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Suppose a system of linear equations has a $3 \times 5$ augmented matrix whose fifth column is a pivot column. Is the system consistent? Why (or why not)?

Inconsistent

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 2

Row Reduction and Echelon Forms

Introduction to Matrices

Campbell University

McMaster University

Baylor University

Lectures

01:32

In mathematics, the absolu…

01:11

Suppose a $3 \times 5$ coe…

08:43

Suppose the coefficient ma…

01:24

If the coefficient matrix …

02:02

Let $A$ be a $5 \times 3$ …

02:48

Apply elementary row opera…

02:39

Construct the correspondin…

01:30

Refer to the system of lin…

03:23

01:10

Apply elementary roes oper…

for this question were asked to determine whether or not given system is consistent. And if it is consistent, if its unique. So the first thing that we're gonna want to do is to find some terms so inconsistent is when no solutions exist for the system consistent you that solutions do exist. And there are two kinds. Ah, there two ways that a system could be consistent. It could be consistent and unique, which means that the system only has one solution or the system can have infinitely many solutions. Okay, so we can tell that a system equations is inconsistent. If we meet, put the Matrix and reduced echelon form, there is a one in the right most column, and that will look something like this will use ass tricks to represent just random numbers. They could be anything. So this entry right here is what we mean when we say that the reduced echelon form has a one in the right first column. And the reason that this is inconsistent is because if we were to write out this fourth row as an equation, we will get zero x one plus zero x two plus zero x three is equal to one. These are these terms. Obviously all cancel out equal zero and we get zero equals one which we know is not true. So this is one of Matrix is inconsistent. When the Matrix is consistent again, we have those two possibilities. We have a unique solution, which means that all variables are leading variables in matrix for so don't look something like this. So here we can see that all variables x one x to index three are leading variables because they're all pivotal column. And no, hear that this does not have a one in the right. Most call that would be having one here and a zero here, which would make it inconsistent. So that's what a matrix that has a unique solution would look like. Now, for infinitely many solutions, we need to have at least one free variable. So in a matrix that would look something like this so we can see that X one is a leading variable. Next to is the leading variable. But X three is a free variable, which means that this major X because and this system of equations has infinitely many solutions. So now we're ready to go ahead and jump into the problem. So here is the first matrix that were given. So let's take a look at the leading variables. So we'll label our columns x one X to an expiry. We can see that X one is a leading variable. X two is a leading variable and x three is a leading variable. So since all variables are leading variables from her definition, we know that this means that this system is not only consistent, but it's also unique. Okay, so now we'll go ahead to the next one. We'll label our columns again, exponents to an x three, and so we can take a look at this and see that this is not yet in reduced echelon form. What will we do know is because we have ones up here. We can very easily do grow operations to knock these out and make them equal to zero, right? We can do that with grow operations, and then once we do that, we can scale these leading variables to make them once again equal to one. So with ro operations, we can get something that looked like this. And since these were all, all the leading variables are all the variables air leaving variables again. By your definition, we know that this is not only consistent, but it's also unique. Okay, now for the next one. Well, go ahead. Label or rose again for this one. Let's do some row operations first. Look, take second row are to set it equal to our two minds are one and this gives us this new matrix. Okay, so we can notice right away that this second row is going to give us a problem because this reads zero x one with zero x two with zero x three is equal to one. This all equals zero. So we have zero equals one. And so we know that this sister of equations is inconsistent. If we wanted to go directly by our definition, we could do so of the more road reduction, and we could twinge these two rows. And so no matter what we would what role operations we would do to reduce the first and second row, we would still have, um, reduce echelon for with one in the right most called. So this is gonna be inconsistent, OK, on to our last one. So again, we're gonna want to do some row operations. Since these two columns are exactly the same, let's see what'll happen if we knock one of them out. So it's that are three equal to our three. Linus are too. What gives us this picture? OK, And from here, let's see if we can get this too equal zero. So we should put in reduced special inform and have this baby only no. 10 injury in the column. So in order to do that will want to take our two. It's that people are too minus or one and that's gonna give us the following matrix. The one minus one is zero zero minus. Some arbitrary number is just another arbitrary number that holds true for this place and one minus an arbitrary number is again just another arbitrary number. And then we have this row of all zeros. So no matter what this number is, we can just divide or two bye done over. So just say let's say our two is equal to our to debunk by store, for example, it will assume that all these air, the same number, they don't have to be And that will give us this new matrix right here. And we can see that this is introduce special on form and for our variables. Excellent X. You next 30. You see, that X one is a leading entry and excuse the leading entry. However, X three is a free variable. So because we have at least one free variable, we know that not only is this that sort of equations consistent, but it also has infinitely many solutions.

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