prove-the-second-part-of-theorem-6-let-mathbfw-be-any-solution-of-a-mathbfxmathbfb-and-define-mathbf

Question

Prove the second part of Theorem $6 :$ Let $\mathbf{w}$ be any solution of $A \mathbf{x}=\mathbf{b},$ and define $\mathbf{v}_{h}=\mathbf{w}-\mathbf{p} .$ Show that $\mathbf{v}_{h}$ is a solution
of $A \mathbf{x}=\mathbf{0} .$ This shows that every solution of $A \mathbf{x}=\mathbf{b}$ has the
form $\mathbf{w}=\mathbf{p}+\mathbf{v}_{h},$ with $\mathbf{p}$ a particular solution of $A \mathbf{x}=\mathbf{b}$ and $\mathbf{v}_{h}$ a solution of $A \mathbf{x}=\mathbf{0} .$

Priyanka Sadarangani · Accepted Answer

okay in discussion. What has given his first question? Question X equals Toby is consistent for some given. Be okay. And a letter PBS solution off this. Okay. And further vector on this is the effect, er people us v edge on for their V edge is any solution for the homogeneous equation. X equals 20 Okay, so these are given and what we have to prove. We have to prove two things. First, any vector off the form. Yeah, people, us. We ash is the solution. Two x equals to be okay. And second thing, we have to prove that every solution toe X equals Toby has deformed p. Plus we h Okay, these two things we have to prove. Okay, so now we have now we will do the first proof. Okay, So the compute the given question is X was Toby for that? He is always a solution. Okay. Solution. So what? We write a X equals Toby for their We can write that a people us we edge. Okay. For solving this, we get a P plus a V edge. And for the discussion about we know that a P p is the solution off e x x equals Toby. Okay, so this will be be here. And V Edge Vector has a solution off a X equals to zero. So this will be zero. So this will be be okay, so we get a P plus we edge. It was to be so every vector of the form people us, we edge people. Us. We&#x27;re satisfied. The equation a x plus b. Okay, so this will be the proof off Approval one. Okay, now we have to prove second and the second waas second that we have to prove that every solution, every solution toe X equals Toby has deformed has the form p plus we h okay. It&#x27;s nothing so difficult. So let the blue be a solution. Let the blue B ISS solution toe a X equals Toby. Okay, now no, on w will be b plus something. Ok, we have to find this something. So on Wayne always that the blue minus p is a solution off homogeneous equation X equals 20 Okay, so when we saw a the blue minus p equals two for elaborated, it will be a w minus a p. Okay. And now the interesting point comes here that a w the W is the solution off X equals toe Be okay. W is the solution off a X equals Toby. So it will be be now, and B is earlier. We have brought that p is also solution off X equals Toby. So this will be also be so this will be zero hair. Okay, so 40 so we can write that w has the form p plus something. And what is something? It is the blue minus b. Okay, so, Doug, end the blue minus b is a solution. Toe the homogeneous equation X equals two zero. Okay, So in result, we can write as a V h equals toe a w minus B. That is a no loo minus a B. That is gentle. Thank you.

Matthew Allcock · Answer

this video, we&#x27;re gonna go to the answer to question number 15 from chapter 9.3. So we asked to prove that if x p s I spies the equation eight times at P B than every solution off the equation eight times X is equal to be is of the form X. Is this some off eh? Solution to the homogeneous branch of the equation, Plus that particular solution to the energy explosion. Okay, so first up less show that a ex actually does satisfy in have a genius for ages, so I x he&#x27;s gonna be a times I&#x27;m a genius Ex eclipse, the particular ex, then matrix multiplied by a vector is distributive, so that could be a witness. A times x h plus a times x p Yeah, eight times X age. We know that that&#x27;s just zero back toe on day of X p with human That that is a solution, as that&#x27;s just gonna be ableto back to be. That&#x27;s just back to bay. Okay, so now we know that the number off solutions thio this equation to the homogeneous equation is exactly the same as the number of solutions. Two the in Virginia&#x27;s question. Ah, so, yeah, we know that it that the number of solutions here and it&#x27;s exactly the same as, or rather the number, um, of yeah, ways that this quick at that. This expression off ex con B is the same as I think it were. X equals X Age. Because having on this x p doesn&#x27;t change the number off this gives you so their father can&#x27;t be any others. Can&#x27;t be any other solutions, whether the other than those that are off this fall. So therefore, any solution Ah, off the this in her Virginia&#x27;s question must be off this fork. It can&#x27;t be any of the form that it could be.

Rakvi . · Answer

fax Plus a is accorded Toby. Then you would subtract, hear from board decides so first step would definitely be during this competition.

