 So one of the reasons that linear transformations are so useful is that once we know what our transformation does to a couple of vectors, we know what our transformation does to many vectors. Of course, this knowledge comes at a price. Fortunately, the price is actually pretty cheap. The price is we have to solve some equations. So let's take a look at an example. Suppose I have a linear transformation and a couple of vectors and I know what the linear transformation does to those vectors. Given this information, I can try and find what the linear transformation does to some arbitrary vector. So let's plan out our approach. It might be useful to recall what a linear transformation is. And the thing we notice is that if I have a linear transformation and I know what the linear transformation does to a couple of vectors, then I know what the linear transformation does to any linear combination of those vectors. So in this case, since t is a linear transformation and we know t of u, t of v, and t of w, then we know t of any linear combination of u, v, and w. And that suggests that my first step is to try and express our vector 3, 7, 1 as a linear combination of u, v, and w. So I'll go ahead and set up my vector equation and then try to solve the resulting system of equations. And we can row-reduce, then back-substitute to find our coefficients. And after all the dot settles, we find that x1 is equal to 2, x2 is equal to 1, and x3 is equal to negative 1. So remember we were trying to express the vector 3, 7, 1 as a linear combination of u, v, and w. And so this shows that the vector 3, 7, 1 is 2u plus 1v plus negative 1w. And because it's a linear combination of u, v, and w, then the transformation acting on this vector is going to be 2 times the transformation acting on u plus 1 times the transformation acting on v plus negative 1 times the transformation acting on w. And I know what those are, so I'll substitute those in and find my result, the vector 1, 0. How about going from the other direction? Suppose I want to obtain the result of a particular vector. What do I have to start with? Once again, we can make use of the fact that we're dealing with a linear transformation and that once I know what the transformation does to some vectors, I know what the transformation does to any vectors at all. So we might start out as follows. We know that T of any linear combination of u, v, and w is going to be a linear combination of the vectors 1, 1, 0, 3, and 1, 5. And since we would like that linear combination to be the vector negative 1, negative 12, this means that we want to express negative 1, negative 12 as a linear combination of these vectors. And this gives us a nice vector equation that we can solve, so we'll set up our augmented coefficient matrix and solve it. Now we get a set of parameterized solutions, and so there's an infinite number of solutions, but we only need one. So we need to pick a value of T and substitute it in to find our values for a, b, and c. So we can pick any value for T that we want, so we'll let T equal 0, and this gives us the solution a equals negative 1, b equals negative 11 thirds, and c equals 0. Now it's important to remember what we actually found here. These values a, b, and c were what we needed to form a linear combination of the vectors 1, 1, 0, 3, and 1, 5 equal to our target vector negative 1, negative 12. However, what we actually wanted was the vector whose transformation would give us negative 1, negative 12. Well remember a, b, and c were the coefficients of the linear combination of the vectors u, v, and w that we applied our transformation to. And so that means we'll use these coefficients for our linear combination of the original vectors u, v, and w, and that will be the vector which, when I apply the transformation to it, will give us the vector negative 1, negative 12.