Linear Systems

Section 1.1 Linear Systems

We start this section with three examples of linear systems. One is about a biathlon with running and biking legs, the second is a traffic model and the last one is an example from chemistry. Examples of linear systems are found from every mathematical subfield and every science.

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Example 1.1.1. Biathlon Example.

Set up the following problem as a set of linear equations.

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Travis runs 6 mph and bikes 18 mph in a race with two events. If the course is 29 miles long and it takes him 2 hours and 10 minutes to complete the race, how long is each segment?

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Solution.

In this case, we need to know how long each segment is. There are two legs to the race, so we will let

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\begin{align*} x_1 \amp = \text{number of miles for the run,}\\ x_2 \amp = \text{number of miles for the bike.} \end{align*}

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The equations come from three statements in the problem above. First, we know that the total course is 29 miles long or

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\begin{equation*} x_1+x_2=29. \end{equation*}

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The remaining equation arises from the time it takes Travis to complete the race. In general recall the relationship between speed (a rate), distance and time is

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\begin{equation*} \text{speed} = \frac{\text{distance}}{\text{time}} \end{equation*}

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or solving for time,

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\begin{equation*} \text{time} =\frac{\text{distance}}{\text{speed}} \end{equation*}

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For example, the time it takes Travis to finish the running leg is \(x_1/6\text{.}\) The total time in hours it takes Travis to finish the race is

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\begin{equation*} \frac{x_1}{6} + \frac{x_2}{18} = 2+ \frac{10}{60} = \frac{13}{6} \end{equation*}

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so the linear system is

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\begin{align*} x_1+x_2 \amp =29, \\ \frac{x_1}{6} + \frac{x_2}{18} \amp =\frac{13}{6}. \end{align*}

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The next example examines the traffic flow in a very limited part of Boston.

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Example 1.1.2. Traffic Flow.

A simple model of traffic flow can be represented by the following graph:

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Figure 1.1.3. A map of a few streets in Boston where the arrows denote the direction of traffic flow (all of these streets are one-way) and the numbers represent the numbers of cars driving down the street in a given time period. The letters \(A\) through \(D\) will be the names of the intersections.
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Write down the equations that balance each of the numbers of cars entering and leaving each of the intersections \(A, B, C\) and \(D\text{.}\)

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Solution.

In this case, we have:

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\begin{align*} x_4 + x_5 \amp = x_1 + 250\\ 350 \amp = x_3 + x_4 \\ 700 \amp = x_2 + x_5 \\ x_3 + x_2 \amp = 600 \end{align*}

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for intersections \(A\text{,}\) \(B\text{,}\) \(C\) and \(D\) respectively.

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Linear equations exist in many different fields, especially the sciences. The following comes from chemistry in which a chemical equation is found by balancing the number of atoms in the equation on both sides.

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Example 1.1.4. Chemical Reactions.

Hydrazine (\(N_2H_4\)) is an important nitrogen-based compound. A chemical reaction to produce it from ammonia and hydrogen peroxide is given by

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\begin{equation*} k_1 NH_3 + k_2 H_2O_2 \rightarrow k_3 N_2 H_4 + k_4 H_2 O \end{equation*}

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Find the equations that balanced the numbers of nitrogen (N), hydrogen (H) and oxygen (O) atoms respectively in the reaction

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Solution.

The values of \(k_1, k_2, k_3\) and \(k_4\) can be found by solving the following linear system which balances the number of nitrogen (N), hydrogen (H) and oxygen (O) atoms respectively or

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\begin{align*} k_1 \amp = 2k_3, \\ 3k_1 + 2k_2 \amp = 4k_3 + 2k_4, \\ 2k_2 \amp = k_4. \end{align*}

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With examples of linear algebra presented, we next move to a few definitions.

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Definition 1.1.5.

