restate-the-last-sentence-in-theorem-2-using-the-concept-of-pivot-columns-if-a-linear-system-is-cons

Name: restate the last sentence in theorem 2 using the concept of pivot columns if a linear system is cons
Uploaded: 2020-07-10
Duration: 2 min 19 s
Channel: Numerade
Description: Restate the last sentence in Theorem 2 using the concept of pivot columns: “If a linear system is consistent, then the solution is unique if and only if

Question

Restate the last sentence in Theorem 2 using the concept of pivot columns: “If a linear system is consistent, then the solution is unique if and only if

Michael Jacobsen · Accepted Answer

In this video, we have augmented matrix given by this matrix here, a B where a is a coefficient matrix to a linear system of equations in and different unknowns. If we like we can re express this matrix a B as follows, we can say a augmented with B would be equal to first calm. One of a column two of a All the way to With There&#x27;s an unknowns will have and calms altogether, then augmented with the right hand side of the equations, which is the vector B. Let&#x27;s hear assume that the system is consistent. Then what that tells us is the following There will be or, let&#x27;s say, then there is no pivot in the far right. Call him in other words, that far right, calm here cannot contain a pivot. So this will mean that there exists because solution And what about uniqueness? We can also say that the solution is unique when, well, what should happen now for uniqueness? Well, if we have a unique solution, that means there is no free variables. But to have no free variables means every one of these columns here is a pivot column so we can say this solution is unique when all columns from the Matrix A. That&#x27;s this portion or this portion here, not the augmented portion. So all comes from the Matrix, say our pivot columns. When all columns are pivot comes, then we have no free variables for a unique solution.

Arden Mccarthy · 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.

Viswesh Uppalapati · Answer

in this video, we&#x27;re gonna be solving problem number 30 from section 1.7 of the book, which is based on a linear independence. So, um, I&#x27;m not gonna write the problem out here, but it has two parts. The first part is a fill in the blank which asked you to fill in the blank if ah, for the following statement. If a is an m by N Matrix, then the columns of a r linearly independent if and only if they has dash pivot Collins so is and by and it is linearly independent. Yeah, okay. As blank pivot columns, then B is the explanation of why it&#x27;s always true. So a the phone The blank is, um is And if a is an n by n matrix its columns, they&#x27;re only linearly independent. If has n pivot columns this because if he has an pickles will say we have a name by and cellos pick up two by three matrix for number 10. So here m is too. I just pick a four by three. Um So m is for, in this case, an n equals three. So here, if you have one euro on star So here, Um, here we have a leading Are you here? We have a pivot position in every column. So, um so this. Make sure that there is no free variable. Final answer. So if let&#x27;s say if let&#x27;s say there were only two pivots and this was zero, that means this column would have been a free variable making the entire column the columns of this matrix b linearly dependent Because, um, these two variables will be dependent on the value of the free variable. So for that snobby, the case, like an M by N. Matrix, has tohave end pivot columns so that there are no free variables possible so that the only solution, um to the matrix equation A X equals zero as a trivial solution, which is, Ah, zero vector. Is there a column vector depending on

James Chok · Answer

okay, in this question, you want to show why, If two rows off a square matrix are equal death, it&#x27;s determinant is equal to zero. So in this case, I&#x27;ve set up a scenario where this one and this one are equal. So all you have to do is you simply just subtract this road from this world. So if you subtract this way, this road is lettuce, all zeros. So you&#x27;re left in matrix. A one a two, a three double talk A N and zero is zero, and these are some of us. So now, Cliff, after expanding by this row, you always get zero times with some matrix zero times was determined and continues on. So then the determinants off a is simply zero to show why this is the case for two columns being equal. I set up this example. So what we&#x27;ll do is we&#x27;ll perform the Rite operations Row two minus be to ever be one road one. And we said that into road, too. Row three, minus B three ever be one row one selected to Road three. Continue on. Go right in minus B and ever be one wrote in and subject to road. And And we also have another one. We&#x27;ll go wrote one minus Being one ever be too wrote Two incidences were one. So to understand what the steps does if you take road to which is this to this road? Yeah, we&#x27;ll take this way and you put in row one here you get B two minus B to ever be one tons of being one. So that&#x27;s a beetroot. Might be, too. So then this number here and this number here and that&#x27;s what this does is what makes this one here and there and these bus make these on starts. And this one awesome exes, It&#x27;s a finally, we have this step, which makes one which makes this one. If it is there. So you&#x27;re less Sister matrix 000 is zero is there are a lot easier. And these little numbers here Now you can keep up to expand by this column of this column and once again will be four zeros. So which implies there determinants off, eh is equal to zero

Michael Jacobsen · Answer

in this video, we&#x27;re handling a transformation. T mapping are in into RM which can be written in this form. T F X equals eight times X because we have a standard matrix here. This implies already that t is linear, which is a very helpful piece of information. Tohave. Now that we know that T is linear, we can ascertain the following that t is one, 21 if and only if all use the bi directional arrows for the F in Olean Statement T or not T but rather the standard matrix A has a pivot in every call him, So it&#x27;s describe why we know there must be a pivot in every column in order for tea to be oneto one. Well, we have a theory that says the following t is one 21 If and only if a has Lee Nearly independent columns. Let&#x27;s say what a having linearly Independent columns is equivalent to best game. It happens if and only if The matrix equation eight times X equals zero has no free variables. Or, in other words, a X equals zero has on Lee the trivial solution. But we know that that can happen if and only if The Matrix A has a pivot in every column. So that&#x27;s all right. Our ideas tied together to give this equivalent statement.

Ankit Gupta · Answer

okay. This particular cushion we just have to answer in just one bird as true involves the question say, is that if I just read out if a system off to linear equations in two variables okay, it&#x27;s consistent than it has exactly one solution, which is completely wrong, that we destroyed the statement. Or let me just tell you that if there are two equations into wearable, like even X, let&#x27;s even bicycles to seven on eight weeks. Let&#x27;s be two bicycles to see toe right on this falls to be consistent as according to the question, This is just pushing. I&#x27;m writing in mathematical form. So let&#x27;s see. Consistent is all about not getting only one solution. This is about many solution as well, because consistent means one solution or consistent means many solutions. But the question is only talking about this case. It is ignoring this fact. So which means this is false. So the correct answer is faults. Okay,

Problem

What would you have to know about the pivot colum…

Problem 27 Medium Difficulty

Restate the last sentence in Theorem 2 using the concept of pivot columns: “If a linear system is consistent, then the solution is unique if and only if

Answer

If a linear system is consistent, then the solution is unique if and only if every column of the
coefficient matrix is a pivot column; otherwise, there are infinitely many solutions.

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Algebra Educators

Anna Marie V.

Kayleah T.

Maria G.

Kristen K.

Algebra Courses

Absolute Value - Example 1

Absolute Value - Example 2

Recommended Videos

Watch More Solved Questions in Chapter 1

Video Transcript

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

Linear Algebra and Its Applications

Anna Marie V.

Kayleah T.

Maria G.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

If a linear system is consistent, then the solution is unique if and only if every column of thecoefficient matrix is a pivot column; otherwise, there are infinitely many solutions.

Linear Algebra and Its Applications

Anna Marie V.

Kayleah T.

Maria G.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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

Anna Marie V.

Kayleah T.

Maria G.

Kristen K.

If a linear system is consistent, then the solution is unique if and only if every column of the
coefficient matrix is a pivot column; otherwise, there are infinitely many solutions.

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

Linear Algebra and Its Applications