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A homogeneous system of twelve linear equations in eight unknowns has two fixed solutions that are not multiples of each other, and all other solutions are linear combinations of these two solutions. Can the set of all solutions be described with fewer than twelve homogeneous linear equations? If so, how many? Discuss.

the set of all solutions be described with less than 12 homogeneous linear equations.

Calculus 3

Chapter 4

Vector Spaces

Section 6

Rank

Vectors

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this question. They ask us to find out how many equations we need, actually, in order to solve the system of equations. They gave us a system with 12 equations, eight variables, large, right? And so that, Solutions said, has eight variables in it as well. And they tell us that it's a homogeneous equation, that which means that it's going to equal a zero vector. So that just means 12 zeros, 12 components all equal to 70 12 rows of zeros. No, uh, this will tell us that there's within this really only two vectors that matter there to independent fixed solutions is the way that they weren't it. Essentially, what they're saying is that any answer is just gonna be some kind of linear combination of these two vectors. You can scale them if you want to. You know, essentially, what that's telling us is that no matter what we do, we have two vectors that matter. You can also jarred out like this. Now you go back and look at this coefficient matrix. You might remember something called reduced the row echelon form, and that's the end up, having the ones with the every single other component in that Colin as a zero. Now, this is a goal of reducing Westerlund form, essentially to isolate one variable on recall that pivot variable. The really. The answer to this question is going to depend on how maney pivot variables we have. If you recall sometimes year, uh, simplifying these and reducing them on the end up with these entire rows of zeros where there's no pivot variable, there's no useful information, really, what we're saying they're these just do not matter. And so that's why it comes down to the number of pivot variables. It's going to tell us how many equations that we need now rank therms helpful year. The rink theory says that the knowledge et of a plus the rank of A is going to equal. And now and, of course, is the dimensionality of our answer. That, for example, if we have you can look at standard form. Here are answers for these kinds of questions is going to be a two dimensional kind of looks like that X and y two dimensions. To use our end right in this situation, we're gonna have something like a h maybe, And those are gonna be eight dimensional. Further down in your book, you'll see that the nullity of a it really is just referring to the number of free variables, the rank of Ayer's air referring to the number of pivot Very bolstered. Remember, that's what we're interested in. So now, if we could just find the number of free variables will be very close to her. Answer. Now, I call that they gave us to fix solutions. Right on. There's these two numbers we could use to scale these These two actually are what are important here. These they're called for you. Variables. They're called free because you can put in any real number for these. Multiply at times the vector. Um, and you're gonna end up with some answers that some combination of these two that is also a solution. And so if we think back to this matrix here you have eight variables. They're really only two options. You can never be free or pivots. And that's essentially the same thing that the rank three missing saying we've got to free variables how maney pivots variables do we have if we have the total eight variables advancers six. That means are reduced or echelon form is going to look like that. Got six of these pivot variables of the rest of these equations air zeros or don't have pivot variables. They don't matter. All you need is the six equations.

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