in-exercises-3136-respond-as-comprehensively-as-possible-and-justify-your-answer-if-p-is-a-5-times-5

Question

In Exercises 31–36, respond as comprehensively as possible, and justify your answer.
If $P$ is a $5 	imes 5$ matrix and Nul $P$ is the zero subspace, what can you say about solutions of equations of the form $P \mathbf{x}=\mathbf{b}$ for $\mathbf{b}$ in $\mathbb{R}^{5} ?$

Gideon Idumah · Accepted Answer

So for this question, say&#x27;s the no space off B is the zero subspace That is, of course, to zero. Then this implies that speedy multiply by extra quarter zero as only the trivial solution as only a trivial solution. What I mean by this is dots only X equal to zero. That&#x27;s what I mean by tree Vo con Mick P. Times X equals zero. No, says p squared. Then we can use the invincible metrics Xyrem. Not since section Superman 30 and we have That&#x27;s hey is invincible. And what this implies is that P eggs equal Toby as unique solution for every be in our five.

Gideon Idumah · Answer

So for this question, since the no space off our it&#x27;s not the zero subspace this implies darts our eggs equal to zero gifts in nontrivial solution. And what I mean by a nontrivial solution is that that sees there is so it is an east. There is there is a ex not 1/4 0 such guts. Our X equal to zero No, since are is square, then by D in vegetable metrics serum. This is found in section Super Intern. In your textbook, it&#x27;s implies that are is not in Vaasa ble on. This would imply that the columns off our ah linearly dependence. Because of that, if it is not invincible, then the columns off our linearly dependence. Now the conclusion is, that is the column space off our is a subspace off our six. But since the columns off our legal independence boats, this would imply that column space off. Our cannot be equal to our six

Joshua Sieverding · Answer

Okay, so we&#x27;ve got a system of equations and were asked to solve. What we&#x27;re going to do is take all of the coefficients in front of our variables X, y and Z, and put them into this matrix so that this first column is all of the coefficients in front exes wise z&#x27;s. And this last one is the answers. So we get the negative one from here, which is here. And, like we get this native three there goes to that, uh, pretty straightforward from here. All that we want to do is roll reductions until we simplify our left chunk of this matrix to the left of the dotted line into reduced echelon form. So the first thing I&#x27;m gonna do is move this, uh, bottom row to the top. Then I&#x27;m gonna take this row and add it to what was already at the top, but will move that to the bottom. So we get a zero a three, a negative one. When we get this five added to zero. So we would get a five here, and then for now, we will leave the middle row alone. Okay, So from this spot, we have a couple options of what we can do. In any case, we&#x27;re gonna leave the Kaprow alone because it&#x27;s in a nice form. Uh, and we&#x27;ll for now we&#x27;ll leave the second row alone, too. If we take this second row, multiply it by a 3/5 and add it to the third row. We&#x27;ll get rid of this three. So let&#x27;s do that. So 3/5 times negative five goes this negative three added the three gives a zero 3/5. Times two gives us 6/5. Added to negative one gives us one this and 3/5 added to negative three. Well, give us negative. 9/5. Five is 25 5th So adding that negative 9/5 to that. Well, give us, um 16 5th 16/5, and then I&#x27;m actually gonna take a second to erase both those. If we multiply this whole row by five, we&#x27;ll get it to nice integers one and six, one in 16. So let&#x27;s do that. This could be at one. This could be a 16. Cool. From here, we&#x27;re gonna subtract Ah, two of row three from row two and we&#x27;re kind of going all over the place. here without work. Sorry for that. Our top row is going to stay the same again. Our bottom row&#x27;s gonna stay the same, but our middle row will become zero negative five. Um and we&#x27;re multiplying by two and subtracting. And so we&#x27;ll get zero. And this would be done. 32 negative three minus 32 would be negative. 30 five. And we can divide both of these by negative five. Uh, and we&#x27;ll get a one and a seven. So let&#x27;s do that quick. One here and seven. So we&#x27;ve got this left side to look like the identity matrix. So what this is telling us is that X equals five. Because remember, this first column was for our exes Wise ces answers. This says that why is equal to seven and Z is equal to 16. So that&#x27;s the final answer

