suppose-a-is-a-4-times-3-matrix-and-mathbfb-is-a-vector-in-mathbbr4-with-the-property-that-a-mathbfx

Question

Suppose $A$ is a $4 	imes 3$ matrix and $\mathbf{b}$ is a vector in $\mathbb{R}^{4}$ with the property that $A \mathbf{x}=\mathbf{b}$ has a unique solution. What can you say about the reduced echelon form of $A ?$ Justify your answer.

Michael Jacobsen · Accepted Answer

in this video, we start out with the four by three matrix, and we&#x27;re going to consider the equation. X equals B. Let&#x27;s start off by making an assumption. Let&#x27;s suppose that the Matrix Equation A X equals B has a unique solution for some vector be in our for So let&#x27;s next. Consider what that would specifically mean for this matrix. Well, it would imply that in the equation, X equals B If we have a unique solution, there are no free variables, so start there. So that&#x27;s last line implies that there are no free variables. How do we know this? Well, if there was a single free variable, we would have infinitely many solutions, not a unique one. So next, knowing that there are no free variables that implies the following. Ah, free variable will always come from a non pivot column of The Matrix, say, so we know that A has a pivot in all Collins, and there&#x27;s three com specifically. So let&#x27;s say all three columns squeezing the three in there. Sophie has a pivot in all three columns, but it is of size four by three. This is our next immediate conclusion. I will say that a must be equal to or rather not equal but row equivalents to a four by three matrix. That would look like this so we can have 100 now. Calm One is a pivot column for row to go 010 Now we have to pivot columns, but we require three. So the third row will be 001 and we&#x27;re looking at the reduced row echelon form, so their final row will be all zeros. So given just this minimal set of information, we now know that the role reduced echelon form of a takes on this form.

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

Donald Albin · Answer

This is a continuation of the problems that that started a problem 11 a little bit surprised that this was so easy compared to have problem. 11. This is determinant of a times B, according Teoh, a postulate nine. That&#x27;s just the determinative of a times the determinant of be So it&#x27;s going to be negative, too. Times three, which is negative six. Just have to maul. Apply those to determine its together.

Runpeng Li · Answer

problem. Um, we want to show that if a X equals zero has only the trivial solution, she will solution. Then way. Want to show Dad? Hey, X, Because be must I have this ocean for every Be okay. Now, The way to do it without the convertible matrix serum is to observe the structure off A. So we consider let&#x27;s consider the reduced row ish in form off, eh? Say a prime. Okay, so scuse me. So that means, um, a prime times X equals zero. This system has has no pre variable because this system has has only has only intriguing solution that it means the system has no free bearable. Because we we know from our previous material we know that, uh, the system has nontrivial solution. It has at least a work repairable. So he&#x27;s just a negation of the statement. So has I&#x27;ll just write down here just for your information, A prime axe times. They&#x27;re a prime axe equal zero has has on Lee Trivial solution. It&#x27;s equivalent to say that this system has no free bearable. No, for you bearable. Okay, So for the Moore Demi&#x27;s matrix A has, um, has an people in positions. Of course. We&#x27;re assuming Matrix a had Matrix A&#x27;s On By and Square matrix. Yeah, this is our assumption. Um, people positions. Excuse me, people. Positions. And this is our conclusion here. Now, from here, we can say, um, since there are n people, people in positions and the matrix structure off, eh, Um, or we consider a prime and B and be seen as some some some values on the diagonal and zero for the lower triangular. An appetite for dinner. We have some numbers. I don&#x27;t. I just, uh, using darts to represent them. So, hey, in terms off a a prime time zags, we can see that of all the terrible here we are using a, uh, using this form of a prime then our nest our last variable se X end must be determined by the value we&#x27;re taking for B. So that means So we&#x27;re saying be one. So let&#x27;s suppose this is our vector B until fiend is equal to a prime times axe. And this is some sort of early on the diagonal times the vector X. So that means we were supposed defector ads his ex Juan x two txn Okay, So that means the last the last term Other day that I&#x27;ll be there diagonal If we multiply, multiply this This value by the nest. Last entry off. This factor acts. We have some. Aye, aye aye. Which is the entry for the last in the center of the diagonal of ape. Right? So? So there&#x27;s a II times the last the last available off the vector acts, which is X end is exactly the end. So we can solve for accent here. So that is bien divided by a II and so on so forth. So on so forth we can solve for everything Go bearable x one Until we were done, we have done for X one because, um, there&#x27;s no free bearable. So that&#x27;s why we can do this. So that means xn. That means acts the vector must exist. So what that means for the more damages the system a axe? It was B has a solution or every single b we&#x27;re taking the knob should be from are in

