given-aleftbeginarrayrr4-6-8-12-6-9endarrayright-find-one-nontrivial-solution-of-a-mathbfxmathbf0-by

Question

Given $A=\left[\begin{array}{rr}{4} & {-6} \ {-8} & {12} \ {6} & {-9}\end{array}ight],$ find one nontrivial solution of $A \mathbf{x}=\mathbf{0}$ by inspection.

Runpeng Li · Accepted Answer

this this problem work even the Matrix or negative six connective eight and 12 and six active at night. Now we need to find the one country solution off X equals zero system. You were considering this homogeneous system. X equals zero. So we first do the Calgary gushing elimination. We observed that the second rule is actually connected with twice next two times, the times the first roll. So I can just, ah, cancel the second rule by by adding the first roll times too. So that&#x27;s for an active six and 00 I also observed that weekend, uh, he was the third role. Here&#x27;s the 33rd World minus, uh, find out of 33 minus 3/2 off the first rule. So that is, uh, 3/2 off for that is six and six minus 60 and 3/2 off. Next six will be next. Nyah, I reckon I minus neck tonight is zero. So finally we have I only we have 34 x one minus six x for X two is zero. So to find a nontrivial solution, we just take X one as three and next to as to so three times worries 12 and two times six is 12 and 12 months minus well, zero. So we&#x27;re done

Michael Jacobsen · Answer

In this example, we have a three by two matrix say that we&#x27;re considering now. It&#x27;s also considered this equation eight times X equals zero vector. So we have a homogeneous system of equations, and we know that we always have the trivial solution, which is X equals zero vector. Now we wonder, can we come up with other solutions? Certainly, if we put all of this information into augmented matrix, we could reduce. But for the sake of becoming more fluent in matrix algebra, let&#x27;s find a solution by inspection. Well, to do that, let&#x27;s expand. This makes tricks equation in terms of a The Matrix, which is negative. 27 negative three for calm, one negative 6 21 negative nine for column, too. The Vector X is equal to x one x two The unknowns of the system of equations and the right hand side is 00 So the next thing we can do to investigate how to determine what the solution is is think about Tell Matrix versus a vector. Multiplication works. If you&#x27;re fluent with matrix algebra, you might see the solution already, but that&#x27;s okay if you don&#x27;t this takes some practice. Let&#x27;s convert this into a vector equation. To see this more easily, we can write x one times the first calm of a negative to seven negative three plus x two times The second column of The Matrix A, which is negative 6 21 negative nine, is equal to the zero Vector. But I&#x27;m noticing my zero vector should have three rows. So let me give it a quick promotion here. There we go. The next thing we can look at to devise the solution without row reduction is to look at equality here. Equality would mean that we have in Row one negative two x one minus six x two must be equal to zero. In other words, negative two x one must be equal to six x two and we could write se X one equals negative three x two. So whatever the solutions are, they must follow this relationship, and that gives us a hint. Already, it&#x27;s effectively telling us if we take the first calm, multiply it by negative three, then added to this column. We should get a zero vector. Let&#x27;s check that out. Let&#x27;s set X one equal to negative three. Here it&#x27;s multiplying negative to seven. Negative three plus x two. I&#x27;m going to leave as a one, so we have here negative. 6 21 negative nine. I&#x27;ll put a one here just to remind myself that that&#x27;s the value I used. Four X two has now said equal to the zero vector. So by inspection, this in fact does work. If we do this killer multiplication, say, in the second row we get negative 21 plus positive 21 giving us a zero, and this works out for every single row. So effectively, what we could have done toe. Find the solution x one x two By inspection, It&#x27;s just decide. How could I multiply on entry here with this entry to get to zero? Then just check if those values for X one x two work for the next two rows. But for our case, we took a few extra steps which is helpful to do as well. And we found that the solution, either by inspection or by looking at the vector equation, is negative. Three for X one and one for X two

Grace Muhihu · Answer

okay to find a here and when it&#x27;s that Q of a equal to cure A. So with that, I&#x27;m going to do six a minus one is equal to negative nine. So then here I can add one to both sides. So the ones dropouts I get that six a is equal to negative E. Then I just have to divide both sides by six. So the six drops now. And if I simplify this native, be over six Aiken, divide the top and bottom by two. So that&#x27;ll give me negative 4/3 so a is equal to negative 4/3.

