Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. A matrix is a rectangular array of numbers. The numbers are called entries or elements of the matrix notation. A sub IJ where I is the row and J is the column. Echelon matrix is going to be a format for a matrix that we're going to use to solve it, and this is going to be called Gaussian elimination. And in an echelon matrix, if there are any rows that are all zeros, they're going to be at the very bottom of our matrices. And if there's some that have zeros in certain locations, we're going to have a basically a diagonal of constants coming down and everything to the left of the diagonal is going to be zeros. The way we number a matrix or the size of a matrix is how many rows by how many columns. So if we look at this example, when I make my matrix, I'm going to have one negative 5/2 and 10/01 negative 7:00 and 5:00. This is going to be two rows by 4 columns. So the size of the matrix is 2 by 4. Now what we want to do with this is we want to figure out what a solution would be. So here we actually already have a diagonal and everything to the left of the diagonal of constants has got zeros. So here's my X1 column. Here's my X2. We don't have a third equation for the X3. So we're going to start by saying let X3 equals some variable, let's call it T. So then we can see that X 2 - 7 X threes is going to equal 5 or that X3 we're going to call to be T So we could actually put X2 as seven t + 5. Then if we go up to X1, we can say that X 1 - 5 of the X twos which are seven t + 5 + 2 of the X threes are going to equal 10. So if we compute this, we can get X 1 -, 30 five t -, 25 + 2 T equal 10. So I want to get X1 on one side by itself. So we get -33 T. So when we add it over, we get 33 T -25 added over we get 35. So what this does is it allows us to write the ordered triplet for these two equations. We only have two equations, but we have 3 unknowns. But there's a relationship between X1X2 and X3, and the relationship is 30 three t + 35, seven t + 5, T. Now there are more than one correct. There will be more than one correct answer. But traditionally we let whatever the unknown variables are be the T and then work backwards. If we look at another one, we're going to have this one being one -2 five, negative 3701, negative 323-0001, negative 4. So when we look at our diagonal of numbers here, we can see that there is no number by itself in that third diagonal location. So we're going to let our X3 be some variable. If we look down here, we can see that X4 is -4. So now our X 2 -, 3 X threes plus 2X fours is going to equal 3. So our X2 is just going to be 3T. This is a -8 we added over, so plus 11. So then our X1 is just right down here X 1 - 2 X twos or three t + 11 + 5 X threes -3 X fours has to equal 7. So X 1 - 6 T -22 + 5 T plus 12 = 7 X 1 is going to equal T -22 and +12 is -10. When we add that over, we get plus 17. So now we have 4 variables, but we only had three equations and that's OK because we can still write the solution as a relationship between those variables t + 17, 3T plus 11, T, -4. So we know that any T that we put into this will make these original equations true. Let's look at another one. This is going to change its directions a little bit because it's not currently in echelon form. So we're going to have to manipulate it around to make it into echelon form before we can do what we were just doing in the other examples. So we're going to have three, one negative 36271, negative 9, 250 negative 5 S The way we're going to deal with this is we want to start with the furthest to the right, furthest to the left column. And I want to get a constant number on the top or an integer, actually any rational number at the top followed by two zeros at the bottom. So one of the things we're going to do is we could start by doing row 1 minus row two. If I do row 1 minus row 2, that's going to give me a one a -6, a -4 and a +15. Now if I left the other two rows alone for just a minute, just so we can see what's happening now, I might take row 1 * -2 and add it to row 2. So I'm going to leave row one alone now. But I'm going to multiply it by -2. So 1 * -2 + 2 is going to give me zero -6 * -2 is 12 + 7 is going to give me 19 -2 * -4 is 8/8 plus one is going to give me 9 negative 2 * 15 negative 30 + -9 negative 39. The next one I might take. We could take row 2 and minus row 3. So 2 - 2 is zero, 7 - 5 is two, 1 - 0 is one -9 - -5 negative 4. Now, there are lots and lots of things we can do here. This is just one possibility. So now we could divide this all by 19, but that's going to give us some icky fractions. So instead, maybe what we'll do is we'll take, we're going to leave that top row alone. If I thought about 19, if I thought about 2 * 9 would be 18. So if I took 19 and I subtracted 18, that would give me a one there. So let's take row two and subtract 9 row threes. So we're going to get zero and we're going to get one. So we get 9 - 9. Nine -9 would be 0 negative 39 -, 9 * -4 so -39 + 36 would give me a -3 there. So now we need to keep going. I need to get the next column over. I'm looking for my diagonal of ones or diagonal of constant numbers, but everything to the left of that has to be zeros, so