Donald Albin · Answer

All right. So we&#x27;re told that supposed that A and B are four by four in vertebral matrices if the determinant of a is negative two and the determinant of B is three than compute the determinant of b squared times The inverse of hey, so first thing the determined of the inverse of a is one over the determinant of a so determinant of the inverse of a is one over the determinant of a, um that is postulate 10 now the determinant of a times B According to postulate, nine is the determiner of a times the determinant of B So we really have here the determinant of be times b times the inverse of a So that&#x27;s the determinant of B times the determinant of B times the determinant of the inverse of a According to postulate nine determined B is three time determining Toby again. Times determine it of the inverse of a And so our answer would be negative. Nine has. So we only used postulates nine and 10 in your book

Donald Albin · Answer

Okay. This is a continuation that started a problem. Number 11. Um so let&#x27;s ah, let&#x27;s work on this one. Um, if the determinant of a is negative two than the determinant of negative A considering that it is a four by four matrix, if you take negative one times a then according to postulate four, the determination determinant of K times A is K to the end power times the determinant of a So we&#x27;re multiplying by negative one the whole matrix by negative one. So negative one to the fourth power because it is a four by four matrix is one. So the whole fact that that you&#x27;ve got a negative here means absolutely nothing. So now we just need to dio According Teoh Um according Teoh order my thinking here. You know what? Let&#x27;s do this. Mina Right this out. We already figured out that negative means nothing, So I&#x27;m just going to write eight of the third power to be and then another be OK, so we&#x27;re multiplying be times two right here. Okay. So, um, determinant of to be So this is the same thing. Um postulate four. That&#x27;s going to be two to the fourth power times the determinant of V. Whoops. Okay, so now we really have to To the fourth power times the determinant of a to the third power be to the second power. All right, so that&#x27;s two to the fourth power times. Um, the determinative A three times. So that would be three times three times three and determine it to be two times. Oh, man. A was negative to my bed back. Um, determined of a It was negative. Two negative, too. Times negative, too. A three times. Okay. And be two times and three. Okay. To to the fourth power times. Negative. Two times negative. Two times negative. Two times, three times three. Give me 1152. But I seem to think that the answer should be negative. The back of the book shows that the answer is positive, and so I&#x27;m wondering where I went wrong. So I&#x27;m gonna pause and think about where my mistake is and ah, I will get back to you. All right. I cannot find the mistake. Now, the fact that this is a four by four matrix would mean that this up here, negative one to the fourth Power is positive. Um, we need to take negative one to the end power and the end powers for because the four by four matrix, that would be negative one to the fourth power, which is one which means that this right here becomes meaningless. So that&#x27;s going to give us a to the third power to be to the second power. And eight of the third power is eight times eight times a. That&#x27;s the only place the negative can come in now. And that&#x27;s gonna be negative times negative have negative, which is positive. I mean, negative. Negative, negative, negative. So I&#x27;m going with the answer in the back of the book is wrong. Um, negative. 1152. Unless I&#x27;m missing something here. Thank you.

Problem

Suppose $A$ is a $3 \times 3$ matrix and $b$ is a…

Problem 33 Hard Difficulty

Suppose $A$ is a $4 \times 3$ matrix and $\mathbf{b}$ is a vector in $\mathbb{R}^{4}$ with the property that $A \mathbf{x}=\mathbf{b}$ has a unique solution. What can you say about the reduced echelon form of $A ?$ Justify your answer.

Answer

$\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \\ {0} & {0} & {0}\end{array}\right]$

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

Linear Algebra and Its Applications

Heather Z.

Alayna H.

Caleb E.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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$\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \\ {0} & {0} & {0}\end{array}\right]$

Linear Algebra and Its Applications

Heather Z.

Alayna H.

Caleb E.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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

Heather Z.

Alayna H.

Caleb E.

Kristen K.

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

Linear Algebra and Its Applications