suppose-the-coefficient-matrix-of-a-linear-system-of-three-equations-in-three-variables-has-a-pivot-

Name: suppose the coefficient matrix of a linear system of three equations in three variables has a pivot
Uploaded: 2019-07-25
Duration: 8 min 43 s
Channel: Numerade
Description: Suppose the coefficient matrix of a linear system of three equations in three variables has a pivot in each column. Explain why the system has a unique solution.

Question

Suppose the coefficient matrix of a linear system of three equations in three variables has a pivot in each column. Explain why the system has a unique solution.

Arden Mccarthy · Accepted Answer

for this question. We want to find the values of a for which this system equations has no solution. Exactly one solution and infinitely many solutions, So we&#x27;ll start with no solution. So first things first. We want to write out our Matrix from the given equation, so we have one. Next one plus two x two plus x three is equal to two to x one minus two x two plus three x three is equal to one and we have one x one plus two x two minus a squared minus three x three is equal to a So before we go have been started. Let&#x27;s remind ourselves of what the Matrix looks like. That has no solution. So the requirement is that the reduce special inform has a one in the right most column. So let&#x27;s do so matrix simplification. Here, our first step is gonna be extinct, are three and said that equal to our three minus are one. So if we do that first column stays the same. Second column. Cesaire Sorry, First road stays the same. Second row stays the same, and the third row become 00 And then we have negative a squared minus three minus one. And then over here we have a minus two. So what we want to dio is we want to figure out a value such that this is equal to zero, that this is not able to see her because that will give us and inconsistent system of equation. So let&#x27;s first simplify this turn over here. So we have negative a square minus three minus one equal to negative. A squared plus three on this one is equal to a negative. A squared plus two right. And we want it to be equal to zero. So let&#x27;s go ahead and do that now and solve for the A that will make this condition true. So a squared is gonna be able to to because we moved the a square to the other side. Then we take the square root but both sides and we get that a has to equal the square root of two. OK, so that&#x27;s our first requirement. Now let&#x27;s go ahead to our second requirement, which states we want a minus two to not be equal zero. So if a minus two can&#x27;t be able to zero, we want any thio not equal to two. These two requirements agree with each other, since two is not equal to the square to. So we know that for no solution to exist, we want a to be equal to the square of two. But for the next one, for exactly one solution, we know that system equations will have exactly one solution if the reduced excellent form of The Matrix has no free variables. So we&#x27;ll go ahead and we&#x27;ll take this simplified matrix here and we&#x27;ll just go ahead and put it on the other page so we don&#x27;t have to repeat our row operations. So we have 1212 two Negative, too. 31 00 negative. A squared plus two and then a minus two. Okay, so let&#x27;s continue so reduction, so we can get a closer to produce a short form. So to do that, uh, let&#x27;s first take our second row and said it equal to the first row plus the second row. And this will help us knock out this term right here. So when we do that, we get this matrix. First row stays the same second row since we&#x27;re out of it, becomes 30 for three and then the bottom row stays the same. Okay, so now let&#x27;s go ahead and do another row operation. So this row operation will help us knock out this turn right here to make it a zero. So to do that, we want to take our one and said it equal to three are one plus. Negative or two not will. Give us this Major X right here. We&#x27;ll get zero six Negative. 13 this rural states. Same. And the last room will also save them. Okay, I know it will Dio is we&#x27;ll just go ahead and swap these two roads so it&#x27;s closer to reduce social reform. So 30 for three 06 negative. 13 00 negative. A squared plus two and a minus two. Okay, so we want no free variable. So what that means is we want to make sure that this entry is not equal to zero. Okay, So in order to that, we set a squared plus two and not the equal to zero We saw for a squared. When we get that a and not be equal to the spirit of two, and if that condition holds, that&#x27;ll make sure that there is exactly one solution and this is sort of operation. Okay, So onto the last word, I don&#x27;t want to find value for a for which there are infinitely many solutions. And this means that the reduce national inform has at least one free variable so greater than or equal to 13 variables. So by the previous implication that we&#x27;ve already done a few times, I&#x27;ll just right are reflex of broad matrix. So 30 for three 06 negative. 1300 negative. A squared was two and a minus two. And we just got that matrix from right here. That&#x27;s where we got it from. Okay, So if we want at least one free variable, that means that we want toe Look at this road right here. And we want to make sure that this is equal to zero and that this is equal to zero. So in order to do that, we sent negative a squared close to you. Well, zero. And we saw for that. And we get that, um well, so yeah, we want to sell for that. And then next up, we won&#x27;t solve Bernie minus two is equal to zero. So we saw this really quickly. We know that we want eight equal to and so we&#x27;re gonna plug this, eh Into this equation will see we have negative a squared plus two equals zero That gives this negative two squared was too equal to zero. And then this gives us negative four plus two has to be equal to zero. And then we send finest and see that negative too has toe equal to zero. But we know that that&#x27;s not true. So we know that this part must be wrong and that part&#x27;s not true, which means that there are no values for an infinite solution.