I need to get rid of this too. So to get rid of that too, we're going to leave the top two rows alone, and we're going to now multiply the new row 2 * -2 and add it to row 3. So 1 * -2 + 2 is going to give us zero 0 * -2 + 1 negative 3 * -2 six plus -4 is going to give us 2. So now we can see X3 equal to X2 equal -3 and X 1 - 6 * -3 - 4 * 2 is going to equal 15. So now we get X 1 + 18 -, 8 equal 15, so X1 is going to equal 5 S This relationship five negative 32. We have 3 equations with three unknowns, so we found a unique solution for that five negative 32. The method we're using right now is called Gaussian elimination, and what Gaussian elimination says is we get a diagonal of numbers with everything to the left underneath the diagonal being 0. So looking at this next example, I went ahead and wrote the matrices, the three negative 611315, etcetera. We're going to take the row 2 and realize that entire row 2 is divisible by three. If we divide that all through by three, we can see that we're going to get some 1 coefficients in the very first spot, which is going to be a nice thing. So now we can interchange our rows. So I'm just going to make row one and row 2 change frequently. We write that with arrows on both ends to just show that we're interchanging the two rows. So one negative 217 and seven three negative 6113152 negative 452623. So now I'm going to take row 1 and multiply it by -3 and add it to row 2. And then I'm going to take row 1 and multiply it by -2 and add it to row 3. So one negative 2177 is going to get left alone. If I think about -3 + 3 is zero -3 * -2 is 6/6 and -6 is zero -3 + 1 is -2 negative 3 * 7 is -21 negative 21, and 13 is going to give us -8 and then -21 and 15 is going to give us -6, we can see that that entire row is divisible by a -2 SO1 negative 217700143. Now row 1 * -2 plus row 3, so we get 0. We're going to get 0 again -2 + 5 is going to give us three. Negative 14 + 26 is going to give us 12 and -14 and 23 is going to give us 9. We can see that row 3 is totally divisible by three, so we get 00143. If we look at our bottom 2 rows, we can actually see that those are identical. So if I said let's take row 2 and subtract row 3 from it, we'd get one negative 217700143 and 00000. So we're trying to get a diagonal with everything to the underneath and to the left being zeros. So we can see that there was no one for our X to the 4th, so let's call it T There was also no single one in the diagonal for our X2, so let's call it S OK, so our X3 plus 4X fours or four TS is going to equal 3. So our X3 is just going to be negative 4T plus three. Our X 1 -, 2 X twos, which we're going to call s + 1 X 3. And we just found our X3 to be negative 4T plus 3 + 7 X fours, which were T is going to equal 7. So now if we solve for X1, we get -2 S minus four t + 3 + 7 T equaling 7. So X1 is going to equal 2 S -3 T +4. So we're going to put in our relationship between these points and we're going to have two S minus three t + 4, S, negative 4T plus 3, T Remember, with variables, we usually put them in alphabetical order with the constant at the end. For this problem, the directions are going to change and it's going to say we want to figure out what the K is for a unique solution, for no solution, and for infinitely many. Now they may not all be possible, but we're trying to figure out what K could be for each of those. So we're going to start with our augmented matrix 32116 K 21. Let's take row 1, multiply it by -2 and add it to row 2. SO32 and 11 zero K -, 4 and -22 + 21 negative one. Now for it to be a unique solution, we need some number here other than 0. If it was 0. If we had 00 negative 10X, 0Y can never equal -1. So if K - 4 equaled 0 or if K equaled 4, there'd be no solution. But if K equaled any number other than 4, five 5 - 4 would say 1 Y equal -1. If K was 2020 - 4, it would say 16Y equal -1. So if K is anything that's not equal to four, we're going to get a unique solution. And in this case, we don't have any a possibility of getting infinitely many because there was no way to make that bottom row turn into 00 and 0. Let's look at one more of this type. If we have 321, 75 and K So if we take the row 2, actually let's do row 1 * -2 and add it to row 2. So we'd get 321 left on the top for a moment, we'd get one -2 * 2. Negative 4 + 5 is 1 and K - 2. So now if we took row 2 and multiplied it by -3 and added it to row one, we would get zero -3 + 2 negative one negative 3K plus 7 11 K -2. Now I can exchange the rows. So let's exchange row one and row 2. Remember, there's lots of ways we could do this, this just happens to be the one I'm seeing. Oh, and I'm going to divide that row 2 by a negative SO. For a unique solution, no matter what the value of K is, we're going to get a unique solution because we have this diagonal. So we can always see that X is going to equal. Actually, let's start with YY is always going to equal three K - 7, so X + 1 three K - 7 is going to equal K - 2, so X is going to equal X + 3 K -7 equal K - 2. So X is going to be negative 2K plus 5 S -2 K plus 5, three K - 7. That relationship, no matter what the value of K is, will always hold for this particular solution set. Thank you and have a wonderful day.