in-exercises-29-32-a-does-the-equation-a-mathbfx0-have-a-nontrivial-solution-and-b-does-the-equati-2

Question

In Exercises $29-32,$ (a) does the equation $A \mathbf{x}=0$ have a nontrivial solution and (b) does the equation $A \mathbf{x}=\mathbf{b}$ have at least one
solution for every possible $\mathbf{b} ?$
$A$ is a $3 	imes 2$ matrix with two pivot positions.

Michael Jacobsen · Accepted Answer

in this example, We&#x27;re dealing with the Matrix A. We&#x27;re told it&#x27;s of size three by two and contains to pivot positions. Let&#x27;s start out this problem by indicating what the format of a would look like using just generic elements. So I have three rows. I&#x27;m going to use a solid bullet to represent an arbitrary real number. We have two columns salt Stop here and here&#x27;s our three by two matrix. We have to pivot positions, and that means, at best will have a pivot here and a pivot here for two rows, which are Pivot rose. Next, let&#x27;s consider the system or matrix equation. A X equals the zero vector. So this is, ah, homogeneous matrix equation, and we know X equals does. Your vector itself is a solution, So this is consistent because it has the trivial solution here. The question then becomes is there were Let&#x27;s Fraser of slightly more dramatically are there nontrivial solutions to determine if there are nontrivial solutions? We go to the columns. We know that column one is a pivot column, as is calling to Let&#x27;s write down here, both columns of a our pivot cones. This is a very important observation in the sense that now that we know that both columns are pivot columns, we can conclude immediately that there are no free variables. So if there are no free variables, we know that the system A X equals zero will have a unique solution. But here&#x27;s one solution. Already the trivial solution. This then tells us there are no nontrivial solutions that&#x27;s phrasing it with a double negative. So if that makes you uncomfortable, you could say there is only the trivial solution. Let&#x27;s consider a different system. Next with the same matrix a size three by two with two pivot positions, consider the system eight times X equals some arbitrary vector B B is unspecified. And so for this particular system, we want to know, Is this always consistent? So is this system always consistent, always being a very important word here, well, to determine if a system is consistent or always consistent in this context. Now we turn to the rose as opposed to the columns when we&#x27;re considering the nontrivial solution. So going row wise, no one is a pivot row. Row two is also a pivot row, but Row three is not so Row three has no pivot, so we can come down here and say, Not all Rose of a Our Pivot Rose. This gives us a set of logically equivalent statements. Now we know now that the columns of a cannot span all of our three or three comes from this dimension we know no. All vectors be in. Our three can be written as a linear combination of a but these air, just different ways to rephrase what we&#x27;re really trying to get at. Since not all rows are pivot rose, we can now conclude. Therefore, a X equals B is inconsistent for some vector be in our three. It&#x27;s not saying it&#x27;s always inconsistent. For example, of B is a zero vector. We certainly have a solution that trivial one. And for non trip for non zero vectors be there&#x27;s likely solutions. What it is saying is we can absolutely find a particular vector B to make the system inconsistent

Runpeng Li · Answer

Okay, So in this problem, work even a is a three bite the remain trees with two people, two positions. Okay, there are two questions we need to figure out here. First question is, that&#x27;s a X equals zero have a nontrivial solution. So we first consider hey, inform off. Reduced a penchant for So that should be 11 one. And with some stuff here because it is a reducing nation form. So all all the entries, um, here will be 00 and zero. Okay, So now we only allow two people positions here. And the matrix I&#x27;m writing down here has has three people in positions, and we have to delete at least one those things, um, we have We have more people to positions here, so a se So this one does not work. So let&#x27;s say they do the new me shakes. They say the new Matrix is so we need to delete at least a warrant road. That means at least one row has to be always ears. So let&#x27;s say the last row is all yours. So right now we have a 11 and all zeros and with some stuff over here, All right. So you can see from in this matrix that that there exists at least one people, one pre bearable. Not people. Just there exist at least one at these one. Pre venerable. Because we have X one times the coefficient here and with some stuff, either. And here should be zero with some cooperation. I say we have some coefficient. Um, you brought off x three say a 33 as the coefficient of X three. He caused zero. And the second row keeps us x to us A. Sorry, uh, let&#x27;s say, um, here the coefficient here that the coalition be one a 13 So that&#x27;s a 13 And this 8 to 3 times x three equal zero So x one x two can be expressed in terms off x three. So ex Tereza free bearable. That means this is stone. We call our effect from over textbook, that is, except a axe equals zero has a nontrivial solution. You can only if there exists of free variable. And here repairable, it&#x27;s extreme. So the system has country view as a non trivial solution. So the first question is down. Now, The second question second question is, Does the equation X equals? B have at least one solution for anybody, so let&#x27;s check. So if we have X one plus a 13 as before, that&#x27;s three course. Some, let&#x27;s say some counseling. Be one and x two plus a 33 Sorry, 8 to 3 a 23 x three. Because at some B two now we still have a heaven entry. I have 1/3 entry for Vector B. No says since we&#x27;re we&#x27;re considering any sector B. So the third entry can be anything but this case. Seems we since the last row is all zero. So that means the last the next entry be has to be so that the three has to be zero. But we&#x27;ve But right now we&#x27;re considering all considering all the bees, that means the three can be anything. So that means we cannot have. That means that the equation here, the system we have here we&#x27;re having here doesn&#x27;t have at least one solution for anybody because we take like we take a B one b 21 So the system doesn&#x27;t work because we have all zeros in the last row, over a and no matter which, no matter which act we&#x27;re taking the NASA rover people wear considering the boat matrix multiplication Here, the last roll we&#x27;ll always be zero. This is the last rule is always zero When we&#x27;re taking the last entry here as one. And this stuff, they&#x27;re not equal. So that means this system doesn&#x27;t have at least one solution for any.

