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In Exercises 21 and $22,$ mark each statement True or False. Justify each answer."a. The echelon form of a matrix is unique.b. The pivot positions in a matrix depend on whether row interchanges are used in the row reduction process.c. Reducing a matrix to echelon form is called the forward phase of the row reduction process.d. Whenever a system has free variables, the solution set contains many solutions.e. A general solution of a system is an explicit description of all solutions of the system.

A. FalseB. FalseC. TrueD. FalseE. True

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 2

Row Reduction and Echelon Forms

Introduction to Matrices

Campbell University

McMaster University

Harvey Mudd College

University of Michigan - Ann Arbor

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for this question were asked to solve the given set of the given linear system using any methods were going to use the Gas Ian elimination method. So that means that we are going to first take our system of equations and put into matrix form and then put that matrix into reduce echelon form and then solve for the variables. So let's first make our matrix from the equations that are given to us in the one your system. So the first equation reads zero x one zero x to because there are no excellent or X two terms but sex three close x four was expired, equals zero. Then we have negative X one. Find his ex too plus two x three minus three x four plus x five is equal to zero, then our third equation. Reed's Ex Gordon, plus X two minus two x three minus x five is equal to zero and lastly or lost question is to x one plus two x two minus x three. But sex five is equal to Sierra. All right, so from there we are going to roll reduction to get this matrix into reduce special informed. So here What we see is ah Siri's of road reductions and we're gonna figure out what happened between each of these matrices. We're gonna put that on top of the arrows. So if we look at the first Matrix versus Second Matrix, we know that the only thing that's different about them is that these two rows are switched. So that means that our role reduction was setting our one equal. Two are four and our current or for evil or our old are for equal to you are one and not give us this matrix here. So now we want to look at the difference between our two and our three. What jumps out is the second row. So those negative ones were eliminated so that this too could be the Onley non zero injury in the rose. That that was the first step in doing so. So in order to do that, we had to take our two and said it equal to 1/2. Our one bus are too. And I gave us our third Matrix. So now we want to see the difference between our Third Matrix and our fourth matrix. We look and we see that again, these ones have become zeros. So in order to do that, what has happened is a change toe. The third rope so are three was set to equal 1/2 or one. We added out to our three. So I just copied, uh, our fourth matrix here to the side. And now we can see that the only difference between these two matrix matrices is the fourth row. So we see that this one was eliminated to be a zero. So there was a change to our four, and it was sent to equal or four was 2/3 are three. All right, it down here so it doesn't run into the Matrix. Okay. All right. So now what we want T. C is the difference between our fifth made drugs here and our six majors major here. So we can see from our fifth Matrix that we have our kibitz. But to get them into produced echelon form, we want to make sure that they're all equal toe one. We can also see that the second row was also scaled to be able to run tau one. So, in order to get the first entry in the first sort of equal to one. We have said Arm One equal to 1/2 are, too, for the second row we had said are two equal to 2/3 partner. Sorry. So for the 1st 1 for the first room, we said an equal to 1/2 or one for the second row. We had to send people to 2/3 times are too third row. We had said unequal to negative 2/3 turns the third room, and since this was already a one, we didn't have to do anything to the fourth row. So that's how we got our matrix here, which is now in produced echelon form. So now we want to solve for variables. We have x one x two x three, explore an x five So we want to figure out what our free variables are. First we see that X one is a pivotal column, so we know that X one is not a free variable. X two, however, is not a pivotal problem. So x two is every variable. X three is a pivotal column as his ex four, so neither one of those different variables. X five, however, isn't so. It's too and I stopped for our free variables. So we want to solve for X one x three the next score in terms of X shoe, the next five. So we'll start with solving for export so we can look Thio, the fourth row and see that we just have explore is equal to zero Hippolyta. So we moved to x three and we can use the third row to see that we have x three plus X pi is equal to zero, So x three is equal to negative X five Now for X one, we use the first row we see that x one about sex too minus 1/2 x three plus 1/2 x five is equal to zero And since we wanted in terms of our three variables which is X two, an x five, we want to but in the substitution for extra that we found here, which is in terms of x five. So we get that X one is equal to negative x two since negative 1/2 x three is equal to 1/2 x five. We know that this is minus 1/2 x five and then moving this to the other side of the equation. We have another minus 1/2 x five, so bring not to a cleaner page. Well, rewrite that really quickly. We have X four is equal to zero. Exterior is equal to negative x five and then we have this. X one is equal to negative x two minus 1/2 x five Linus another 1/2 x five We combine these like terms to find that X one is equal to negative X two minus expire and then we can write this altogether x one x three and explore which one right? He's in terms of x two. Tom's a vector plus x five times a vector so we can see that X one has a coefficient of negative one for x two and negative one for X five. We could see that X three has no value for X two. So there's a zero there and negative one is the coefficient for X five. Since export is equal to zero, we know that there is no X to term and there is no x five

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