturned. Determinant, change the column with lines We find the opposite to when we split. To the contrary it is available where? For example; we had a numerical equation. Ax is equal to b. What did we say? x equals b, we say to the split. When the matrix is of course no such thing as a department, we have the opposite, minus one. If we multiply this equation from the left by a minus; Minus once A. This unit data matrix. Once the cons on the right side b. It remains. According to give the identity matrix times the identity matrix minus one times x, feature of the unit matrix this already, multiply what you hit the a matrix, a matrix of neutral neutral. Vector Clone, giving again without change. The reverse circulation with the x's as b multiplied. So, if such a place as the only unknowns single equation Ax b is equal, including the number, we were hit with the back of the x. We obtained by multiplying the inverse matrix like that. Example; again, one, two, three, four Let the matrix. We have seen that the lower matrices, we find the cofactor. As we do this, the sub-matrices. Where one of the columns from the opposite line and threw four, also threw two opposing rows and columns found that two of the three, three where three of the rows and columns found to be similar to that of threw two is staying, all four lines also found that two of the four location you're writing, When I opened a column. Using the determinant of the matrix obtained by this sub-matrix. To do this cofactor plus signs, minus a change. plus; minus, minus; It comes as a plus. Because here the number of rows, as a line, A double row of columns when we collect a number of them. Minus number one because of the relative strength of the couple's sign does not change. But we have come to this as a line three times, Two rows of columns, plus two; because the three single number minus one minus sign comes from the third force in the same way as others. Determinants we can find right now. The first line of the cofactor We found just hit the first line of the determinant. Otherwise we found it. The first column to the first column of the second line, second line to the third line, all give the same value. We find any of them. It is cofactor expansion. We're getting ousted the cofactor. We are changing the numbers on the diagonal others where the same mold. And it split the determinant by cofactor that he also was -2, we find the opposite. We split 4 -2, -2. -2'yi -2'ye Böldük 1. -2, -3, We split it into three split supply. 1, we divide to -2 minus one half. Now we find here on this question. I wonder if we did the right thing because our account. A'yl to minus 1 hit the unit should give Mattis. Similarly minus 1's AI also gets hit again should give the identity matrix. In fact, we wrote this order, wherein the A. Minus 1, minus we found one. These are indeed hit the first column we received, we hit. -2 It was +3 1. We got the first column we hit the second-line -3 divided by 2 plus, see here are 12 divided by two 6. -6 Plus 6 gave 0. Similarly others, minus one forward similar this type of product from the unit off again martisor back to writing. If we right side of the equation 2, That we take the number 4 b Minus 1's were going to hit the Find x that b. We found this really challenge multiplied by minus 1 here. If we multiply by b minus 1 after 2, 4 and 0, we find one of x1 and x2 values. Indeed we put in place x1 0 x2 2 equals 2 instead of 1 we put out. The second equation 0 instead of x1, x2 instead of 1 Putting x1'l term that will fall to 0. 1 4 4 times seeing that we provide equal equation. 3 of 3 for x has been made While we're getting the matrix again. We find the determinant of the matrix of 2 x 2 matrix of an opposing it. Minus 20, minus 30 Pardon, 5 times 6, 30 plus 16 -14. In this way we achieve this matrix M. -2 To -2, where the milk line and threw back 12 minus 4, going 8. In this way, the 3 x 3 sub-matrices, we find the matrix. We need to change the signs for cofactors. The same sign has changed, it has changed, it has changed. Plus or minus, plus or minus, we find increasingly cofactor. The numbers on the diagonal ousted cofactor others remain the same, changing place, according to the diagonal. Around here as if you reversed the draw this line, these around something like a hinge comes in when you return it to the place. To find the determinant of the cofactor in the first line to the first line for example, it çarpsa was -14 plus 16 + 2, minus 3 times 1 -3. So was the determinant -1. We find the AI. So, this is the inverse of the transpose of the cofactor He's dividing the determinant. This gives us the inverse matrix. And we can do to provide it. That's when we hit on the right minus 1'd this, I recommend that you practice hitting a hand. Take the first column, we stood deposit. 3 plus 14 minus 16, it was one. We turn first column stands. 28, 14 times 2, 28, 5 times 8 40, 12, 4 times 3, 12, 28, plus 12, 40, minus 40, 0. Likewise, others are obtained by calculating the unit matrix. Again, if you do the same again in reverse order unit matrix multiplication You can easily see what you find. Although we use the equations that, for example the right side of 0, 1, 2 Even though we have received, martisl to b to hit the opposite, We find here the reverse Martis, 14, -8, 3 and so on he goes, right side, b 0, 1, 2 taken by, has hit the -2 and -1. So, x1, -2, x2, x3 -1 and 0. After we put them in place of the equation, the equation of time x and here, to the times x, plus minus 2 x1 x2 x3 x1 is 0 and 2 times, Plus 4 times to 5 times x2 x3 The second element is equal to b here again. Third equation x1 minus x 4 plus 6 times x3, this is the third equation. His right side 2. Indeed, x1 -2, -1 x2, x3 If data is also 0, x3 fell here. x1 minus 2, minus 2 was minus 1 to plus 1 gave here. Minus 2 plus 1, 0. Similarly, other equations are also provided. Now I want to take a break here. We can find the same inverse matrix of the Gauss-Jordan elimination. We also see it as better to handle, There was a determinant factor accounts mainly cofactors way. You can see it in three different ways: Formula bump with cofactors Or, come to recognize that the business proposition but in the end the same way. Calculations with cofactors. The second way in Gaussian It was determinant in calculating qualifying. We have seen what the determinant as it plays an important role in finding the opposite, We have seen how important a role the cofactor. Here likewise determinant As to calculate the inverse matrix to find the Gauss-Jordan We will see that the important role of screening.
