Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Determinant of a two by two matrix A equaling the determinant ABCD is just AD minus BC, sometimes thought of as the major diagonal minus the minor diagonal, so AD minus BC. Notations are determinant of the matrix A equal determinant. Instead of writing the letter A, we could actually write the matrix. Remember, matrix is used with notation of brackets. A different notation is straight lines, and that actually tells us we're looking at the determinant of a matrix. So if we had straight lines of ABCD, that notation tells me I'm looking for the determinant Cramer's rule. Given two equations, a 11X plus A12Y equaling B1 and A21X plus A22Y equaling B2. Remember the 11 and the 1-2 and the two one and the two two really are just telling us row and then column. So first row first column, 1st row, second column, second row, first column, second row second column. So then the determinant of A is just taking the coefficients and then doing the major diagonal minus the minor diagonal A11, A2, 2 -, A two 1812. If the determinant of A doesn't equal 0, then there's a unique solution where we take this B1B2 and replace it for the X values. And find a new determinant. So now we have B1A22 minus B2A12 for the X value, and then we divide it by the determinant of A. For the Y value, we're going to take the original determinant of A instead of the Y value coefficients. We're going to put in what the equation equal B1B2. So the Y is just the determinant A11 times B2 minus. A21B1 all over determinant of A. This gives us a unique solution, minors and cofactors. Let A of Aij be an N by N matrix. The IJ TH minor of A is the determinant mij of the n -, 1 times N -1 sub matrix. That remains after deleting the ith row and the J column of A. The IJ cofactor or AI sub J of A is defined to be aij equaling -1 to the I + J power times mij. I'm going to show an example of that in just a moment. Cofactor expansions of determinants. The determinant of an end by end matrix A equaling A sub IJ can be obtained by expansion along any row or column. The cofactor expansion along the ith row is determinant of a equaling AI one capital AI1 plus the coefficient AI two times the matrix AI2 plus... The cofactor expansion along the JTH column is determinant of a equal aij plus the matrix aij plus A2J times the matrix A2J plus... Once again, I'm going to show an example in a moment. We have a bunch of properties. These are row and column properties for matrices who have greater dimensions than two, so N greater than or equal to three property one. If the N by N matrix B is obtained from A by multiplying a single row or column of A by a constant K, then determinant B equal K times determinant A. Basically, this just means that I can take a factor out of any row or any column property two, if the n * n matrix B is obtained from A by interchanging 2 rows or two columns, then the determinant of B equals the opposite of the determinant of A property three if two rows. Or two columns of the end by end matrix A are identical. Then the determinant of A = 0, property 4. Suppose that the end times end matrix A1A2 and B are identical except for their ith rows. That is, the other end -1 rows of the three matrices are identical, and that the ith row of B is the sum of the ith rows of A1 and A2. Then we can find the determinant of B just by adding the determinant of A1 and the determinant of A2. This result also holds if the columns are involved instead of rows. Property 5 if the end by end matrix B is obtained by adding a constant multiple of one row or column of A to another row or column of A. Then the determinant of B equal the determinant of A, and the determinant of a triangular matrix is equal to the product of the diagonal elements. Talking about transposing a matrix, so the transpose of the M by N matrix A equaling the elements aij. Is the N by M matrix, so we're changing the dynamic the dimensions. Instead of the row column, it's now going to be column row dimensions. AT is defined by AT or the transpose matrix of A equaling the elements aji. If the matrix A is square, then we get A, the transpose of the matrix A by interchanging elements of A. That are located symmetrically with respect to the main diagonal. Thus the transpose of the matrix A is the mirror reflection of A through its main diagonal. We've got a lot of properties for this. If we have the transpose of matrix A that's then transposed, we get back A, IE if we take a mirror reflection along the diagonal and then do the mirror reflection again, we get the original back. If we add 2 matrices and then take the transpose of it, that's the same thing as taking each matrices separately, transposing it, and then adding them together. A constant times the original matrix transposed is the same thing as the constant times the transpose of the matrix, and then the product of 2 matrices transposed is really just the product of the B. Transpose times a transpose. Once again, I'm going to show some of these properties in a minute. The last property is. If A is a square