Baby Linear Algebra

Previously, we tried to earn some money selling fidget spinners. To do that, we will sell $x$ fidget spinners, where $x$ is given by the $x$ -coordinate of the intersection point of the two graphs below.

Remark 1. In economics terms,

the graph of $y = 2x + 1$ is called the supply curve,
the graph of $y = -\frac 12 x + 11$ is called the demand curve.

In more mathy terms, we need to find a pair $(x, y)$ of numbers that satisfy both equations:

$\left\{ \begin{aligned} y &= 2x+1, \\ y &= \textstyle -\frac 12 x + 11. \end{aligned} \right.$

We call this process solving a system of simultaneous linear equations.

Example 1. Solve the system of simultaneous linear equations:

$\left\{ \begin{aligned} y &= 2x+1, \\ y &= \textstyle -\frac 12 x + 11. \end{aligned} \right.$

Solution. If $(x,y)$ satisfies both equations, then the $y$ terms on the left-hand side equal. Subtracting the first equation by the second:

$\begin{aligned} y - y &= (2x+ 1) - \left(\textstyle -\frac 12 x + 11 \right) \\ 0 &= 2x + 1 + \textstyle \frac 12 x - 11 \\ &= \left( 2x + \textstyle \frac 12 x \right) + 1 - 11 \\ &= \textstyle \frac 52 x - 10. \end{aligned}$

Therefore, we just need to solve for $x$ :

$\begin{aligned} \textstyle \frac 52 x - 10 &= 0 \\ \textstyle \frac 52 x &= 10 \\ x &= \textstyle \frac 25 \cdot 10 \\ &= \textstyle \frac 25 \cdot 5 \cdot 2 \\ &= 2 \cdot 2 = 4. \end{aligned}$

Hence, $x = 4$ . What is its matching $y$ -value?

Since this $x$ -value must satisfy both equations, we can back-substitute $x =4$ into either equation:

Supply curve: $y = 2 \cdot 4 + 1 = 9$ .
Demand curve: $y = -\frac 12 \cdot 4 + 11 = 9$ .

Therefore, our solution is $x = 4$ and $y = 9$ .

In ordered-pair notation, the solution is $(x, y) = (4, 9)$ .

Whether we calculated the intersection point $(4, 9)$ , or read it off from the graph, we obtained the same answer. This observation is massive—it connects pure symbols with familiar pictures!

This connection is the simplest building block behind linear algebra—which uses calculations to represent “straight-ish” objects like lines and planes (see Theorem 1 below). Its humble beginnings arise from what we just did—solving linear equations simultaneously.

In higher-level mathematics, the elimination method we used is given the fancier title Gaussian elimination.
There is an alternate (and more general) strategy to solve simultaneous equations, called the substitution method. We will explore this method when investigating quadratic curves.

Example 2. Let $L$ denote the line given by the equation $x + 2y = 3$ . Determine the intersection point between $L$ and the $x$ -axis, called the $x$ -intercept of $L$ .

Solution. Let $M$ denote the $x$ -axis. Since $M$ is horizontal and passes through the point $(0, 0)$ , it has equation $y = 0$ . Therefore we need to solve the system of simultaneous linear equations

$\left\{ \begin{aligned} x+2y &= 3, \\ y &=0. \end{aligned} \right.$

Multiplying the second equation by $2$ :

$\left\{ \begin{aligned} x+2y &= 3, \\ 2y &=0. \end{aligned} \right.$

Subtracting the first equation by the second:

$\begin{aligned} (x+2y) - ( 2y) &= 3 - 0 \\ x &=3. \end{aligned}$

Back-substituting $x = 3$ into the second equation, since there are no $x$ terms, we just have $y = 0$ . Therefore, $L$ has an $x$ -intercept $(3, 0)$ .

Example 3. Using the definition of $L$ in Example 2, determine the $y$ -intercept of $L$ .

Solution. Let $M$ denote the $y$ -axis. Since $M$ is vertical and passes through the point $(0, 0)$ , it has equation $x = 0$ . Therefore we need to solve the system of simultaneous linear equations

$\left\{ \begin{aligned} x+2y &= 3, \\ x &=0. \end{aligned} \right.$

Subtracting the second equation from the first:

$\begin{aligned} \textstyle (x+2y) - x &= 3 - 0 \\ (x-x) + 2y &= 3 \\ 2y &= 3 \\ y &= \textstyle \frac 32. \end{aligned}$

Back-substituting $y = \frac 32$ into the second equation, since there are no $y$ terms, we just have $x = 0$ . Therefore, $L$ has a $y$ -intercept $(0, 3/2)$ .

Our examples aren’t special; here’s the more general theory:

Theorem 1. If $ad - bc \neq 0$ , then there exists a unique pair $(x,y)$ of real numbers that solves the system of simultaneous linear equations

$\left\{ \begin{aligned} ax+by &= p, \\ cx +dy &= q. \end{aligned} \right.$

In fact, $\displaystyle x = \frac{1}{ad-bc} (dp - bq)$ , $\displaystyle y = \frac{1}{ad-bc} (-cp + aq)$ .

Proof Sketch. Multiply the first equation by $d$ and the second equation by $b$ :

$\left\{ \begin{aligned} adx+bdy &= dp, \\ bcx +bdy &= bq. \end{aligned} \right.$

Subtract the first equation by the second:

$\begin{aligned} (ad- bc)x &=dp - bq. \end{aligned}$

Assuming $ad - bc \neq 0$ , we can obtain a unique value for $x$ , then back-substitute to obtain a unique value for $y$ .

Corollary 1. If $a$ and $b$ are both nonzero, then the line $ax+by = p$ has $x$ -intercept $(p/a,0)$ and $y$ -intercept $(0, p/b)$ .

Proof. Use Theorem 1:

Set $(c,d,q) = (0,1,0)$ to obtain the $x$ -intercept.
Set $(c,d,q) = (1,0,0)$ to obtain the $y$ -intercept.

Theorem 1 gives us a taste of the theory of linear algebra, and our discussion on supply and demand curves illustrate one (over-simplified) real-world use case of said theory. We can summarise our findings with a simple cliche:

X marks the spot!

Next time, we will discuss some simple ideas about lines and angles, which can be described using lines, and be precisely formulated using—you guessed it—linear algebra. 90% of my blog was dedicated to ensure that what we do at the O-Level really is legitimate, and thankfully, I can confidently say that it is!

—Joel Kindiak, 4 Oct 25, 1516H

KindiakMath

Baby Linear Algebra

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