A linear combination of \(x_1, x_2, x_3, \ldots, x_n\) has the form
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\begin{equation*} a_1 x_1 + a_2 x_2 + \cdots + a_n x_n, \end{equation*}

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where the constants \(a_1, a_2, \ldots, a_n \in \mathbb{R}\) are the combinations coefficients.
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A linear equation has the form
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\begin{equation} a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = b\tag{1.1.1} \end{equation}

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where \(b \in \mathbb{R}\) is a constant and \(a_1, a_2, \ldots, a_n \in \mathbb{R}\text{.}\)
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The \(n\)-tuple \((s_1,s_2,\ldots,s_n)\) satisfies or is a solution of (1.1.1) if this point satisfies (1.1.1) or
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\begin{equation*} a_1 s_1 + a_2 s_2 + \cdots + a_n s_n = b \end{equation*}

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A system of linear equations or linear system is a set of linear equations:
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\begin{align} a_{1,1} x_1 + a_{1,2} x_2 + \cdots + a_{1,n} x_n \amp = b_1 , \notag\\ a_{2,1} x_1 + a_{2,2} x_2 + \cdots + a_{2,n} x_n \amp = b_2, \tag{1.1.2}\\ \vdots \amp = \vdots \notag\\ a_{m,1} x_1 + a_{m,2} x_2 + \cdots + a_{m,n} x_n \amp = b_m, \notag \end{align}

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and this linear system has \(m\) equations and \(n\) unknowns (variables).
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The \(n\)-tuple \((s_1,s_2,\ldots,s_n)\) satisfies or is a solution of (1.1.2) if this point satisfies every equation of (1.1.2).
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Example 1.1.6. Linear Equations.

The following are linear equations:

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\begin{align*} 2x + 3y \amp = 6, \amp y_1 -y_2+y_3-y_4 \amp = 10, \\ 10x_1 - x_3 + 5x_5 \amp = 9, \amp \sum_{i=1}^{10} i x_i \amp = 0 \end{align*}

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where summation notation has been used in the last one and note the variable names can vary. The following equations are not linear:

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\begin{align*} x_1^2+x_2 \amp = 6, \amp y_1y_2 + y_3 \amp = 5, \\ \frac{x_1+x_2}{x_3} \amp = 6, \amp \sin(x+y) \amp = z \end{align*}

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Each of the equations in the latter group have multiplications, squares division or other functions between variables.

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The next two examples give a way to determine if a point or \(n\)-tuple is a solution to a linear system.

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Example 1.1.7. Showing a Point is a Solution to a Linear System.

Show that the point \((2,3)\) is a solution of the linear system:

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\begin{align*} 3x_1 - x_2 \amp = 3 \\ 2x_1 + 4x_2 \amp = 16 \end{align*}

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Solution.

Substitute \(x_1=2\) and \(x_2=3\) into both equations and check.

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\begin{align*} 3(2) - 3 \amp = 3, \\ 2(2) + 4(3) \amp = 16. \end{align*}

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Since each equation is satisfied at the point \((2,3)\) is a solution to the linear system.

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Example 1.1.8.

Recall that the linear system in Example Example 1.1.4 can be written:

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\begin{align*} k_1 \phantom{+2k_2}- 2k_3\phantom{+2k_4} \amp = 0, \\ 2k_2 + 2k_3 -2k_4 \amp = 0, \\ 2k_2 \phantom{+2x_3}- k_4 \amp = 0, \end{align*}

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Is \((8,4,4,8)\) a solution to this linear system? Is \((4,2,2,4)\text{?}\) Is \((6,1,3,2)\text{?}\)

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Solution.

For \((8,4,4,8)\text{,}\) we need to substitute this point in and check all the equations:

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\begin{align*} 8-2(4) \amp = 0 \\ 2(4)+2(4)-2(8) \amp = 8+8-16 = 0 \\ 2(4)-8 \amp = 8-8 = 0 \end{align*}

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so it is a solution.

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For \((4,2,2,4)\text{,}\) again check all of the equations:

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\begin{align*} 4-2(2) \amp = 0, \\ 3(4)+2(2)-4(2)-2(4) \amp = 12+4-8-8=0, \\ 2(2)-4\amp = 0 \end{align*}

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so this is also a solution.

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For \((6,1,3,2)\text{,}\) we substitute this into the equation:

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\begin{align*} 6-2(3) \amp = 0 \\ 3(6)+2(1)-4(3)-2(2) \amp = 18+2-12-4 = 4 \neq 0.\\ 2(1) - 2 \amp = 0 \end{align*}

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and it satisfies 2 of the equations, but not all, so this is not a solution.

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