 In a previous video, we introduced the following linear transformation, t equals r cubed to r squared given by the following rule, t of x1, x2, x3 is equal to x1 plus 2x2 in the first entry, x3 minus 3x2 in the second entry. We've shown that this is a linear transformation. We've also computed the kernel of this linear transformation. In this video, I want to consider the image of this transformation, t. What is the image? One could approach this by asking the following question. Let's take a specific vector here. Let b be the vector of 1, 2. If b is the vector of 1, 2, we could ask, is the vector b inside of the image of t? Is this specific vector inside the image? How would one actually make such a decision here? What we're trying to do is we're trying to solve the equation t of x equals b for some x. Can we find a vector x that when you plug it inside of t, you'll get out the vector b? Can b come out of this transformation? What is t of x? t of x is this formula right here. You're going to get x1 plus 2x2. You're going to get x3 minus 3x2. This is supposed to equal b, which is specifically 1 and 2. We considered a similar type problem here when we were considering the kernel of the map. The only way that two vectors could equal is that component by component, they are the same entry. The first components have to equal x1 plus 2x2 has to equal 1. x1 plus 2x2 has to equal 1. The second component has to also equal each other on the left and the right. x3 minus 3x2 has to equal 2. Then we get x3 minus 3x2 is equal to 2. You'll leave those two equations just like we did with the kernel. We see that this is a linear system. It's a non-homogeneous system as opposed with the kernel, which is homogeneous, but we could still solve it nonetheless. We could solve this using elimination. Last time we did substitution. I think I'm going to do substitution again because when we did it with the kernel, we found there was a free variable in the system of equations. We're going to see that's also the case as well. If you solve this by substitution, let's solve for x1 in the first equation and x3 in the second equation. We're going to get that x1 is equal to 1 minus 2x2 and we're going to get that x3 is equal to 2 plus 3x2. Again, in this situation, we see that x2 can be treated as an independent variable and the x1, x3, these are dependent variables in terms of solving the system of linear equations, which tells us that we could choose whatever we want. We could choose whatever we want for t. Let's just do something simple. x2, we can pick any number we want, t. Let's just make our lives easier. Let's pick it to be 0. If you choose x2 to be 0, that means x1 will be 1 and x3 will be 2. This is just a specific example, right? But let's look what happens to the transformation when you do that. t of 1, 0, 2. By the formula you get, you're going to get 1 plus 2 times 0 and 2 minus 3 times 0, like so. You just get 1, 2 again, that's great. What was so special about the vector 1, 2? We could have done maybe some arbitrary vector, right? What if b just had the form b1, b2? In retrospect, we can see with this transformation here, it's like, you know, I could always just set x2 to be 0. If you had just some generic vector over here, like b1, b2, you could always just set x equals 2 to be 0, in which case that would kind of tell you that x1 should equal b1 and x3 should equal b3. And so what have we learned here? First of all, we have learned that the vector 1, 2 is in fact inside the image of t, so the answer to that question was yes. But we also learned along the way that for any vector, right, so for all vectors b inside of the co-domain are squared here, for any vector b here, we saw that t of b1, 0, b2 will equal b1, b2, which is our vector b. So we've also now computed that the image, the image of this transformation is all of r2. Every vector in the co-domain is actually hit by this function because you can just kind of basically ignore the second component if you want to. Now, you don't have to ignore the component. You could also get away with setting x2 to be 1 or 2 or 3 or pi, if you really are feeling exotic, it doesn't matter. But we can actually get any vector coming out of this, out of this transformation. And so that tells us that the image of this function is all, all vectors in r2. And this technique can force can be generalized to any linear transformation. We can compute the image of any transformation using these techniques. We can decide is this vector inside the image of this transformation. It comes down to solving systems of equations. It's kind of interesting here that these questions about the linear transformation boil down to solving linear transformations. It ain't a quink-a-dink, it's called linear algebra. And we'll again, of course, talk some more about this in the future.