Payton Asch · Answer

Hello. So I&#x27;m gonna be looking at the question of if we&#x27;re given a three by five matrix with three pivots. Is it a consistent system? Eso first of all, a consistent system has to have at least one solution. Ah, and an inconsistent system, therefore, is a system which has no solutions. So let&#x27;s take a look at this three by five matrix, uh, so we can see the pivots are all highlighted in red and it&#x27;s in a row National in form, um, and so we can see that since there&#x27;s through pits and there&#x27;s three rose, which ah relates to the number of equations that we have in our system that there must be unleased. One solution because you can see here in the system, uh, of equations, that there is a solution if we solve for Z, and then we can substitute that in for the top two equations and we can solve for y and W. Since there&#x27;s infinitely many solutions to that, uh, system and then plug those in to our top equation and we will have at least one solution. But there should be a lot more than that

Viswesh Uppalapati · Answer

So, uh, in this in this video, we&#x27;re gonna be solving problem. Number 40 from the book from Section 1.7 is on linear Independence. So it gives us a statement and tells us to explain why it&#x27;s true. So it tells us that there&#x27;s an M by N matrix a wood, um, that has n pivot columns. So this means that and must be equal to M or must be greater than equal then because, uh, if there&#x27;s more if N is greater than our greater than M, then that means we cannot have end to the columns because there is. There aren&#x27;t enough Rose and the Matrix for, um all of every end toa have its own pivot. So where these Criterion line it tells us that, uh, the vector equation X equals B for this particular n by N matrix, which is a has must have a unique solution for every B. Um, this is true because, uh, if you set sent this equation equals zero X equals zero. This, uh, this problem or this n by N matrix will always have a trivial solution. If it has, Ah, if it has a unique solution for every B because being linear independence basic means that any vector that is made out of a linear combination of the columns of A has a unique representation, which means that there&#x27;s only one X so one solution for every, uh, X equals for every B and the equation X equals B. In other words, if we if you were to write this out, that means that, um, if and has a pivot every column, that means that, um so that means that there will always there there will always be ah, no free variable, so there will never be a free variable. So, for example, if you take, let&#x27;s say, uh, four by three matrix with a ah, give it and every every column because there are three columns. But for Rose, so the last row can just be zeros, and these can be a non zero numbers. We can see that if we label each column with a variable cause, um, because solving an augmented matrix is just solving a set of system equations, Um, where each column represents a different variable. You know that every call him having a pivot makes it so that there is no free variable. Um, which means that which means that every variable has one single unique solution. That is why this statement is true.

Arden Mccarthy · Answer

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&#x27;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&#x27;t look something like this. So here we can see that all variables x one x to index three are leading variables because they&#x27;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&#x27;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&#x27;re ready to go ahead and jump into the problem. So here is the first matrix that were given. So let&#x27;s take a look at the leading variables. So we&#x27;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&#x27;s also unique. Okay, so now we&#x27;ll go ahead to the next one. We&#x27;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&#x27;s also unique. Okay, now for the next one. Well, go ahead. Label or rose again for this one. Let&#x27;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&#x27;re gonna want to do some row operations. Since these two columns are exactly the same, let&#x27;s see what&#x27;ll happen if we knock one of them out. So it&#x27;s that are three equal to our three. Linus are too. What gives us this picture? OK, And from here, let&#x27;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&#x27;s that people are too minus or one and that&#x27;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&#x27;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&#x27;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.

Neel Faucher · Answer

and this question, we&#x27;re told a consistent system. Three linear equations and three variables. So that means that three equations Andrea three. When you plug in system creations in a matrix, the number of equations corresponds to the number of rows. Variables respond to close, so we know that the Matrix B three by three number froze and the number okay, So in part, a told to consider there&#x27;s one meeting look, so that would mean that their to free variables fear it&#x27;s just used for Serbs and variables. This is because the number of free variables is equal to the number off tables. Honest number. England&#x27;s So it&#x27;s three because of columns is number variables minus one Barbie. Still, we have to leading ones. One free variable seems as three new ones is you. Free fours because The Matrix would look something like This is your my dear told for the well, that&#x27;s impossible because three by three matrix maximum number ones history. So no

Runpeng Li · Answer

All right, so in problem 34 suppose A&#x27;s of three by three matrix and A&#x27;s B&#x27;s a vector in our three with the property. That Exodus p D has a unique solution. And we want to explain why the column Self Bay must spend our three. So Okay, so in this case, since we know the leaner system they X equals B is has a unique solution. So that means if we consider the or commended matrix and furthermore, if we considered the rule reduced form off this matrix, it must has has to ruin form like this 81 1 a two to something on the diagonal, diagnose and have some stuff there some stuff here. But it doesn&#x27;t matter. What are they? And we have zeros here, and I&#x27;m on the right hand side. We have B one b two B three as a constant. Okay, So this is the only impossible for this system to have a unique solution because only in this way we can solve a unique by the By the last row, we can solve the unique, unique x three, and furthermore that from the second road that X two will be determined by X three and x one would be determined by X two and x three. So this the only impossible way that we can have a unique solution for this system. So that means by observing this matrix, we have three people, two columns and three people. Columns means, uh, means axes. Um, yeah. So three people columns actually gives me, gives us the information that all the columns because A said three by three matrix And since every every single column is, is that people column So these columns are linearly independent, very independent, and furthermore, they must spend our three because in this case, we actually have three vectors and our three estimation three and those three Rector&#x27;s Arlene Early independent so thes three directors must spend our free so

Problem

Restate the last sentence in Theorem 2 using the …

Problem 26 Hard Difficulty

Suppose the coefficient matrix of a linear system of three equations in three variables has a pivot in each column. Explain why the system has a unique solution.

Answer

See Explanation.

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

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Absolute Value - Example 1

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Linear Algebra and Its Applications

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David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Grace H.

Catherine R.

Anna Marie V.

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications