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What would you have to know about the pivot columns in an augmented matrix in order to know that the linear system is consistent and has a unique solution?

In order for the solution of a system to be unique each column of the coefficientmatrix must contain a pivot. The last column of the augmented matrix cannotcontain a pivot.

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 2

Row Reduction and Echelon Forms

Introduction to Matrices

Campbell University

Oregon State University

Harvey Mudd College

Idaho State University

Lectures

01:32

In mathematics, the absolu…

01:11

09:31

Suppose a system of linear…

08:43

Suppose the coefficient ma…

If the system represented …

00:46

Explain how you can recogn…

01:29

00:54

What characteristics of a …

00:52

Describe what is meant by …

02:11

When solving a system usin…

02:18

What can you conclude abou…

03:43

THINK ABOUT IT

(a)…

for this question. We want to find the values for A B and C, for which we have a consistent system. So let's remind ourselves of what consistent needs consistent Ian's that there exists a solution. So our first step is going to be to put the Matrix that we can create from our linear system and to reduce Echelon four. So right here we can read off the Rose and see how we got them from the equation. So the first row reads excellent. Blustery X two plus extra is equal to a second row reads negative X one minus two x two plus x three is equal to be the 3rd 1 reads three excellent plus seven x two minus X three is equal to see. So now we want to put this matrix and reduced echelon form. So let's figure out the steps that we had to take to do so so we can see the difference between these two matrices matrix born matrix to is that this term is now zero. So we had to have done something to row two to get that answer so we can see that we said row to equal to rogue one plus room, too. And I gotta start zero there. Now we want to see the difference between Matrix to and matrix three. You can see here that another zeros knocked out. So that means that we had to have done something to row three. We had to use row one to do it. What we do is we said Road three equal to road three plus negative one. And that goddess Todd knocked out zero right here. So now we want to find the difference. Finally, between Matrix three and Matrix for So you can see that this is the big difference here. So we had to do something to our third row. Amused are three was too are too to get that answer. So now that we have our beatrix and reduce special on before I just copied our final matrix onto this cage, we can start trying to solve for values of A B and C, for which this has a concern for which this sister of equations is consistent. So let's first see what we can find from the third row. So if we want this to be consistent, we want to make sure that this is not equal to zero. Er sorry. We want to make sure that it is equal to zero because we have our variables x one x to an x three, but we can see the zero x one plus zero extremes with zero X three has to be equal to zero for this to be consistent, otherwise being consistent. So let's go ahead and write that down. You won't see him on this three A was two times b plus, eh, people of zero. Well, go ahead and simplify that we get C was to be minus A is equal to zero. Now go ahead and see what information we can get from our second row. So this reads ex to was to x three equals B was a So we have that. We can see that if we solve for X two, we get X two is equal to B plus, eh? Minus two x three. It's extra is a free variable. So we might as well see what happens when we solve for X in terms of its three variables. Lastly, let's see what we get from this first equation. Here we get X one was three x two plus x three is equal to a So what we first want to look at is we want to go back to this equation right here. So let's take a moment and look at that again. We have X two plus two. X three is equal to B plus, eh? So since we have a value for a up here, let's go ahead and plug that into this a right here. So x two was two x three is equal to be waas X one plus three x two was x three. So now let's go ahead and try and solve her beat because we won't again to find the values of A B and C, for which the solution is consistent. So let's go ahead and silver beast that we get x two plus two x three minus x one by this three x two minus x three is equal to be and this simplifies to negative X one minus two x two plus X three is going to be equal to be okay. So now we have our values for both A and B. Okay, so now we're going to go ahead and try and sell for I receive value. So let's go back. So what we have here we have C was to be minus A is equal to zero. We have our values for a and we have our valleys for beat. So let's go ahead and start looking in those to solve for C in terms of x one x to a next three. So here we have C is equal Thio a minus to be and we'll continue up here. We'll put it in black and so we differentiated and we get that this is equal to negative Two tongues will be found for being which is negative exploited by his two x two plus x three. And then we're adding that to our value up here for a so plus x one those three x two plus x three. So we don't want to go ahead and simplify this So we have. This is equal to two x one plus four x two buttons to x three plus x one plus three x two plus x three We get that this is equal 23 x one plus seven Next to awareness X three So now we have all the values for a for B and first see so we can write those out together. And we can see that for this solution to be consistent. A has to equal x one was three x two less x three a pass to equal negative x one by his two ex too plus x three and lastly, see has to be equal to three x 47 x two minus the x three.

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