Anna Jones · Answer

So when you first look at this problem in number 30 you may recognize that the new dominators look very similar, but they&#x27;re not the exact same expression. So these first steps here that you can do are actually to rewrite the fraction on the left hand side to match the denominator of the right hand side. So in this expression, we could rewrite this denominator by factoring out a negative one. On when we do that, this becomes negative, multiplied by a minus eight. So multi taking out that negative one and re writing this expression inside with the opposite signs. Um and then we know that fractions can be written with the top negative or that bottom negative. So this middle fraction is equivalent to bringing that negative one up to the numerator, so becomes negative for over eight minus eight. So now I can rewrite the given expression with negative for over a minus eight equals four minus a over a minus eight. At this point, with both denominator is being equal. We know that the denominator cannot equal zero, so a minus eight can never equal zero, which means a cannot equal eight because eight would make that denominator zero again. That&#x27;s because we cannot divide by zero. So now we can multiply by the denominator a minus eight, the whole expression whilst pie by both sides of this equation. When we do that, all the denominators were actually cancel with that a minus eight. And so we&#x27;re left with just negative. Four equals four minus A. Our final step here is to simplify, to be have a equal to something. So we subtract four and then divide by a negative one, and we find that a would equal eight. Well, we know that they cannot equal eight based on the domain of our function, So this equation has no solution.