Michael Jacobsen · Answer

here. In this example, we have a matrix. A. It&#x27;s going to be a square matrix of size three by three. We&#x27;re also told that there will be three pivot positions. Let&#x27;s write down what they could look like using a generic bullet point to denote any real number. So it&#x27;s of Size three by three, and each of these bullets can be any particular number. But if we were to really roll, reduce this to echelon form because there&#x27;s three pivot positions, we would find a pivot here, here and here for this particular size matrix, which is three by three. Now, given this matrix, let&#x27;s consider the equation. After I grab a pen eight times X equals zero vector. So we know that X equals zero Vector is the trivial solution. We want to know if this has nontrivial solutions. So to determine if there are nontrivial solutions, we return to our Matrix. A. We know that the Matrix A has no free variables because every Colin is a pivot column. If, for example, we didn&#x27;t have this cognitive it column, we would have had one free variable. So since we have no free variables, we can answer this question in the negative, a free variable would give non trivial solutions. So here there are no nontrivial solutions, and this is because there are no free variables. Let&#x27;s change the situation just a little bit. And consider a slightly different matrix equation. Se X equals Justin generic Vector B, which would be in our three. The question then, is this system always consistent? In other words, no matter what we change over here for the vector B, can we still solve this system? The way to answer this question now is once again to return to the original Matrix A. Here, we&#x27;re talking about a matrix equation that we&#x27;re trying to solve. So what we&#x27;re most interested in are where the pivots are row wise. We have enrolled one a pivot, a pivot in Row two and a pivot in Row three, so we can determine that there is a pivot in every row of a and this gives us immediately many different conclusions. We know that having every a pivot in every row of a is logically equivalent to the columns of a spanning all of our three since its size three by three. It means every vector B we pick from our three would be a linear combination of the columns of A but in particular we also know that therefore, a X equals B is always consistent. In other words, even if we change this vector, be as much as we like to make the system as terrible as we can, it&#x27;s okay. We can always still solve the system. Looks like I forgot a t here.

Runpeng Li · Answer

All right, so in this problem, we&#x27;re given a as a four by two matrix. And again, we need to consider two questions here versus does. That&#x27;s the equation. A X equals zero have a nontrivial solution. So that&#x27;s check. So right now we have a, uh, by our assumption, that is a two by four matrix end. It has two people positions. So that means the first role has to be something like this. Must be leading by one. Uh, let&#x27;s say it has the form like this say one and zero and something in the third untrained force entry. And the second rule will be 01 Some stuff here. So right now, if we&#x27;re considering system A X equals zero, then we can find us. Uh, the system we have we&#x27;re having here has has to re variables at least. So this case I&#x27;ll write it down here. So 23 variables will be first, we have x one plus some stuff say a 13 and x three us a. 14 x four, and the second rule gives us X two plus he again a 23 a 23 x three and a 23 x for these two equations and x one x x two can be determined by x three, I guess on exports. So there are 23 barrels and we are because it&#x27;s the same for other cases we can&#x27;t. But we have to keep in mind that there are two people in positions. So we cannot have any roll up all zeros. We can only have one some some gross leading by one. It is. It is either like the first position, second position or the third position as well as as the last it is leading by one. And no matter in which case, we always have a well, we always have a solution on trivial solution. So we always because we always have, ah, free bearable. So we&#x27;re done in this case so that the answer is yes. And it has nontrivial solution in this case. So the second question is, um, that&#x27;s the question They actually will be have at least one solution for every possible. Be so again, we need to check. So that is considering the axe equals B uh, whether it is head, whether it has a decent one solution. Brandon be is so it&#x27;s very clear that we have two equations, but we have three on nose in this case, if we are considering the case like like we&#x27;re dealing with for the first time. So that is so x one extra rehang export. We have three on those, but we have only two equations, which means three on North three unknowns and to your questions, which means there are There are actually even in many solutions in this case. And even you were considering different different matrix. We still like, uh, we still have have the information have the same result because the unknown the unknowns is always greater or equal to the the equation here. So that means we have at least one solution for any branding be branded Beach will be to be chosen. So that&#x27;s our guy. No answer. So in this case, still Yes, remember, the unknowns is always greater or equal to the number of equations, So that means we have to We have at least one solution for anybody

