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Find the general solutions of the systems whose augmented matrices are given in Exercises $7-14$ .$\left[\begin{array}{rrrr}{3} & {-4} & {2} & {0} \\ {-9} & {12} & {-6} & {0} \\ {-6} & {8} & {-4} & {0}\end{array}\right]$

$x_{1}=\frac{4}{3} x_{2}-\frac{2}{3} x_{3}$$x_{2}$ is a free variable$x_{3}$ is a free variable

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 2

Row Reduction and Echelon Forms

Introduction to Matrices

Missouri State University

Campbell University

University of Michigan - Ann Arbor

Idaho State University

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all right, so just all of this problem. You take this mattress e and turned into reduced row echelon form. So reduced for OS will inform is when you have a major C like going we have there, it's gonna be a three by four. So that means reduce or echelon form means that you have diagonal wine of ones from the top corner. And then you have everything else in every single row. Lo the warns is zero, and then hopefully we can also get all of these other space is to be zeros as well. And if we have three free variables will be able to do this. But if we have variables that are not free than some of these upper corner zeros will actually be numbers. But that's all right. So then what will be left in this last row will be our answers to our systems of equations. So to get this major see up here So first of all, for this system of equations, all you need to understand is that this room is the first variable Will call it X one. This room is the second variable will call it x two and then this last one will call x three and then this last last row is gonna would be is representing what is on the other side of the equal sign. So this top run, for example, is saying Negative three X one minus four x two plus two X three equals zero and that's what this top liners representing. So now we need to translate this into reduced, really, Rashwan for which is this sort of form All right, here. So to do that, we're gonna be able to do what's called Roe operations, which basically means that you can add they're subtract anyone row from any other row, for example, wrote to you can subtract for a one from 02 or row one from row three. You can also multiply that row by a constant. So you could, for instance, what we're gonna do. This one is we're going to so we know we're going to start with the first rail. We know what we want to get this nine negative nine in this negative six to be zeros. So we have a negative three. So negative nine negative dine ship. Oh, I actually wrote this down row my apologies. This negative three should actually a possibility. So negative. Nine plus three times three, which I'm gonna represent by row one. He's going to give us a zero there and then right on this one to plus negative six plus two times Rohan, which is a three. Well, give us serum. So now we're going to go through and we're going apply these rail operations toe every single column in the Metris E. So we're not performing any real operations on our top row, so we're just That's gonna remain the same. So we'll go ahead and fill that in three for to zero. Now, this we already decided already. Know that's gonna be a zero. We'll move on to the next com. So 12 plus three times negative for is 12 minus 12. She's gonna be zero and negative. Six plus two times three is also gonna be zero, and that's all zeros. That all speech. Then on the final row, you know that's gonna be a zero eight plus two times negative for is eight minus eight, which is zero. This one's gonna give you negative four plus four, which is zero and is here, so Now we're into this for so immediately. You can see that there's no way that we're gonna get that diagonal row of ones. But that's OK. So since sees both went to zero, these two bottom to rose went to zero. That means we know we're gonna have to free variables within our system, meaning that one very one variable will have an equation that defines its value. And that equation will be based on the two values of the other two variables, which will be able to be any any value in the book, any number ever. So to get that one equation that will define our system of equations for us, we need to take one more step to get it closer to reduce Romash one for, and that is to get a one in this top form here. So to do that, But it's gonna divide this top row by three, and that's going to give us a one negative, for there's 2/3 and this girl and then the rest of these air also still Syria because we didn't do anything to now to get our equation that don't define our system for us, you take this top row right here and write it down an equation for him. Like I mentioned before with these three variables Unequal sign something on the other side of the equal sign. So the first row has a one in that row represents X one. So it's one times x one, and we have a negative 4/3 negative. 4/3 on that row is for X two, and then our last row is for X three when we got a 2/3 in that one. So to Parents X three and then we're gonna have an equal sign and equal zero. So now you have this with equation. This is the equation that governs the relationships between our three variables. And so now you could define anyway wanted. You could solve it for any of the valuables. But since we set this one x one upto have it Oh, have a constant of one on the outside. It's gonna be easiest for us to solve for explain. So all we're gonna do is just move the x two and X three terms over to the others upside by adding and subtracting, we will get 4/3 next to minus 2/3 brooks you, I urge x so x one will always equal 4/3 times x two minus 2/3 the next three. And since we got these bottom two rows equaling zero, we know that X two and next three. The other two variables are equal, are equal to whatever they want to be. Whatever you decide that can be taken people to anything, and the system will still hold true. And X one was to have be able to have any value depending on what the values R of X two and X three and that means the next to the next three are what we call free variables, meaning that their value it's not constrained. That's free to be whatever it wants, and we're ready only do is use those to free variables to define the last variable, and that is your final answer.

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