matrix, then the determinant of the square matrix transpose is the same thing as the determinant of the original matrix. The inverse of the invertible matrix A is given by the formula A inverse equaling the transpose matrix of Aij divided by the determinant of A. Whereas usual Aij donates the IJ TH cofactor of A, that is, iij is the product of -1. I to the I + J term and the IJ minor determinant of A. OK, let's look at some examples. So the first thing we're going to do is we're just going to find a determinant. So if we're given the original matrices 210121012, we're going to choose a row or column. Some of our key things we want to look for are rows and columns with zeros. The more zeros the better. So if we look here and we see a zero in the first row, we're going to do the determinant going across the 1st row. So the first row, first column, so -1 to the first row plus first column, one plus one, the number that's located there, so the two. And then we're going to actually eliminate the row and the column that the two is in and write a new two by two matrices. SO2112 And we're going to find its determinant, then we're going to add to it going to the next location. So now we're going to look at this location here. And this is the first row, second column. So first row, second column -1 to the first row plus second column, the number actually that we have in that location. Times the new determinant that doesn't include the row or column that that number's in SO1012, then we're going to add that to the next location. So if we look here, this zero is in the first row, third column, so -1 to the 1 + 3, the number that's in that location. And then we're not going to use the row or the column. And we're going to have a new two by two matrices 1201. So -1 to the one plus one is 1 * 2. And now we're going to do our diagonal 4 - 1 -1 to the 1 + 2. So that's going to be a -1 * 1 or a -1. Our major diagonal to minus our minor diagonal 0 + -1 to the one plus 3-4 times zero major 1 -, 0 1 -, 0. So when we compute this, we get our determinant to be 4. Now we could have used. A row or a column instead. So if I did 210121012, let's say this time I want to go by this column instead of the row. So we'd have -1 first row, first column. So -1 first row, first column times the number that's in that location, in this case 2. And then we're not going to use the row or the column that the Two's in, and we're going to get a new square matrix and we're going to find its determinant SO2112. Then if we add that to the next number in that column, so let's say this one here, which is in the 2nd row, first column, that's where this 2 + 1's coming from. The actual number that's in that location, which in this case is one. If we eliminate the row and the column that it's in, we get left matrices of 10121012. We're going to find the determinant of that square matrix. Plus, if we go to the next location, this zero is in the 3rd row first column, so 3 + 1. The zero's the number in that actual location. If we eliminate the row in the column, we're going to find the determinant of the new square matrix 1021. So then just simplifying it up -1 to the 1 + 1 * 2, the major minus the minor 4 -, 1, simplifying it up to -0. Major minus minor 1 -, 0. Simplifying it up, we can see we got four, so it didn't matter if we used a row or a column. Now, sometimes it's nice to use those seven properties to get the matrices to be easier and smaller. So in this case, my original matrix was this thing on the left. And I'm going to multiply row one by a two and add it in to row four. And what that does is it gives me a new row that says 0001. So now I'm looking for rows and or columns with lots of zeros. So I'm going to start with looking at this row 4. I can see that 00 and 0 is just going to give me zeros when I multiply them. So the only one I really care about is this one down here. It's in the 4th row, 4th column. So I'm going to have -1 to the 4 + 4. The number in that location is 1. And then I'm not going to use the row or column and I'm going to write a new determinant based off the matrices 2000111005. That I then can evaluate using a row or a column, and I'm going to use this row this time. So the two is in the first row, first column, so -1 to the one plus one, the number two. And now we're going to use the new determinant of the matricy that doesn't use the row or column that the two was in, so 11105. Now it's just a matter of the major minus the minor times the coefficients. So we're going to end up with 10. So now let's actually look at something that has some meaning beyond just finding a determinant, which is solving an equation. If I have 5X plus eight y = 3 and 8X plus thirteen y = 5, I can find the determinant or the delta of the coefficients, the 58813. So remember it's the major minus the minor, so 65 -, 64 in this case, which is one. Now to get our X, it's going to equal 1 over our delta. And then we have it times a new determinant, the new determinant, we're going to take out the X coefficients and we're going to put in what the equation equaled, so 