Michael Jacobsen · Answer

in this example, we&#x27;re going to let a be a three by two Matrix. Then let&#x27;s consider the Matrix equation eight times X equals zero vector. Let&#x27;s also give this a particular condition. Let&#x27;s suppose that this equation has on Lee the trivial solution. Let&#x27;s say next what this would mean specifically about our Matrix A. So we have if we have only the trivial solution that would then imply that the equation has no free variables. But if there are no free variables, that would then imply that the columns of a are all pivot Collins. So let&#x27;s write down an example of such a matrix. A. It&#x27;s going to have three rows two columns, and we have want to have only the trivial solution for this particular matrix. A when we consider the Homo genus equation. But this is really the key. We just have to make sure the columns of a are all pivot columns. So we could say, Put a 100 Making this matrix a little bit simpler is a decent way to go. We could also go 0 to 0 if we want to get a little fancy and not use ones everywhere And for this particular matrix, this has to pivot columns so it will satisfy the original condition. That X equals zero has only the trivial solution. Notice that that means immediately the columns of a are linearly independent. Now let&#x27;s change this up a little bit and now consider a matrix be of the same size. Let&#x27;s convert to a new paid and we&#x27;ll say here, let the Matrix B be a three by two matrix. Well, consider again the Matrix equation that not a but to be Times X, equals the zero vector. Let&#x27;s suppose here that there are nontrivial solutions this time, So if there are nontrivial solutions, let&#x27;s say next. What that would imply that would imply that are matrix equation be X equals zero has a free variable. At least one. We&#x27;re going to go for the case of exactly one when we write down the example for this matrix. So if be X equals zero has a free variable that would also imply that one of the columns of Be is not a pivot column. So now there&#x27;s various ways of writing down this particular Matrix B. It&#x27;ll still be of size three. By true three by two. But I want one of its columns to not be a pivot column. We could. If we like just zero out a column now, there&#x27;s no way calling to could ever be a pivot column. Let&#x27;s also go for a fancier solution. If you want a different kind of way of solving, this will make my combi 111 and now I will make these columns linearly independent. As follows multiply com one by two so calm to is 2 to 2. Then this will apply immediately that there will be a pivot here and nowhere else. That means the Matrix equation be X equals zero. We&#x27;ll have non trivial solutions, and the reason we&#x27;re getting nontrivial solutions immediately is because these columns are linearly dependent, since Column two is a multiple of a non zero column. So these are two different cases where we&#x27;ll have no solutions or just the trivial solution and nontrivial solutions

Noah Duncan · Answer

Okay, So in this problem, we want to rewrite this system of linear equations as a matrix equation. Um, a X equals B where a will be our coefficients Matrix X will be our variables, and B will be our Constance vector. So what that&#x27;s gonna look like is something like this, So I always write out the outline of each of these things before actually filled them in. So a is going to be a big matrix. X and B are just gonna be, um, like one row vectors. Um and, um, I&#x27;ll usually start out by just putting the variables in here. Um, you want to put them in the order that you feel most comfortable with, but if you put them in a certain order, you just want to make sure that your columns line up with whatever, Um, order you put them in here because you the order of the X, y and Z and the coefficients matrix is important. If you wrote them in a certain way in this matrix here, um, and over here in the B vector is going to go all of the constants in the way that we do This is we just go row by row? Um so if you look at this first row right here, we can represent this row as the first row of our Amy tricks and are constants. Matrix just is the coefficients on everything. So there&#x27;s a one on the X. We don&#x27;t see it, but it&#x27;s implied is a four on the Why and there&#x27;s a negative one on the sea in the constant term is three here and it&#x27;s a three. Um, it&#x27;s important that, uh, we take note of the equal sign. The three should be on the other side of the equal sign in order to put our constant there on the second row. Looks very similar. So it&#x27;s going to be a one on the X A three on the why and a negative two on the sea in the Constance Matrix will put five because that comes after the equal sign. And then during the third round, the same way there&#x27;s gonna be a two on the X seven in the wind, the negative five on the Z and then a 12 in the constants of me trick. So this right here is the maid

Problem

In Exercises $29-32,$ (a) does the equation $A \m…

Problem 31 Hard Difficulty

In Exercises $29-32,$ (a) does the equation $A \mathbf{x}=0$ have a nontrivial solution and (b) does the equation $A \mathbf{x}=\mathbf{b}$ have at least one
solution for every possible $\mathbf{b} ?$
$A$ is a $3 \times 2$ matrix with two pivot positions.

Answer

a.) No
b.) No

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Algebra Educators

Catherine R.

Anna Marie V.

Caleb E.

Michael J.

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

Catherine R.

Anna Marie V.

Caleb E.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

a.) Nob.) No

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Caleb E.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Catherine R.

Anna Marie V.

Caleb E.

Michael J.

a.) No
b.) No

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

Linear Algebra and Its Applications