Grace Muhihu · Answer

Okay, so for this problem, number one is given, So we&#x27;re gonna move on right to to a So for two A. We use the Iran 6.3 0.2, which is the second theorem in section 6.3. So it says that if a quadrilateral is a parallelogram than its opposite sides Arkan grew in. So that&#x27;s what them that we&#x27;re gonna use for to A to show that e f is congruent to h G now for three b. So for three b, we want to use the reflexive property of congruence. So with this, since we&#x27;re not really working with any angles just yet, we just have sides. So we want e age to be concurrent to E. H. And the reason for that is we want we want some sides of of the angles to be congruent to each other. So this is gonna be the reflexive one that we use okay Now for for C So for four c, we want the triangles e f age and h g e to be congruent. But how can we say that well, we have e g is can go into a chap an E f is converting to G H and e h is concurrent to e h smell means we have to use a side side, side congruence, triangle congruence there. Um, so that&#x27;s how we&#x27;re gonna That&#x27;s the reason that we&#x27;re going to use for foresees A side side, side triangle congruence here, Um, now, for five D. So for five d, we need to figure out which angle is f e age congruent to. Well, if we use corresponding angles with the course with the congruent triangles, we&#x27;ll see that we get angle f e age has to be congruent to angle G h e okay, And how How can we show that? So by what reasoning? Well, we were just saying that the because the triangles are congruent, That means we need, um, Anglos to be corresponding. So we&#x27;re gonna I&#x27;m gonna abbreviate this reasoning, but it&#x27;s basically the corresponding parts of congruent triangles are congruent, so I&#x27;m just going to abbreviate thought to see p c t. C, which basically says if the triangles air congruent than its corresponding parts are also growing. Okay, now for six f, so for six f, we want angle F e h and G h e to be supplementary. And this is the case. Because, remember, we&#x27;re working with the parallelogram, So same side, interior, same side angles of ah, parallelogram are supplemented or they add up to 1 80 So So that&#x27;s basically, um, the reason that we&#x27;re going to use for this one. So consecutive consecutive angles off a parallelogram are supplementary. And that&#x27;s the reasoning for six up now for seven G. So for seven g, we want to use the congruence. Supplementary angles are right angles. So again, if those two angles f e h and G h er supplementary. So that means that either one of them has to be 90 degrees. So we&#x27;re basically to say that the measure off angle F E H is equal to 90 degrees or is a right angle. And for that, um, we&#x27;ll see that we can refer to, you know, someone from 6.5. But we&#x27;ll get to that in a second. So now, for eight h So now we want toe finalize that e f g h is a rectangle and again from serum. So this is what I was saying before your, um, 6.5 point one so that was going to allow us to say that E F G h is a rectangle by definition of rectangle. So definition off directing, which basically, is it the idea that if one angle of the parallelogram is a right angle than that, parallelogram has to be a rectangle, and that&#x27;s how we complete this proof.

Ankit Gupta · Answer

so Hellestrae Nowhere. But understand this question. It is given Let bxb poor normal with real co person and let be positive you was the division. So be X. If you do, I&#x27;d say, uh, X minus b explain it to be delighted and cute X magical sent. And Audie that a window So are is always a good rifle. Did you go to the window? Cannot be zero. The winter can be zero demanded Radio one Liwen. The Parliament BX is completely ritual by expense. P r can say expense be the fact European, then are zero in other cases are greater than projected, though no, it is given that it is also giving that there is one number Jed which is a good within B. So PGE it to award the P Genesis option A beach it this just hopeless exited Jed minus B on cue. Jed, blessed are not us Checked the sign off a few weeks. Eso you feed it so it is given that jade is greater than be it with this positive is this It is also given the art is greater than or equal do you? And if you exceed the question that is always funny. So therefore, this all number, all right inside the Polish what you do so we can we can conclude that b J it Did you connect with them? Do you to be not realistic? B x The first part B exit is given that x minus beat q x. For this. This is Gerard X Acquired will be their hope. Upper limit Upper limit execute It&#x27;ll be and support be X explicit a Q X plus part. So here he is. Positively. I rendered acts. So here the lower limit Lord limit is X equal minus it left a deposit agent and for the sea, the 2nd 2nd part B j equals toe Jed minus B and huger. Blessed Art. So now it&#x27;s telling that used the first part to find the lower limit hobbling out this What not just the bliss degenerate ex missus acquired Toe X minus B and que x. Let&#x27;s part here. You&#x27;ll find that a problem. It I just wanted to be B minus six, which is minus X. Let&#x27;s be Q X. Let&#x27;s our bodies apology to and here the lower limit. You&#x27;ll find that Lord limit, which is the goal of the Cejudo at Xel minus P. So a lot of women minus B

Vikash Ranjan · Answer

the bulls grass be crossing? Will it are saying B minus a B industry. Therefore equal if it went is it me meals? You, uh, visit? Huh? We have last cities see equal to Sears. Going sees it. Therefore, we have first of all, be frosty equal to be one Caesar minus. Be there. See what? My message e and beings senior minus beer sayings Glass Keep cap being C minus. Things being Finally, we have a cross being girls see on this equal to God a one B s B C s. So this is less here is in being cesium minus. Is there these air? Sea X minus Jake back if the x the women x to be wealthy ex My last video be while seizure plus it in these in glass. Get God, We have this. You got to do my guess. Hey, X being See there Glass here? Yes, These n c x man be one season left a way bigger now police to have on the subject. That is, uh, some drag and being seems in all the dump. That is the terms of I that comes off the end of the terms of K therefore will pay for standard is I? He won&#x27;t being see why. Minus 31 b one c x place is a B C there. My last Asia busier seeks. Yes, we have. You think Well, yes. Yeah, I see. What? Plus in a season. My last c x, it won&#x27;t be y plus is in He is a plus and bx c x This this? Yes, my nasa iss Bs six way Have this equal. Yes. You want to see when last his in civilian clothes and seeing my CS, You might be right. Is there a busying class airs? Yes. So far we have being into headed up seeing my ask six into a B for I&#x27;m similarly full day and maybe have be in our embassy My house to see why it r b for the end. So this isn&#x27;t on the day. I don&#x27;t see my nest season. It are so summing up. This&#x27;ll equal eight cross big, grassy equal dude a Darcy into B minus a doctor. Be embassy and stuff. It is boom thing Is that

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Suppose $A \mathbf{x}=\mathbf{b}$ has a solution.…

Problem 25 Easy Difficulty

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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

Anna Marie V.

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Absolute Value - Example 1

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

Anna Marie V.

Heather Z.

Alayna H.

Kayleah T.

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

Linear Algebra and Its Applications