In this example, we're starting off with a set B, and we're told that this set forms a basis for our to let's label these two basis vectors as B one and B to for this sake of convenience. Next, we're going to introduce a vector X and said X equal to, let's say, negative two and one. Then, given this vector X, our goal is to find the cornet vector of X relative to the basis be that's given. So if we could determine what this is equal to, our work is going to be complete. But let's start by writing by definition, that this corn in vector would be equal to see one. Let's use a different color here. See one times c two. My eyes will close at the blue bracket. So this is the situation that we start with, and by the definition of a coordinate vector, see one times be one plus C two a times B two must result in the vector X, so that's by definition of having a coordinate vector. And now that we have a vector equation, notice that we can immediately skip to a an augmented matrix where the first column comes from B one, so it's one negative. Three. The second column of this matrix equation comes from B to, so it's too negative. Five. And we're augmenting with the Vector X, which is given to be negative 21 So going to this augmented matrix is going to allow us to solve for the scale. Er's C one and C two weakened by row reduction and I'll copy Row one, which is one too negative, too. Multiply this role one by three at the results of Row two, and we obtain one or excuse me 01 negative five. Then if you copy Row two down, which is euro one. Negative five. Then visualize multiplying row to buy negative two and added to row one we'll obtain one zero and eight. But this then tells us the following vital information the echelon form. Recalling that we've augmented with X and that this court column corresponds to see one and this to see to tells us C one is eight and C two is negative five. So we're ready to stay our solution. The cornet vector of X relative to the basis B is C one, C two, but it solved to be eight n negative five. So what we're saying is, if we put on a tear and a negative five here we have a linear combination that multiplies through resulting in this vector X.