in-exercises-5-8-find-the-steady-state-vector-leftbeginarrayccc7-1-1-2-8-2-1-1-7endarrayright

Question

In Exercises $5-8,$ find the steady-state vector.
$$
\left[\begin{array}{ccc}{.7} & {.1} & {.1} \ {.2} & {.8} & {.2} \ {.1} & {.1} & {.7}\end{array}ight]
$$

Michael Jacobsen · Accepted Answer

in this example, we have a three by three probability Matrix P that&#x27;s provided and Arkle here is to find the steady state vector. The first step in this process is going to be to solve the equation that p of X is equal to X. We could rearrange this equation. So vets of the form P minus the identity matrix Times X equals the zero vector. Now, from this particular matrix equation, we can augment as follows First P minus. I is going to cause a subtraction by one just along the main diagonal. So I&#x27;m going to be copying the entries otherwise So for the first entry, take 10.7 minus one for a negative 10.3, then copy 0.1 and 0.1 for row two will copy point to then take the entry 0.8 minus one so that we have a negative point to here. Now back to copying points to 0.1 point one. Then take 0.7 minus one so that we have another negative 10.3 that will go here. So this is P minus I and what we would like to do is augment this with zero vector. So let me place this zero vector here and now Finally closed this matrix to reduce this matrix weaken taken initial step where we do the following I&#x27;m going to multiply row three by 10 So that will have integers and make it a row one So now we have a 11 negative 30 in row one, I&#x27;ll make the old row one here Row three and also multiplied by 10. So we have a negative 100.3 110 Then multiply row to buy 10 So that will have to negative 22 and zero. Now with this matrix weaken, begin road reduction as usual and the arithmetic Matic operations won&#x27;t be as bad Now that we have integers will find that this is row equivalent to 10 negative one zero in row one row to is 01 negative to zero and the third row contains only zeros. This that implies that the solution for the vector X indicated here will have first next one equals x three, which came from the first row From the second row we find that X two equals two x three and x three is the free variable soldiers right X three equals X three. Now we have infinitely many solutions, but we just need one. So it&#x27;s set X three equal toe one. Then the solution X is going to be equal to 1 to 1. We have to be careful, though, since this is not the steady state vector, since it&#x27;s not even a probability vector. But it&#x27;s very close to being on the right track. If we take the some of the entries in this vector x one plus two plus one, we have all together four. And now we can claim that our answer is Cue the steady state vector for in spite, dividing this vector X by four So we&#x27;ll have 1/4 2 4th or 1/2 if we reduce and then 1/4. So this vector Q is the steady state vector for the Matrix P.

Amy Jiang · Answer

Okay. So if p X is equal to X, we can rewrite that as people in his eye times exited to Joe. Well, that&#x27;s this P. Marcus. I is 0.3, 2.20 negative, 0.8 point four, 3.6 and 0.6. Okay, if we wrote this this with our Joe&#x27;s. Oh, those You get that? This reduces too. 10 noted one. So there are one negative 1/2 0 Joe Zozo, though. Okay, so we have exes are matrix x one x two, the next three, and we get that. Oh, x three times 1 1/2 and one. So one of our solutions is just to one and two. Okay. Since you sum up to five, we divide the sorting by five. So he gets our solution is just to over 51 over five and two over five.

Amy Jiang · Answer

Okay, so we have P X is equal to X, but we can rewrite that. As P one is. I was ecstatic with your help. So we have people eyes. I is equal to negative 0.2 point five points to an anchor 50.5. Okay, If we&#x27;re over this, we can solve our matrix. We get one *** five over too thorough until Joe&#x27;s up. Okay, so we know that X physical to X one x two, which is equal to x two times, um, five over to print one. Right. We said our X one wrecker is this. Okay, so we have one of our solutions. That&#x27;s just five over, too. Okay. And since the entries some up to seven, he does not mind sets. Q is equal to five over 7 to 7. And that&#x27;s his equal to right. Actually, that&#x27;s decision

Ryan Sander · Answer

Hello. In this video we want to find a vector a unit vector in the direction of the given vector and are given Vector, which will call B is equal to it has seven force in the first component, one half of the second component and one in the third component. So if we were to draw this out in three D Um, and this is the X one component X two x three We&#x27;ll just call these x one x two and X three, uh, three D could be kind of hard to visualize, but we&#x27;ll do our best year. Uh, that would look something like this where? Well, project it down and this will be our point. So be will point in this direction. And in linear algebra and a lot of other, um, stem fields, it&#x27;s really important to find not just vectors but unit vectors, which point in the same direction as vectors. I will call this be hat, but an important constraint. Um, in addition to be hat pointing, being in parallel with B is that the magnitude of the hat is equal to one. And to do that to find this be hat vector How do we confined this by simply taking our original vector B and dividing it by its magnitude? So this is going to be equal to the Vector seven Force one half one, provided by the magnitude of this factor, which is the square root of the first component squared, which is 7/4 squared. Plus the second component scored, plus the last component squared, which on the denominator here gives us the square root of 74 squared is 49/16 plus 1/4 plus one, which is equal to the square root of If we get a common denominator here of 16 it&#x27;ll be 49 plus four plus 16/16 which is equal to the square root of 69/16 I believe. Ah yes, 69/16 which is equal to the spirit of 69. And then we&#x27;ll take the denominator out of the square root, which is the score to 69/4. And so this is our denominator. And in the numerator we have our original vector and since we are solving for a vector, we do want to divide a vector by a scaler, which gives us another vector. So we&#x27;re on the right track and, uh, then if we just drove right this vector here we get, um if we factor in this denominator, terminate our numerator, we get the skirt of 16 I ever. Four. The force here will cancel out and so will get seven divided by the square root of 69 And then, uh, two divided by the spirit of 69 uh four, divided by the screw to 69. And this will be our behead vector. And to verify you wondering how do we know this is a unit vector? Well, is the magnitude of b had equal to one. We could check that right now. What eso now, Paul Sulfur magnitude The same way we did before. Bye. Taking the square root of the first component squared Plus this square root are the square second component scored, plus the last component scored. And this gives us the square vert of 49/69 plus for over 69 plus 16/69 which gives us just adding these all together since they have a common knowledge. Now that gives us 69/69 which gives a spirit of one which is one so yes, indeed. This behalf actor has a magnitude of one. And so now we found a unit vector be had in the same direction as B. And that&#x27;s given right here. Thank you.

