in-exercises-5-8-find-the-steady-state-vector-leftbeginarrayccc7-2-2-0-2-4-3-6-4endarrayright

Question

In Exercises $5-8,$ find the steady-state vector.
$$
\left[\begin{array}{ccc}{.7} & {.2} & {.2} \ {0} & {.2} & {.4} \ {.3} & {.6} & {.4}\end{array}ight]
$$

Amy Jiang · Accepted 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.

Michael Jacobsen · 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 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

Supratim Roy · Answer

Hello, everyone. So we have been provided with two vectors in their component forms. So we will have to find the CO sign off the angle between them first. Then we have to draw a conclusion from the result. We opt in in the previous question, and then we will verify this conclusion using by making the sketch of this vector sports directors. Okay, So for our convenience, let us Asian, the 1st 1 to be vector A and the 2nd 1 to e vector B. Okay, so now let us see how to find Costinha. So the formula for costed a or co sign of today&#x27;s rectum e dot vector b divided by the product off there. Magnitudes. Okay, so we have a and B here, so let us take this down work. It would be easier to compute. Okay, so what would because teeter so cost it would be It would be it on the new new miniter. So, in a component from when we want to find a tractor product Sorry, The skill A product that is the dot product. Then we have to multiply the X component of the first vector with the X component of the 2nd 1 so 16 to minus four, then added with the productive the Y component of the two vectors. That is negative. Eight multiplied with negative three. Okay. And on the denominator, we have the magnitude off each vectors. So it is the square root off the some of the squares of the components. So for factory that would be square of competently six square and there some is. So therefore six square plus everywhere square. And for B, it is nobody Foursquare plus negative three square. Okay, so now we will calculate the sales. So six multi bladed negative for this negative 24 then eight with negative three is 24 positive train before and here on the denominator we have six square plus h squared. That is 36 plus 64 and again 36 0 And that would be 16 plus nine. Okay, so now you can see on the new miniter. We have to oppose it numbers. That means the result will be zero. Okay, so we have costed er equals 20 So from the techno metric table, we can find that the value of teeter is actually man did it so we never costed a zero the veil of today&#x27;s 19. Okay, that is the accused one. Okay, that is another one. Ah. Okay, So which is the are more than one into degree? That would be 70 Vegas. Okay, so this is this is the Cuban today close to 90 degree for cost. Today goes to zero so we can conclude, or B, that the factors, uh, our pindi cooler school, each of it do each other. OK, so there&#x27;s the conclusion. Now we will verify by sketching whether or conclusion is true or not. So they&#x27;re perpendicular or not. Let us check. So we have the coordinates X y plane, gruff. So let me check the and we dig this. Ah, well off nb downwards so that we can easily Florida growth. Okay, so we have our a and to be so you can see a will be in the four Texas four quadrant. Okay, Because why is negative excess positive and we will will be in the third quarter. So we will have to make the herd in the fourth quarter in only. Okay, so that is negative. X positive X is the origin on we have you could do. Wait. Okay. So to draw the vector was Director of P. We will have to join the origin to the coordinates. Six comma, minus six. Okay, so six comma minus eight. So that is on the positive sixties. On the positive XX is so I will go six units straight. 123456 and negative. Ladies, I don&#x27;t want 1234 5678 So this is eight, and this is six. So this is six comma negative. Okay, so now we will join from the origin to this point. Okay, so we got the vector. Hey, so this is victory. And for victory, we have to join a line from the origin to the warrant. Negative four common. Negative three. So negative four is on the negative. XX is 1234 and negative three here on every way exists. So this is negative for comma negative three. So we will join this point with the origin, So this will give us the factor. Be So this is Victor. Be, as you can see from the graph, the angle between this two vectors, our 19 clearly this 90 degree. So We have verified that these two vectors are Piper new, collective each other through this bag Him. Okay, so these are the two factors A and B and they&#x27;re perpendicular to each other. That was our conclusion on cash number B. So we have verified 80 using this sketch. Okay, so there&#x27;s all for this problem. Thank you.

