📂Matrix Algebra

Simultaneous Linear Equations

Definition¹

For constants $a_{1}$, $a_{2}$, $\dots$, $a_{n}$, $b$, we define a linear equation for variables $x_{1}$, $x_{2}$, $\dots$, $x_{n}$ as follows.

$$\begin{equation} a_{1}x_{1} + a_{2}x_{2} + \cdots + a_{n}x_{n} = b \label{lineq} \end{equation}$$

At least one of $a$ is not $0$. In other words, not “all $a$ are $0$”. A finite set of linear equations is called a system of linear equations or simply a linear system, and the variables are called unknowns. In Korean, linear and first-order mean the same thing. A linear system consisting of $m$ linear equations for $n$ variables $x_{1}$, $x_{2}$, $\dots$, $x_{n}$ is represented as follows.

$$\begin{equation} \begin{aligned} a_{11}x_{1} + a_{12}x_{2} + \cdots + a_{1n}x_{n} &= b_{1} \\ a_{21}x_{1} + a_{22}x_{2} + \cdots + a_{2n}x_{n} &= b_{2} \\ &\vdots \\ a_{m1}x_{1} + a_{m2}x_{2} + \cdots + a_{mn}x_{n} &= b_{m} \end{aligned} \label{linsys} \end{equation}$$

When represented as a matrix, it is as follows.

$$\begin{align*} \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \end{bmatrix} &= \begin{bmatrix} b_{1} \\ b_{2} \\ \vdots \\ b_{m} \end{bmatrix} \\ A\mathbf{x} &= \mathbf{b} \end{align*}$$

Explanation

The values of $x_{1}$, $x_{2}$, $\dots$, $x_{n}$ that make the linear system true are called solutions. When a linear system is given, it must satisfy one of the following three cases. There are no other cases. The proof is introduced at the bottom of this article.

• There is exactly one solution.
• There are infinitely many solutions.
• There is no solution.

If there is at least one solution, the linear system is said to be consistent. If there is no solution, the linear system is said to be inconsistent.

Specifically, in the case of two variables, a linear equation means an equation of a line. If there is exactly one solution in a linear system with two variables, it means the case where the lines meet at one point. If there are infinitely many solutions, it means the case where the lines meet at infinitely many points, i.e., they overlap. If there is no solution, it means the case where there is no point where the lines meet.

A linear equation with three variables means an equation of a plane, so depending on the solution of the linear system, it signifies how the planes overlap.

Example

Let’s solve the following linear system.

$$\begin{align*} 4x -2y &= 1 \\ 16x -8y &= 4 \end{align*}$$

Multiplying the above equation by $-4$ and adding it to the below equation results in the following.

$$\begin{align*} 4x -2y &= 1 \\ 0 &= 0 \end{align*}$$

Then, since the lower equation does not display any information, let’s only represent it with the upper equation.

$$4x -2y = 1$$

In this case, it geometrically means that the two lines coincide. In such a case, after arranging $x$ for $y$, it is denoted as $x = \dfrac{1}{2}y + \dfrac{1}{4}$, and then some arbitrary number $t$ is substituted into $y$ to represent the solution.

$$x = \dfrac{1}{4} + \dfrac{1}{2}t, \quad y = t$$

Such $t$ is called a parameter, and the above equation is called parametric equations.

Proof²

A system of linear equations has no solution, one solution, or infinitely many solutions. There are no other cases.

When there are two different solutions, the proof is completed by showing that there are infinitely many solutions. Let $\mathbf{x}_{1}$, $\mathbf{x}_{2}$ be two different solutions of the linear system $A\mathbf{x} =\mathbf{b}$. And let’s designate $\mathbf{x}_{0} = \mathbf{x}_{1} - \mathbf{x}_{2}$. Since $\mathbf{x}_{1}$ and $\mathbf{x}_{2}$ are two different solutions, $\mathbf{x}_{0} \ne \mathbf{0}$. Moreover, the following equation holds:

$$A \mathbf{x}_{0} = A (\mathbf{x}_{1} - \mathbf{x}_{2}) = \mathbf{b} - \mathbf{b} = \mathbf{0}$$

Let’s designate $k$ as an arbitrary constant. Then, by the above result, the following equation also holds:

$$\begin{align*} A (\mathbf{x}_{1} + k\mathbf{x}_{0}) &= A\mathbf{x}_{1} + A(k\mathbf{x}_{0}) \\ &= A\mathbf{x}_{1} + kA\mathbf{x}_{0} \\ &= \mathbf{b} + \mathbf{0} \\ &= \mathbf{b} \end{align*}$$

Therefore, $\mathbf{x}_{1} + k\mathbf{x}_{0}$ is also a solution of the linear system $A\mathbf{x} = \mathbf{b}$. Since this holds for any arbitrary constant $k$, there are infinitely many solutions.

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