Math / Algebra

QuestionIs $\left[\begin{array}{l}\frac{0.8}{2.5} \\ \frac{0.3}{2.5}\end{array}\right]$ the steady state vector for the transition matrix $P=\left[\begin{array}{ll}0.7 & 0.2 \\ 0.3 & 0.8\end{array}\right]$ ? (Hint: You don't need to solve, there's an easier way.) [This question is based on your assigned pre-reading/prep for the upcoming Assignment] True False

Studdy Solution

STEP 1

What is this asking? Does multiplying the given transition matrix by the given vector result in the same vector? Watch out! Don't solve for the steady state vector the long way, there's a much faster method!

STEP 2

1. Multiply the transition matrix by the vector.
2. Compare the result with the original vector.

STEP 3

Let's multiply our transition matrix $P$ by our vector $v$ : $Pv = \begin{bmatrix} 0.7 & 0.2 \\ 0.3 & 0.8 \end{bmatrix} \begin{bmatrix} \frac{0.8}{2.5} \\ \frac{0.3}{2.5} \end{bmatrix}$

STEP 4

Remember, we multiply rows of the first matrix by columns of the second.
So, we get: $\begin{bmatrix} 0.7 \cdot \frac{0.8}{2.5} + 0.2 \cdot \frac{0.3}{2.5} \\ 0.3 \cdot \frac{0.8}{2.5} + 0.8 \cdot \frac{0.3}{2.5} \end{bmatrix}$

STEP 5

Let's calculate those values! $\begin{bmatrix} \frac{0.56}{2.5} + \frac{0.06}{2.5} \\ \frac{0.24}{2.5} + \frac{0.24}{2.5} \end{bmatrix} = \begin{bmatrix} \frac{0.62}{2.5} \\ \frac{0.48}{2.5} \end{bmatrix}$

STEP 6

Now, let's compare our result with the original vector $v$ : $\begin{bmatrix} \frac{0.62}{2.5} \\ \frac{0.48}{2.5} \end{bmatrix} \neq \begin{bmatrix} \frac{0.8}{2.5} \\ \frac{0.3}{2.5} \end{bmatrix}$

STEP 7

Since the resulting vector is not equal to the original vector, the given vector is not a steady state vector for the given transition matrix.

STEP 8

False.

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Studdy Solution

STEP 1

What is this asking? Does multiplying the given transition matrix by the given vector result in the same vector? Watch out! Don't solve for the steady state vector the long way, there's a much faster method!

STEP 2

1. Multiply the transition matrix by the vector.
2. Compare the result with the original vector.

STEP 3

Let's multiply our transition matrix $P$ by our vector $v$ : $Pv = \begin{bmatrix} 0.7 & 0.2 \\ 0.3 & 0.8 \end{bmatrix} \begin{bmatrix} \frac{0.8}{2.5} \\ \frac{0.3}{2.5} \end{bmatrix}$

STEP 4

Remember, we multiply rows of the first matrix by columns of the second.
So, we get: $\begin{bmatrix} 0.7 \cdot \frac{0.8}{2.5} + 0.2 \cdot \frac{0.3}{2.5} \\ 0.3 \cdot \frac{0.8}{2.5} + 0.8 \cdot \frac{0.3}{2.5} \end{bmatrix}$

STEP 5

Let's calculate those values! $\begin{bmatrix} \frac{0.56}{2.5} + \frac{0.06}{2.5} \\ \frac{0.24}{2.5} + \frac{0.24}{2.5} \end{bmatrix} = \begin{bmatrix} \frac{0.62}{2.5} \\ \frac{0.48}{2.5} \end{bmatrix}$

STEP 6

Now, let's compare our result with the original vector $v$ : $\begin{bmatrix} \frac{0.62}{2.5} \\ \frac{0.48}{2.5} \end{bmatrix} \neq \begin{bmatrix} \frac{0.8}{2.5} \\ \frac{0.3}{2.5} \end{bmatrix}$

STEP 7

Since the resulting vector is not equal to the original vector, the given vector is not a steady state vector for the given transition matrix.

STEP 8

False.

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Studdy Solution

STEP 1

What is this asking? Does multiplying the given *transition matrix* by the given *vector* result in the same *vector*? Watch out! Don't solve for the steady state vector the long way, there's a *much* faster method!

STEP 2

1. Multiply the transition matrix by the vector.2. Compare the result with the original vector.

STEP 3

Let's **multiply** our *transition matrix* PPP by our *vector* vvv: Pv=[0.70.20.30.8][0.82.50.32.5]Pv = \begin{bmatrix} 0.7 & 0.2 \\ 0.3 & 0.8 \end{bmatrix} \begin{bmatrix} \frac{0.8}{2.5} \\ \frac{0.3}{2.5} \end{bmatrix}Pv=[0.70.3​0.20.8​][2.50.8​2.50.3​​]

STEP 4

STEP 5

STEP 6

Now, let's compare our *result* with the *original vector* vvv: [0.622.50.482.5]≠[0.82.50.32.5]\begin{bmatrix} \frac{0.62}{2.5} \\ \frac{0.48}{2.5} \end{bmatrix} \neq \begin{bmatrix} \frac{0.8}{2.5} \\ \frac{0.3}{2.5} \end{bmatrix}[2.50.62​2.50.48​​]=[2.50.8​2.50.3​​]

STEP 7

Since the *resulting vector* is *not* equal to the *original vector*, the given vector is **not** a steady state vector for the given transition matrix.

STEP 8

False.

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What is this asking? Does multiplying the given transition matrix by the given vector result in the same vector? Watch out! Don't solve for the steady state vector the long way, there's a much faster method!

1. Multiply the transition matrix by the vector.
2. Compare the result with the original vector.

Let's multiply our transition matrix $P$ by our vector $v$ : $Pv = \begin{bmatrix} 0.7 & 0.2 \\ 0.3 & 0.8 \end{bmatrix} \begin{bmatrix} \frac{0.8}{2.5} \\ \frac{0.3}{2.5} \end{bmatrix}$

Now, let's compare our result with the original vector $v$ : $\begin{bmatrix} \frac{0.62}{2.5} \\ \frac{0.48}{2.5} \end{bmatrix} \neq \begin{bmatrix} \frac{0.8}{2.5} \\ \frac{0.3}{2.5} \end{bmatrix}$

Since the resulting vector is not equal to the original vector, the given vector is not a steady state vector for the given transition matrix.