Tim Strang · Answer

Everyone said the problem we&#x27;re looking at today gives us a system of linear equations written in this form. All right, well, well, technical difficulty, but rhythm this form. So where ah, ex prime and X or two column vectors of length to and A is a two by two matrix that describes their relationship. And, um, we&#x27;re going to use Jordan canonical forms to sell this and show you how so it all is based off the transformation of letting X equal s y for some some s. And basically what that gives us is we couldn&#x27;t rewrite the problem in tow. Why Prime being s inverse added, We&#x27;re letting SP in veritable because we&#x27;re choosing s, um A s, Times X. Okay, Now, if this seems familiar, it&#x27;s because well, if we choose the correct s which we can solve four based on a uniquely we can get this equal to J, which is the Jordan canonical form of sorry, that&#x27;s not supposed to be the next. That&#x27;s supposed to be a why it&#x27;s all just why Why is it that call infected applying to again so we can get it into the for my prime equals? Ah, joint economic reform times. Why? And basically with these two going back and forth solving for y and then ah, knowing s and plugging it in for ah, this first equation, we can solve the system of linearly independent upside the system of differential equations. I don&#x27;t know. I said linearly independent. But you&#x27;ll see there are two things we need to solve or four, right? So it&#x27;s Thean vertebral Matrix s such that this holds and it&#x27;s the Jordan canonical form of A so well, let&#x27;s start with the Jordan canonical form and s will follow. So first we need to find the Eigen values of a. So we get the characteristic polynomial that a minus slammed. I set that to zero. What does that look like for two by two? Something like this determinant. And ah, well, that&#x27;s that&#x27;s you multiply across and subtract the, uh, off off diagonal terms. So this will look something like this is taking the regular determinant on minus minus one, and that is going to go to the multiply it out six times for his 24 plus one is 25 25 plus 10. Lambda slammed a squared and we want to set that all cereal, so you could probably fact that it&#x27;s pretty easy. Looks pretty meat, but we&#x27;re going to use the quadratic formula, so Lamda equals negative B plus or minus the square root of B squared minus four. A. C L. A is one and C is 25. That&#x27;s gonna be 100. And we see that this cancels out all over to a So there is Onley one lambda because it&#x27;s, ah, land equals negative five and then we have this plus or minus zero. So negative five plus Syria is negative. Five minus euro. It&#x27;s all the same. So it has a multiplicity of to Okay, it&#x27;s gonna mark that for later. That&#x27;s one finding that we have. But we still don&#x27;t know how many Jordan blocks there are. Even if we know that the Lambda equals negative, five will be the diagonal. Um, so we have a theory that states that the number of Jordan blocks in a matrix is going to equal the number of linearly independent Eigen vectors. And since we have ah one Eigen value, there&#x27;s going to be at least one and it has multiplicity of two. So there&#x27;s gonna be a maximum of two. So we have to separate options. We need to know which one we&#x27;re working. So in order to do that, we need to calculate the Eigen vectors. So we have our again vector equation where V is the Eigen Vector and weakness. 1,000,000,005 78 plus five i b. And we want to set that equal to zero. Right? That&#x27;s that&#x27;s just the definition of an Eigen vector. So a plus five is gonna be something like That&#x27;s wrong. It&#x27;s 11 negative one that one and augmented matrix Self owns a V well, weaken very quickly. Reduce this by just adding the first throw to the 2nd 1 Well, uh, so what this tells us is that one plus the first element of E what, one times 11 of the first element of B plus one of the second element envies an equal zero. So the first element is negative. The second element. So we because, well, we can just write arbitrarily one negative one right? And because this rare was all zeros, we have a free variable and we can multiply this by whatever we want. The only restriction is that one is negative of the other. So that is our Eigen vector. And that is actually the only hide in vector. Right, Because there&#x27;s no way we can split that up anymore. There&#x27;s only one free variable. Yes, So that&#x27;s Ah, that&#x27;s the vector. Um, so now we are poised to write what? We have one vector so we can write J. Jordan. Canonical form of A is ah, Eigen. Values on the diagonal, zero below. And Jordan canonical form, uh, one Jordan block. But to buy to We&#x27;re just gonna sort of glue these fives together with this one to get this one block right here. Perfect. Okay, that&#x27;s her Jordan canonical form. Excellent. So we have found our J, but a little check Founder Jay. Now all we need to do is find our s and will be ready to solve. So what is Rs? Well, we have, um, we we we know how to do this regularly. If it&#x27;s a diagonal Izabal matrix, a vase, diagonal, izabal. But it&#x27;s not because it&#x27;s Jordan. Canonical form is not diagonal. So, um, normally you would find the Eigen vectors and put them Ah, based on which Eigen value they correspond to into a matrix where the vectors are the columns. But in this case, we need the generalized Eigen vectors. So we need to write out, um, the elements of this cycle and you recall a cycle is some set of the form A minus slammed I That&#x27;s very common. Recurring theme times. Well, savy prime for a generalized Eigen vector. Um, this is the end minus one. Right? So it&#x27;s just a collection of vectors multiplied by successively greater powers of a minus Lambda I and we need, um Well, first of all, this is a two by two matrix. So s is logically, only going to be, ah, two by two matrix itself. Otherwise they wouldn&#x27;t multiply and has to be square. We can see from this, um and we see that that&#x27;s confirmed by the fact that we have a theory. Um, that says, thea length of the cycle for a generalized Eigen vector is equal to the size of the Jordan block, which is a size two, right there. Two elements that the one is going together, and, um so we need to find a cycle of length to so basically just cut it off right here, Right. And, ah, for a cycle to be a cycle, the&#x27;s cannot be zero, right? That wouldn&#x27;t really give us anything interesting. So, um, as opposed to an Eigen vector where this is definitional e zero Generalized. Eigen vector actually has the next one up is equal to zero, So Ah, a minus slammed I squared V Prime does equal zero. Right? And that would be the next one in the theoretical set. But it is hero, and, uh, we&#x27;re letting a minus slammed ivy prime not be zero. So, um, is there a to constraints? You re read him over here. Perfect. Okay, so we can go about solving this. We know what a minus lambda eyes, but we do not have yet a modest fund. Eyes squared. Um, so many minus lambda I for five. Negative five. That&#x27;s going to be again, Uh, 11 negative one negative one. And we just wanna multiply that by itself. Says some just quick matrix multiplication here. That&#x27;s going to be That&#x27;s terrible. I made it worse. That&#x27;s going to be in the first slot. Zero second slot is zero. There&#x27;d slot 00 Okay, so ah, any generalized Eigen vectors such that this condition holds, um will actually follow that this condition holds because, well, a minus slammed I squared is too serious. So it times any vector is going to be zero itself. So we just need this condition and that&#x27;s pretty easy to find. Basically, just can&#x27;t be Anak Chua Lagon vector. So it can&#x27;t be this V. It has to be linearly independent. So Well, what&#x27;s something that&#x27;s just sort of easily linearly Independent from V. Let&#x27;s keep it on the frame. V um, he prime pick your favorite victor. Perfect, Right, That&#x27;s pretty easily linearly independent from the, uh, lovely. And then we need to find for, uh So the columns. They&#x27;re going to be this in this of rs. So we need to multiply a minus lambda I times v prime to find our first column is the V prime is gonna be your second columns. Well, a minus slammed I one negative one one negative. One times 10 That&#x27;s going to just be one negative one, right, because it&#x27;s just taking the first column, and that is going to be our second our first column in the S. But our second sort of element in this cycle. Okay, so we are ready. We have found s we have found J. Now, let&#x27;s set this up. Let us just write down s on its own. Just your future reference One negative One on zero. Perfect. All right, now going back up. So we have, um Well, let&#x27;s solve this equation first. And then we can convert the wise that we found. So why prime equals J times? Why? Why Prime equals J. Why? So, uh well, what&#x27;s J? So if we right, Why prime is why one? Why to prime crime. Why is why one y two just sort of quickly multiply it out. Um, so why one prime is going to equal and why to crime is gonna be just Ah, negative five. Why? To write. And ah, this isn&#x27;t as good. Normally, if if it were a diagonal Izabal matrix, if a was then we wouldn&#x27;t have this term. And I would just be too unmixed differential equations. But by putting it in Jordan Jordan canonical form, you get at least one on mixed differential equation which we can solve. We know that Ah, derivative That equals a constant times. It&#x27;s it&#x27;s self right. Like y two Prime Nichols Negative five white to Well, you know why two is just gonna equal you to the needed the negative five times tea Because tea, Zahra Traditional. Ah, independent variable. I don&#x27;t know. I said by square, it&#x27;s physically a subscript, so that&#x27;s gonna be why two equals. That&#x27;s gonna be y two y two of tea. And you can just check real quick that this this is solved. But ah, anyways, that&#x27;s taken care of. So now we need to find, uh, why one So why one prime plus five? Why one? Oh, and we can Ah, at a constant of integration, right, Because this is differential equation. So we&#x27;re basically a roundabout way of taking an integral so always add, this is very important. If you want to think of dividing both sides of this by why two and then taking the derivative taking it taking ah, okay, integral with respect to t, you can understand why this is multiplied rather than ah added. But that&#x27;s that&#x27;s basic calculus. I&#x27;m not gonna go every in depth here, So this is gonna equal why to t and that c one e negative. Five TV. Okay. All right. So, um, you remember from earlier in the book, Let&#x27;s just ah, take this down and get it a little more separated so they can take a better look. Remember from earlier in the book, this equation. Uh, well, if we multiply through by an integrating factor which we didn&#x27;t know, I t and we let that equal eat of the five T. If you multiply that through in here, we get speaking, multiplying both sides. Yes. Why? One prime e to the five. T plus five by one. You did the five t equals C one because he&#x27;s cancel out right each of the five t sometimes eat a negative five g. Okay. And, uh, if you already know where this is going, that&#x27;s great. But if you if you just look here, you can check me on this, but it&#x27;s it&#x27;s pretty straightforward. Ah, you might not see it on your own, but it&#x27;s it&#x27;s you can tell that I works. Um, that&#x27;s just going to be equal to why one either the five t crime. So the whole the derivative of that. And you do? Ah, with respect to t and you do chain rule and at sorry, product rule. And you get this. And that&#x27;s gonna equal C one so we can just integrate and why one each of the five t equals C won t plus C two. So we get that. Why one is going to be Ah, perfect. Okay, so now we can write our why. And remember, this is ah, column Vector is going to be sorry. That&#x27;s supposed to be Why one it&#x27;s gonna be Mr Why one and why? To Excellent. OK, so this is important, but it&#x27;s not quite we need to solve for X. Right. Um, so we&#x27;ve done this equation. Let&#x27;s just cross it right off. And now we need to focus right here. So X equals s why. Well, what&#x27;s rs right? Were a better earlier Go come into play. So X equals, that&#x27;s why we just multiply it out. So X equals, was it zero times? And I&#x27;m just gonna right? Why? Because we have it right here on. So we get that X equals this that get any more space for this? Sorry, guys. Right. And you can simplify that first equation. Uh, but it&#x27;s not important, not super important. You&#x27;ve solved the differential equation. And basically, that&#x27;s how you saw pretty much everything in this form of problem. And if you see that the joint economical form is diagonal, then it just reduces to irregular, non defective argument. But ah, this is just generally, it will always work for you. Uh, every matrix says it or economical form. So this Yeah, this is a very powerful tool. Um, hope this video is helpful. And, uh, thanks for watching.