Taylor Shimono · Answer

So for this problem, we want to test whether our three points are Colin here, and we can go ahead and do this using vectors. Um, if we label our first point point Pete, our second point point Q and R third point are we can go ahead and turn these into Victor&#x27;s P Q and A P p. R. And so when we do this, we know that we are going to find that component form of these two vectors, Um, by subtracting the terminal point minus our initial point. So for a peek, you two is going to be our terminal point and P is our initial point. So our extra boat, it will be negative for minus negative, to which becomes plus two our why component will be eight minus seven and ours a component will be one minus four, so we can simplify this to be a negative. Four plus two will be negative, too. 817 is one and one minus four is negative three and then we can do the same thing for vector p. R. We know that our is our terminal point and P is our initial point, so we could go and get our exponents by subtracting zero minus negative, too will become zero plus two six minus seven for our why component and seven minus 44 hours, he comported. So this will simplify into two negative 13 and next. We know that if our three points are Colin here that are vectors P Q and PR must be parallel, so we can go ahead and test whether they are parallel. Um, by setting up three different equations, I&#x27;m one of them will be our X component of p. Q is equal to some constancy times the X component of our vector p r. Our second equation will be the why component of P Q is equal to that same constancy times of the y component of PR. So times negative one. And our third equation will be the Z component of P Q, which is negative. Street is equal to our skill or times three. And I found these three equations by essentially saying we know that two vectors are parallel. For instance, two vectors U N V are parallel if you is equal to some scaler seat time sector V. So when we multiply our scaler by Vector V. We know that see our vector V is going to be equal to see Time&#x27;s our X component. See, Time&#x27;s our why component and see times are Zeke. Got it. And since we know that this has to be equal to vector you, we know that the X component of you will have to line up and be equal to the X component of C ten&#x27;s V and the same thing for our y and Z components. And so we can go ahead and solve for arms killer seats in our first equation. And when we do that, we will divide both sides of our equation by two. To get C is equal to negative one. And so, if our two vectors are parallel, then we should be able to plug this killer into our other two equations and get true equalities. So, uh, we plug it into our second equation we get one is equal to negative one time. The negative one which simplifies to one, is equal to one which we know is true, and we could go and plug it into our third equation. Now negative three is equal to see which is negative one times three. So we get negative. Three is equal to negative three, which is true. So we can see that because our two vectors are parallel, then our lines are Our points are indeed colon ear.

Doruk Isik · Answer

in this problem, we&#x27;re given the basis to factors B one and B two and the Vector X and whereas to find recording it better, uh, we know that Rex Rex believe to see one. So the first elemental, coordinated vector most like first factor plus C two second element of quarters record most liked by the second phase is Victor. So from the first row, we know that forests see one plus five C two and from the second row, we know that&#x27;s your musical See one minus six C two. So from the second row, we can see that C one is they could do *** three c two and a few years This. And if you like this in the first question we find that negative tree See two plus five C to visit before so to see two isn&#x27;t comfortable for S C to Z two. And using this we find C one to be negative sense. So then I recorded it back to you will be negative six and two

Problem

Determine if $P=\left[\begin{array}{cc}{.2} & {1}…

Problem 8 Medium Difficulty

In Exercises $5-8,$ find the steady-state vector.
$$
\left[\begin{array}{ccc}{.7} & {.2} & {.2} \\ {0} & {.2} & {.4} \\ {.3} & {.6} & {.4}\end{array}\right]
$$

Answer

$\mathbf { q } = \left[ \begin{array} { l } { 2 / 5 } \\ { 1 / 5 } \\ { 2 / 5 } \end{array} \right] = \left[ \begin{array} { l } { 4 } \\ { 2 } \\ { 4 } \end{array} \right]$

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

Linear Algebra and Its Applications

Catherine R.

Caleb E.

Kristen K.

Samuel H.

Vectors Intro

Vector Basics - Example 1

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$\mathbf { q } = \left[ \begin{array} { l } { 2 / 5 } \\ { 1 / 5 } \\ { 2 / 5 } \end{array} \right] = \left[ \begin{array} { l } { 4 } \\ { 2 } \\ { 4 } \end{array} \right]$

Linear Algebra and Its Applications

Catherine R.

Caleb E.

Kristen K.

Samuel H.

Vectors Intro

Vector Basics - Example 1

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

Catherine R.

Caleb E.

Kristen K.

Samuel H.

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

Linear Algebra and Its Applications