 We can extend the use of transition matrices through the use of homogeneous coordinates. So ordinarily, we represent a point in Rn as an n-tuple, the point 5, 3 in R2, the point negative 4, 1, 6 in R3, the point 3, 1, 4, 5, 6 in R5. We can then use matrices to represent geometric transformations like rotations and reflections. However, the simplest of the geometric transformations is a translation where we move something from one place to another. Can we use matrices to represent translations? Well, let's try it out. Let's see if we can find a transition matrix T where T sends the point x, y to the point x plus h, y plus k. So our transition matrix gives us the coefficients of x and y in the formulas that allow us to compute the new values of x and y. So here, our new value of x, x prime is equal to x plus h and y prime is equal to y plus k. So we want to find coefficients that satisfy the formula x plus h is something times x plus something times y and y plus k is something times x plus something times y. But in these formulas there's no way to get the constants h and k and that means that no transition matrix exists. And that's somewhat upsetting because translations are the simplest of the geometric transformations. So what can we do about it? To be able to represent translations and all geometric transformations using matrices, we introduce homogeneous coordinates. The point x1, x2, and so on in Rn has coordinates x1, x2, and so on, comma 1. It's as if we've added an extra coordinate, but here's the important feature. This coordinate value is always going to be 1. Later on, in other classes and for other purposes we can actually even wave that requirement that this extra coordinate always have a specific value. But that's in another course. So let's see if we can use homogeneous coordinates to represent a translation. So a little bit of analysis goes a long way. We see that m acts on vectors with three components and produces vectors with three components. And that means that m takes vectors in R3 to vectors in R3. So it must be a 3 by 3 matrix. Suppose the entries of the first row of this matrix are a11, a12, and a13. Remember that every row of a matrix gives us the coefficients of a formula that tells us how to find the new values from the old values. So these are the coefficients of the formulas that act on the components of x, y, 1 to produce x plus h, y plus k, 1. Well, paper is cheap. Let's write down that formula. So we'll have a11 times x plus a12 times y plus a13 times 1, and that's supposed to give us x plus h. And so now we want to find the values of a11, a12, and a13 that will make this true for all values of x. In order for this to be true for all values of x we need a11 to be 1, a12 to be 0, and a13 to be h. And so this gives us the first row of the matrix. By the same argument, if the second row of the matrix is a21, a22, and a23, then these are the coefficients of the formula that give us the new value of y from the old coordinates. And so that gives us the formula a21x plus a22y plus a23 times 1, and that's supposed to give us y plus k. And so we can use this to find a21, a22, and a23. And so we find a21 is 0, a22 is 1, and a23 is equal to k, and there's the second row of the matrix. And finally, is the third row of the matrix is a31, a32, a33. Then we can write down a formula that will give us the new third component from the old third component, and we find that a31 has to be 0, a32 has to be 0, and a33 has to be 1. And these will be the third component of our transformation matrix.