John Fullerton · Answer

number 4 56 We have two and 5/8 minus one and 7/8. Changes into improper fractions eight times two is 16 16 plus five is 21 eight times one a 77 plus Sorry, eight put times one is 88 plus 7 15 tsunamis going to subtract. Keep the eight 21 minus 15. It&#x27;s six eights. I can reduce that two three force.

Bridget May · Answer

Okay, So today I&#x27;m gonna show you guys how to use the dot product to solve this type of problem. So the problem says usually given vectors to find v times W and v times re. So your your first thing you want to look at is what is the dot product. So if we define the dot product, the DOT product says that if you have a times would be that&#x27;s gonna be equal to a one. Be one plus a two be too. So you might be wondering, what exactly is a one? What&#x27;s B one? What&#x27;s a two would be, too on. So, you know, if you have even more numbers than that, it&#x27;s gonna be greater. You can have a one B one plus a to B to pull a three b three. It all depends on the vectors you&#x27;re given. And when we look at these two vectors here V and W, we have, we can see that we have only two numbers for each of them that we&#x27;re gonna be working with. So to define what a one and be one is gonna be, let&#x27;s put it in terms of V and W. So if we put it n v and w terms, we&#x27;re gonna look at a dot product that looks like this be times w equals V one, w one plus V two w two. So if we wanted to find what the woman w won our first, we&#x27;re gonna look, let&#x27;s look at V and Viner v terms. So this seven right here is gonna be the V one because it&#x27;s the first number that comes with a variable. Our second variable is J. Again. We have only two variables, which is why we define the equation as V one w one plus three to W two. If we had three, three variables say we had plus to Z, then we would have two more. But for this we only have two. So we&#x27;re gonna have V one and V two. When we look at the W, it&#x27;s the same thing. So negative three is our first value. So this is gonna be W one and it says negative J. And we&#x27;re gonna pretend there&#x27;s a one there cause we know that it&#x27;s negative one times j and that&#x27;s going to get us w two. So if we put this in we do the terms. W. And we defined V one at seven, which we&#x27;re gonna put there, and w one we said was negative three. So we&#x27;re gonna plug in negative three, and then V two, we said was negative, too, because it&#x27;s really important to not forget that negative sign. So negative, too. Times W two, Which is gonna be negative one. So if we multiply this out, we&#x27;re gonna get V W. He&#x27;s cool. It&#x27;s a negative. 21 plus negative two times negative one. We know when we have two negatives multiplying, it cancels out. So we get a positive and two times one is two. So negative 21 plus two is equal to negative 19. So we&#x27;re halfway done solving this equation. So first, we&#x27;re gonna go look at the second part we have to solve for which is V times be so we&#x27;re gonna define it in the same terms. So we&#x27;re going to say the times V, just like the dot product is equal to V one, and V is the same thing in Savi one plus V two v two. So if we plug this in, we&#x27;re going to get the times V to equal seven times seven plus negative to again. Don&#x27;t forget that negative times negative too. So if we simplify this, we are gonna see that the Times three equals seven times seven. So that&#x27;s 49 plus negative two times negative to the the negatives. Cancel out and you get plus four. So 49 plus four is equal to 50 three.

Problem

In Exercises $5-8,$ find the steady-state vector.…

Problem 7 Medium Difficulty

In Exercises $5-8,$ find the steady-state vector.
$$
\left[\begin{array}{ccc}{.7} & {.1} & {.1} \\ {.2} & {.8} & {.2} \\ {.1} & {.1} & {.7}\end{array}\right]
$$

Answer

$\underline { x } = \left[ \begin{array} { c } { \frac { 1 } { 4 } } \\ { \frac { 1 } { 2 } } \\ { \frac { 1 } { 4 } } \end{array} \right]$

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

Linear Algebra and Its Applications

Heather Z.

Kayleah T.

Caleb E.

Kristen K.

Vectors Intro

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$\underline { x } = \left[ \begin{array} { c } { \frac { 1 } { 4 } } \\ { \frac { 1 } { 2 } } \\ { \frac { 1 } { 4 } } \end{array} \right]$

Linear Algebra and Its Applications

Heather Z.

Kayleah T.

Caleb E.

Kristen K.

Vectors Intro

Vector Basics - Example 1

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

Heather Z.

Kayleah T.

Caleb E.

Kristen K.

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

Linear Algebra and Its Applications