Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Reduced row echelon matrices. Each leading entry of the matrix is 1 and each leading entry is the only non zero element in the column. Also, every all zero row lies beneath every row that contains a non zero element. The leading nonzero entry in each row lies to the right of the leading nonzero entry in the preceding row. Basically we're going to get a diagonal of ones, but it may not be a straight diagonal, It may be kind of like a stair step diagonal. The Gauss Jordan elimination. We're going to transform the matrix by the Gaussian elimination. Recall the Gaussian elimination means that we're going to have everything to the left of something matrix wise having zeros in it, something like that with other numbers up here. So what's going to happen with the Gaussian Jordan elimination is we're going to get all these ABCDEFGH to turn to 0 and we're going to get that two to become a one. So we're going to divide each row to have a one as the leading entry and then we're going to proceed to eliminate any non 0 numbers in the column. This matrix will be unique by the Gaussian elimination. It was not necessarily a unique matrix, but this one. There will only be one possibility when we get done computing things. A linear system of equations has either a unique solution, or no solution, or infinitely many solutions. Those are only three possibilities. A homogeneous linear system of the form A 11X1 plus A12X2 plus... Plus A1 NXN equals 0, where this first one represents the row in the second number represents the column. SO11 means first row, first column, first row, second column, second row, second column, MTH row, second column. If all of these terms equaling equals 0, then we can see that every homogeneous system has at least the trivial solution of X1 equaling 0 and X2 equaling 0 and X3 equaling 0, because 0 + 0 + 0 = 0. Thus, this kind of a system has either one solution, the trivial one, or infinitely many. Every homogeneous linear system with more variables than equations has to have infinitely many solutions. The reason is we're going to let one of those unknown variables be T or S or some new variable, and then all the other locations, all the other XS would be in terms of it. So when we look at 37 negative 1528, what this equation really might look like is 3X plus 3X1 plus 7X2 equaling -1 and 5X1 plus 2X2 equaling 8. So what we've really done is we've just taken the coefficients on the variables, and we've also taken whatever it equals down here on the right and put it in a matrices. So our goal is to get it in the form where we have 1001 with some numbers at the end, because then we could see that 1X1 would equal A and 1X2 would equal B. So to do this problem, we're going to take row 1 and multiply it by two and subtract row 2. So 3 * 2 is 6 - 5 is 1 two 7 * 214 - 2 is 12 negative 2 - 8 negative 10. So now we've got a one in a column, we want to make all the other entries in that particular column go to 0. So now we're going to take row 1, multiply it by -5, and add it to row 2. So 112 and -10. When I multiply row 1 by -5 and add it -5 + 5 is 0 -60 + 2 is -58 negative 5 * -10 is 50 + 8 is 58. Now remember, our goal is to get 10 and then we're going to go to the next column and I want a one right here. So now we're going to take row 2 and divide it by -58 1/12 negative 10 01 negative 1. So now we're going to take row 2 * -12 and add it to row 1. So 1 zero -12 * -112 + -10 two. So the solution is going to have our first variable being 2 and our second variable being -1 So 2 negative 1 is going to be a unique solution for this system. If we look at something like 5/2 negative 594 negative, 741 negative 7, we're going to use this to build upon how to solve. If we thought about these three, this could be one of two situations. These three could be variables where we don't currently know what each of those equations we're going to equal. And in reality, this is the one that is most standard. And what will happen is once we understand this matrices and come up with the 100010001 format, then we're going to be able to interchange the AB and C and make them anything we want. So we're going to do the Gaussian Jordan elimination here. The ABC may change based on the criteria of whatever we're testing. So our first concept is we need to get a one in this upper left corner. So let's do row 1 minus row three, 5 -, 4, one, 2 -, 1, one, negative 5 - -7 two. We'll leave the others alone for just a minute. Now we can see we have this one here. So now I need to get zeros in all the rest of the columns. So I'm going to take row 1 and multiply it by -9 and add it to row 2. And then I'm going to take row 1 and multiply it by -4 and add it to row 3. So we have 112. If I multiply by -9 and add -9 and add -18 and -7 is going to give us -26 -18 No, it's not -18 and -7 how about -25 S here -4? We're going to get zero. We're going to get -3 negative 8 and -7 is going to give us -15. So now we're going to take row 2 and divide it by -5 and we're going to take row 3 and divide it by negative 3112 015015. If we took row 3 minus row 2, now we'd get 112015000. Now this would be Gaussian elimination, but we want Gauss Jordan. And Gauss Jordan says that I need everything else to be zeros above in a column and below wherever the 1 is. So our first column was OK because we had 100, but this one is the next column over is not OK. I need to get the one gone in the 1st row second column. So we're going to say take row one and minus row 2. So when we do this, we're going to get 102 minus five -3 015 and 000. This is now a unique matrices. It's the only one for the simplification. So now we can see the fact that the last column is all zeros. We can see that one is five and one zero -3. So for this one, this would be our unique matrix. Let's look at another one. One -4 Negative 2/3, negative 1212, negative 85. The first thing we're going to do is we already have our one here. So now we're going to take row 1 * -3 and add it to row 2 and row 1 * -2 and adding it to row 3. So we get the first column being left alone. Row 1 * -3 so -3 and 3 positive 12 and -12 6:00 and 1:00 009 This one is going to be a no solution because when we look here 007 and 009 we're going to not be able to do. We can't get a 0 here and have a diagonal of ones. So this is going to be no solution. Now, we could divide row 2 and row 3 by 7 and 9 respectively, but the problem is that even though those last two rows are the same, I can't get my diagonal of ones with zero everywhere else. There's no way to eliminate that -4 Thank you and have a wonderful day.