We left off last time asking whether we could find two numbers, C_1 and C_2, that makes this equation true. We would need C_1 times this column vector, plus C_2 times the second column vector, and we want it to equal this other column vector, 7, 4, negative 3. How do we do this? This idea, linear combinations in general is really powerful. We're going to use it for the next bunch of lessons going forward. You're going to definitely need to know this if you ever want to do more advanced math, more advanced data science, especially more advanced linear algebra. It is a foundational concept. How we do this is we say that in general, we can just multiply this out. We'll have C_1, negative 2C_1, negative 5C_1 plus 2C_2 plus 5C_2 plus 6C_2 equals 7, 4 negative 3. This is just a scalar times a matrix. We learned this when we originally learned matrix algebra. We can multiply a number times a matrix. It just multiplies that number by each entry. This is perfectly valid, so is this. Adding them is perfectly valid because we're just adding two vectors. We have this equals this. Generally, we have it in matrix form, but either one is completely fine. Now, this should look familiar. This is essentially 1, 2, negative 2, 5, negative 5, 6, 7, 4, negative 3. How do we solve something like this? Well, we pretty much just saw it a few lessons ago when we finished up the matrix module, which is Gaussian elimination. We want to row reduce this. Let's go ahead and do that. A little bit of a review never hurt anybody. I don't think. Excuse me. Let's take the first row times two and add it to the second row. New row 2 will be equal to old row 2 plus 2 times row 1. Hopefully, you've done enough practice that you can do these really smoothly. You don't need to be able to do two operations in one go, but maybe you've done extra practice and can keep up. If not, it's okay. You can just go back and do it step-by-step and hopefully you get to the same matrix. We're going to say new row 3 is row 1 times 5 plus the old row 3. We have row 3 plus 5 times row 1. We're only changing row 2, row 3. Row 1 will stay the same, 1, 2, 7. There's a little note here. If this pivot, right here is a one. I had said during the matrix module that when you set up Gaussian elimination, the easiest way to do that is to get a very convenient number in this pivot. One or negative 1 will be the most optimal thing you could do here. If it wasn't originally in this top row, I would have swapped rows many times as needed, most likely one, but again, if the five was up here, I probably would have gotten it in order. The matrix came in a nice order, one, then two, then five. If it wasn't in that order, I would have just swapped rows until it was, and then I would have started Gaussian elimination. New row 2 is equal to old row 2 plus 2 times row 1. Here, we're going to have zero. New row 2 is old row 2, which is 5 plus 2 times 2, which is 4. So 5 plus 4, we get 9. Then new row 2 was old row 2, which is a 4, plus 2 times 7, which is 14. So 4 plus 14 will give me 18. Now row 3 is the old row 3 plus 5 times row 1. Here, we're going to get a zero. That's the entire reason why we picked that five in the first place. Hopefully, this is coming back to you if it's been a little bit since you watch some of the older videos. Here, new Row 3 is old Row 3, which is a 6, plus 5 times 2, which is 10. 10 plus 6 is 16, and then new Row 3 is old Row 3, which is negative 3, plus 5 times 7. 5 times 7 will give me 35, minus 3 will give me 32. Now, I'm going to do two more operations at the same time. Well, let's do one operation first actually just to make sure we're understanding what's happening here. I'm going to multiply new Row 2 equals old Row 2 times 1 divided by 9. I'm not changing rows one or three. I have one, two, seven, and I have zero, 16, 32, 1/9 times this row will give me 0, 1, 2. Now I was going to do both last time, but now maybe it's a little bit easier to see what I would've done, which is if I multiply the second row times some number and add it to the third row, it's going to cancel out because these proportions are identical, one to two is the same proportion as 16 to 32. These rows, Row 2 and Row 3 are saying the same thing, these are equivalent equations. Is saying that 1 times C2 will give me 2, 16 times C2 will give me 32. What do I actually want to do? I want to say Row 3 and this is just to be rigorous about it, equals the old Row 3 times negative 16 times Row 2, plus positive 16 will give me 0. Negative 16 times this 2 added onto 32 will give me 0 as well. I'll end up with one, two, seven, zero, one, two, zero, zero, zero, which is completely fine. Let's finish this up. I'm going to want to get rid of this two. Remember, we want to get rid of the lower triangles. We want to keep the diagonals and get rid of the triangles on both sides of that. This two is in its own little triangle here. We want to get rid of that. Let's take Row 2 times negative 2 and add it. We're going to do new Row 1 equals the old Row 1. We want the old Row 1 plus negative 2 plus negative 2 times Row 2. That will give me, let's say Row 1 is only one changing, this entry right here won't change because Row 2 is zero here, so it'll just be a Row 1 equals the old Row 1. I kept that one, and I get to keep all of my other rows. Now, Row 1 in this entry equals the old Row 1, which is 2, plus negative 2 times 1. That will cancel the zero, and here Row 1, which is 7, plus negative 2 times 2. Negative 2 times 2 is negative 4 plus my original 7 will give me 3. What this is saying is, that in my original equation, we now have a solution for it, which is nice. My C1 equals 3, my C2 equals 2. If we want to know about linear combinations of vectors, if we want to know if there exists a linear combination of a set of vectors that equals another vector, all we have to do is transfer it into matrix notation and eventually from here, at this point, once we've gotten it into this vector notation, we can do Gaussian elimination and solve it, assuming that any solution even exists. Values of C1 and C2, in this case, they existed, but they don't always exist, so this is the way to figure it out if they exist and if they do what their values are.