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

$q = \left[ \begin{array} { l } { 5 / 7 } \\ { 2 / 7 } \end{array} \right] = \left[ \begin{array} { l } { 714 } \\ { 286 } \end{array} \right]$

Calculus 3

Chapter 4

Vector Spaces

Section 9

Applications to Markov Chains

Vectors

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

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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're over this, we can solve our matrix. We get one *** five over too thorough until Joe'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'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's his equal to right. Actually, that's decision

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The number 2 is also the smallest & first prime number (since every other even number is divisible by two).

If you write pi (to the first two decimal places of 3.14) backwards, in big, block letters it actually reads "PIE".

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