35813. Now we have the major minus the minor, so 39 -, 40 that delta was one. So 1 / 139 -, 40 is -1 the Y is going to be one over the delta times. Back to the original delta instead of the Y coefficients, now we're going to put in what it equaled. So 35 major 25 minus minor 24. So 1 / 1 * 25 - 24 is one. So the coordinate that works in this ordered pair system is -1 one. So the linear equation has a solution of -1 one for the. Those ordered solutions, those two linear equations -1 one's our solution. We can do this for bigger matrices. This is where some of the real power comes. So if we take our coefficients 5, four negative 22032, negative 11, remember, I'm always looking for rows and or columns with lots of zeros. So we're going to use. The 2nd row, so the -1 second row, first column, that's where that 2 + 1 comes from. The two is the number that's actually in the location. We're going to cross off the column and the row and get a new matrices for -2 negative 11 and find the determinant of it plus. -1 second row, second column. That's where this 2 + 2 is coming from times the number in the location, which is 0. If we cross off the row in the column, we get a new matrix E5 negative 221 that we're going to find the determinant of. Plus -1 second row, third column, that's where this 2 + 3. The number in that location was the three here. And then if we cross off the row in the column, we get a new matrix and we're going to find its determinant 542 negative 1. So then we get -2 the major 4 minus the minor -2 The zero times anything is 0 and then major -5 -, 8. When I compute that, I get the determinant being or my delta, sorry, the delta being 35. So it is a determinant. It's just the determinant of the main coefficients, not looking at the 421. So to find X1, we're going to do the exact same steps, but instead of using the X1 coefficients, we're going to put in the 421, so 42140 negative one. Negative 231. I'm going to use the 2nd row again because it's got a zero in it. Oh no, I'm not. I'm going to use the column. Just trying to show a variety of methods. So if I use this column here, that four is in the first row, second column, that's where that 1 + 2 is coming from. The four is in that location. And then we multiply it by the new matrices determinant 2311. Remember, we don't use the row or the column the number is in. Plus this next number is in the 2nd row, second column. So -1 to the 2 + 2 times the number that's in that location, 0 times the new matrices that's not using the row or the column. So four negative 21 ones. The determinant of that matrix plus the -1 here, which is third row, second column. That's where that 3 + 2 is coming in -1's the number that's in that location times the determinant of the new matrix four negative 223. When we compute this, we get our coefficient of -4 here major minus the minor, so 2 -, 3 the zero and then the major diagonal minus the minor 12 - -4, we get 20. So what this tells us is the solution for X1 is just 20 / 35, the 20 / 35 or 4 sevenths. Now we can do the same thing for X2 and X3. This next problem we're actually going to find the inverse of the matrix. So when we want to find the inverse of the matrix, we're going to find the determinant of the original matrix. So in this case, I just chose the 1st row and I did the procedures I've shown previously. So the two and the -1 was in the first row, first column. So 1 + 1 * 2 times the new determinant of the new matrix -4 two negative 11, etcetera. Now the part that's new here is we're actually going to find the cofactors of each of the locations. So we're going to get a 11, so first row, first column, and we're going to have it be -4 two negative 11. Now we're not taking into account the actual number in that location. This time we're going to do our a 12, same thing, first row, second column, first row, second column. So we get negative 5221, and then 1st row, third column, so -5 negative 4/2, negative one. Now the coefficient's important, whether it's positive or negative. We're going to do that for all nine locations, and we're just going to find the major diagonal minus the minor diagonal in each location. Once we get all nine of those, we're going to put them into a matrix. We're going to get negative 2913, negative three, negative 4/2/12, negative 19, negative 8. So we're now going to transpose this. So AIJ transposed. So our rows are going to turn into columns. So instead of negative 2913 as the row, it's going to turn into the column negative 2913. Instead of -3 negative 4/2 is a row, it's going to turn into the next column. So basically what we've done is we've looked at the diagonal and we've made a mirror. So the nine and the -3 change positions, the 13 and 12 change positions, the two and the -19 change positions. To finish this up, The inverse is just this transposed matrix C times one over the determinant, which we found a moment ago to be 35 S Our A inverse is just one over 35 negative two, negative 312 nine -4, negative 1913 two -8. And that's a method defined an inverse matrix. Thank you and have a wonderful.