Jose Avalos · Answer

determine whether the given number is a solution to the equation. To determine if a given number as a solution, we&#x27;re going to substitute the value into the equation. So we have four times 18 equal 72 multiply 18 times for and we get 72. 72 is equal to 72 so a equals 18 is a solution.

Problem

Construct a $3 \times 3$ nonzero matrix $A$ such …

Problem 34 Medium Difficulty

Given $A=\left[\begin{array}{rr}{4} & {-6} \\ {-8} & {12} \\ {6} & {-9}\end{array}\right],$ find one nontrivial solution of $A \mathbf{x}=\mathbf{0}$ by inspection.

Answer

Inspect how the columns $a_{1}$ and $a_{2}$ of $A$ are related. The second column is $-3 / 2$ times the first. Put another way, $3 a_{1}+2 a_{2}=0 .$ Thus $\left[\begin{array}{l}{3} \\ {2}\end{array}\right]$ satisfies $A x=0$

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

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

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Absolute Value - Example 1

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Inspect how the columns $a_{1}$ and $a_{2}$ of $A$ are related. The second column is $-3 / 2$ times the first. Put another way, $3 a_{1}+2 a_{2}=0 .$ Thus $\left[\begin{array}{l}{3} \\ {2}\end{array}\right]$ satisfies $A x=0$

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

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

Anna Marie V.

Heather Z.

Caleb E.

Michael J.

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

Linear Algebra and Its Applications