Matrices Modeling Contexts — AP® Precalculus Notes

Beyond the exam. The AP® Precalculus exam does not test this unit. Many classes teach it, and it prepares you for calculus.

Matrices model situations where things move between a few states in steps: bikes between stations each day, customers between phone plans each month, a town's weather from one day to the next. If the chance of each move stays the same from step to step, one matrix predicts the whole future.

The transition matrix

Each day a bike at station A stays there with chance $0.8$ and is ridden to B with chance $0.2$ . A bike at B goes to A with chance $0.3$ and stays with chance $0.7$ . Put each station's chances in a column, "from" across the top and "to" down the side:

	from A	from B
to A	$0.8$	$0.3$
to B	$0.2$	$0.7$

T = \left[ \begin{matrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{matrix} \right]

This is a transition matrix. Each column adds up to $1$ , because every bike ends the day somewhere. (Some books put "from" in the rows and multiply on the other side; the numbers are the same, read the other way. Check which way a matrix is set up before using it.)

Figure 1. The bike-share model as a diagram. LevlPrep original diagram.

One step forward

A state vector lists how many are in each state, here $\mathbf{x}_0 = \left[ \begin{matrix} 400 \\ 100 \end{matrix} \right]$ : $400$ bikes at A and $100$ at B. The expected numbers one day later are $T\mathbf{x}_0$ .

Worked example 1. Predict the bikes at each station after one day and after two days.

$\mathbf{x}_1$ $= \left[ \begin{matrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{matrix} \right]\left[ \begin{matrix} 400 \\ 100 \end{matrix} \right]$ $= \left[ \begin{matrix} 0.8(400) + 0.3(100) \\ 0.2(400) + 0.7(100) \end{matrix} \right]$ $= \left[ \begin{matrix} 350 \\ 150 \end{matrix} \right]$ .

Row 1 collects everything that ends at A: $320$ bikes that stayed plus $30$ that came over from B.

$\mathbf{x}_2$ $= T\mathbf{x}_1$ $= \left[ \begin{matrix} 0.8(350) + 0.3(150) \\ 0.2(350) + 0.7(150) \end{matrix} \right]$ $= \left[ \begin{matrix} 325 \\ 175 \end{matrix} \right]$ . The total stays $500$ .

In general $\mathbf{x}_n = T^n\mathbf{x}_0$ . A calculator with matrix operations finds $T^{10}$ or $T^{30}$ in one step.

One step back

If today's state is known, yesterday's solves $T\mathbf{x}_{\text{yesterday}} = \mathbf{x}_{\text{today}}$ , so $\mathbf{x}_{\text{yesterday}} = T^{-1}\mathbf{x}_{\text{today}}$ . Here $\det T$ $= 0.56 - 0.06$ $= 0.5$ , and $T^{-1}\left[ \begin{matrix} 350 \\ 150 \end{matrix} \right]$ $= \left[ \begin{matrix} 400 \\ 100 \end{matrix} \right]$ , the starting state, as it should be.

The long run

Keep stepping and the numbers settle: $350$ , $325$ , $312.5$ , ... at A, approaching $300$ . A steady state is a vector that one more step does not change: $T\mathbf{x} = \mathbf{x}$ .

Worked example 2. Find the steady state for $500$ bikes.

At a steady state the bikes leaving A balance those arriving: $0.2a = 0.3b$ , so $a = 1.5b$ . With $a + b = 500$ : $2.5b = 500$ , $b = 200$ , $a = 300$ .

Check: $T\left[ \begin{matrix} 300 \\ 200 \end{matrix} \right]$ $= \left[ \begin{matrix} 240 + 60 \\ 60 + 140 \end{matrix} \right]$ $= \left[ \begin{matrix} 300 \\ 200 \end{matrix} \right]$ .

The steady state does not depend on how the bikes start; with $400$ or $100$ at A, the system drifts toward $300$ and $200$ .

Shares instead of counts

The same matrix works with shares that add up to $1$ . If today is rainy, the weather state is $\left[ \begin{matrix} 0 \\ 1 \end{matrix} \right]$ (chance of sun $0$ , chance of rain $1$ ), and $T^n\left[ \begin{matrix} 0 \\ 1 \end{matrix} \right]$ gives the chances of sun and rain $n$ days later.