Brian Ketelobeter · Answer

Okay, Um, let&#x27;s try this again. It&#x27;s still doing the same weird thing. Okay, so I have matrix a here. This is a problem. Number 11 from section 2.8. I&#x27;m given a matrix A, um and I&#x27;m after the null space and the cone space. Okay, so on drily, this is just the null space is just a question of matrix multiplication. So we&#x27;re after all the vectors accession X equals zero and just examining matrix multiplication. You can see that the vectors X would have to be in our four. So that means that the, um uh the, um que ah que would have to equal to four. Um, this is a subspace of our four. I&#x27;m sorry. I don&#x27;t know what&#x27;s going on here all of a sudden, um, this is contained in, um are sore. Okay. On dso Similarly, when we look at the column space So we&#x27;re looking at the span of the columns, and just by looking at the columns, you can see that there three, they have three coordinates. So that means that the column space, um, is contained in our three. So the column space is a subspace of our three And that means that he would have to be, um uh in P would have to be three p. Just being the thes subspace are the dimension of the subspace in which the Colin spaces in. So just by looking at that, you can see that the Colin space is contained in our three. So, uh, yeah, so the null space eyes all the vectors acts such an ax equals zero. And just by looking at the the number of columns in a, you can see that it&#x27;s four. So that means that it would have to be a subspace of our four. And the column space is just the subspace spanned by the columns. And when you look at the columns, you can see that they&#x27;re three dimensional vectors. So that means that it would have to be contained in art. So that&#x27;s it for this problem here. Thank you very much and have a nice

Ankit Gupta · Answer

artist quit eggs. It was to be white plus five. Or this can be returning out of square X minus B ways because to fight the form off the commission is off X plus b white. It was to see that implies the question is lenient here the value off a Is it close to our square BZ cause to minus B and see it because to fight?

Viswesh Uppalapati · Answer

in this video, we&#x27;re gonna be solving problem number six of section 4.1. And it asks us to determine if the given set is a subspace of piece of men for inappropriate for any appropriate value of n. Um, And the given polynomial we have is p f t equals a plus t squared where a is any real number. So this is pretty simple. Um, for example, it could be any real numbers. So let&#x27;s just choose four plus t squared as our polynomial for now. Um, And for for this particular set of calling almost to be in peace of end, you have to be closed under scaler multiplication. So the determine if it is closed under Scott wealth creation, we can just multiply by like like any number. Let&#x27;s say three. So we got 12 plus three t square. Um, for for this example polynomial, um, we see that 12 is still in still a part of all real numbers, because is an all real numbers. But three t squared is doesn&#x27;t fit fit our original definition of pft because gift, he is a plus t squared where he has a coefficient of one. But here. He has a quick, efficient of three. So this set is not close under skill multiplication. So it&#x27;s not a subspace of piece of in yeah, and that&#x27;s that&#x27;s basically it.

Problem

In Exercises 31–36, respond as comprehensively as…

Problem 34 Medium Difficulty

In Exercises 31–36, respond as comprehensively as possible, and justify your answer.
If $P$ is a $5 \times 5$ matrix and Nul $P$ is the zero subspace, what can you say about solutions of equations of the form $P \mathbf{x}=\mathbf{b}$ for $\mathbf{b}$ in $\mathbb{R}^{5} ?$

Answer

If $\mathrm{NulP}=\{0\},$ then the equation $P x=0$ has only the trivial solution. Since $\mathrm{P}$ is square, the IMT shows that $P$ is invertible and the equation $P x=b$ has a solution for each $\mathrm{bin} R^{5} .$ Also, each solution is unique, by Theorem 5 in Section $2.2 .$

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

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Alayna H.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

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If $\mathrm{NulP}=\{0\},$ then the equation $P x=0$ has only the trivial solution. Since $\mathrm{P}$ is square, the IMT shows that $P$ is invertible and the equation $P x=b$ has a solution for each $\mathrm{bin} R^{5} .$ Also, each solution is unique, by Theorem 5 in Section $2.2 .$

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Alayna H.

Caleb E.

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.

Alayna H.

Caleb E.

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

Linear Algebra and Its Applications