* So it's y-intercept is going to be at minus 2. We can kind of call the system as being overdetermined. There is no intersection of all three of these points.* So if we go plus 1/2, that's right there, and then the slope is minus 1/2. They all intersect the other two, but they don't all intersect each other in one point.The value of Legendre's method of least squares was immediately recognized by leading astronomers and geodesists of the time.

* So it's y-intercept is going to be at minus 2. We can kind of call the system as being overdetermined. There is no intersection of all three of these points.* So if we go plus 1/2, that's right there, and then the slope is minus 1/2. They all intersect the other two, but they don't all intersect each other in one point.The value of Legendre's method of least squares was immediately recognized by leading astronomers and geodesists of the time.

The reader may have noticed that we have been careful to say “the least-squares solutions” in the plural, and “a least-squares solution” using the indefinite article.

This is because a least-squares solution need not be unique: indeed, if the columns of and that our model for these data asserts that the points should lie on a line.

Of course, these three points do not actually lie on a single line, but this could be due to errors in our measurement.

How do we predict which line they are supposed to lie on?

So the first one is 2x minus y is equal to 2, the second one is x plus 2y is equal to 1, and the third one is x plus y is equal to 4. So A transpose A is going to be equal to-- We have a 2 by 3 times a 3 by 2 matrix, so it's going to be a 2 by 2 matrix. So what do we get, we get 2 times 2 which is 4, plus 1 times 1, plus 1 times 1. And then finally, we get minus 1 times minus 1, which is positive 1. And this'll be a little bit more straightforward to find a solution for. That's my first row operation that I choose to do, just because I like to have that 1 there.

So let's first just graph these, just to have a visual representation of what we're trying to do. Plus 2 times 2, which is 4, so we're now at 5.

The accurate description of the behavior of celestial bodies was the key to enabling ships to sail in open seas, where sailors could no longer rely on land sightings for navigation.

The method was the culmination of several advances that took place during the course of the eighteenth century: The technique is described as an algebraic procedure for fitting linear equations to data and Legendre demonstrates the new method by analyzing the same data as Laplace for the shape of the earth.

they just become numbers, so it does not matter what they are—and we find the least-squares solution.

So I've got three lines in R2, and I want to find their intersection. So let's find the vector x this is our least squares solution. A transpose looks like this, you'll have 2 minus 1, 1, 2, 1, 1. And then of course A is just this thing: 2 minus 1, 1, 1, 1, 1. And then we have 2 times minus 1, which is minus 2, plus 1 times 2, so those cancel out. So the minus 2 plus 2 is 0, plus 1 times 1, so we get a 1. Which is just 6, 1, 1, 6 times my least squares solution-- so this is actually going to be in the column space of A --is equal to A transpose times B, which is just the vector 9 4. So to find a solution, let's create our little augmented matrix: 6, 1, augmented with a 9. Actually, first I'm going to swap these two rows. And then let me replace my second row with the second row minus 6 times the first row. And then let me just put this in complete reduced row echelon form.

## Comments Solving Least Squares Problems

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