Linear Systems as Equations and Geometry, linear algebra I lesson

Two unknown numbers add to $3$ , and the first exceeds the second by $1$ . Each condition alone allows infinitely many pairs, but together they allow only one. Replace the second condition by a different one and there may be no pair at all, or a whole line of them. This lesson sets up the language for such questions, reads every system in two ways, as crossing lines or planes and as a combination of matrix columns, and explains why a linear system always has no solution, exactly one, or infinitely many.

Key terms

Linear system

Definition A linear equation in the unknowns $x_1, \dots, x_n$ is an equation that can be written as

a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = b,

where the coefficients $a_1, \dots, a_n$ and the constant $b$ are fixed real numbers. A linear system is a finite collection of $m$ linear equations in the same unknowns $x_1, \dots, x_n$ . A solution of the system is a list of real numbers $(s_1, \dots, s_n)$ that makes every one of the $m$ equations true when each $s_j$ is substituted for $x_j$ . The set of all solutions is the solution set. On this page an inline tuple such as $(s_1, \dots, s_n)$ also stands for the column vector with the same entries.

In plain words Each equation is one requirement, and in it every unknown is only multiplied by a fixed number and added. A solution has to meet all of the requirements at the same time. Meeting some of them is not enough.

Example Consider the system

\begin{aligned} x + y &= 3 \\ x - y &= 1. \end{aligned}

The pair $(2, 1)$ is a solution, because $2 + 1 = 3$ and $2 - 1 = 1$ . The pair $(1, 2)$ is not, even though $1 + 2 = 3$ , since $1 - 2 = -1$ rather than $1$ . An equation such as $xy = 2$ or $x^2 + y = 1$ is not linear, because it multiplies an unknown by an unknown, $x$ by $y$ in the first and $x$ by itself in the second.

Matrix form of a system

Definition The system of $m$ linear equations in $n$ unknowns

\begin{aligned} a_{11} x_1 + \cdots + a_{1n} x_n &= b_1 \\ &\;\,\vdots \\ a_{m1} x_1 + \cdots + a_{mn} x_n &= b_m \end{aligned}

is the single matrix equation $A\mathbf{x} = \mathbf{b}$ . Here $A = [a_{ij}]$ is the $m \times n$ coefficient matrix, $\mathbf{x} = (x_1, \dots, x_n)$ is the column of unknowns in $\mathbb{R}^n$ , and $\mathbf{b} = (b_1, \dots, b_m)$ is the right-hand side in $\mathbb{R}^m$ . Entry $i$ of $A\mathbf{x}$ is the left side of equation $i$ , so a vector $\mathbf{x}$ solves the system exactly when $A\mathbf{x} = \mathbf{b}$ . The matrix $A$ is a map from $\mathbb{R}^n$ to $\mathbb{R}^m$ , and solving the system means finding every input that this map sends to $\mathbf{b}$ .

In plain words Collect the numbers that multiply the unknowns into one matrix, the unknowns into one column and the constants into another column. One matrix equation then carries all $m$ equations at once.

Example The system $x + y = 3$ , $x - y = 1$ becomes

\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}.

Multiplying out the left side gives the column with entries $x + y$ and $x - y$ , which recovers the two equations. The coefficient matrix is $2 \times 2$ because there are two equations and two unknowns.

Augmented matrix

Definition The augmented matrix of the system $A\mathbf{x} = \mathbf{b}$ is the $m \times (n + 1)$ matrix $[\,A \mid \mathbf{b}\,]$ obtained by placing $\mathbf{b}$ after the last column of $A$ . Row $i$ lists the coefficients of equation $i$ in the fixed order $x_1, \dots, x_n$ , followed by its constant $b_i$ . An unknown that is missing from an equation gets the coefficient $0$ . The vertical bar only separates the coefficients from the constants, so the last column is not the column of any unknown.

In plain words The augmented matrix is the system with the variable names, plus signs and equal signs erased. Nothing is lost, as long as everyone agrees on the order of the unknowns.

Example The system $x + y = 3$ , $x - y = 1$ has augmented matrix

\left[\begin{array}{cc|c} 1 & 1 & 3 \\ 1 & -1 & 1 \end{array}\right].

The system $x + 2z = 5$ , $3y - z = 2$ in the unknowns $x, y, z$ has augmented matrix

\left[\begin{array}{ccc|c} 1 & 0 & 2 & 5 \\ 0 & 3 & -1 & 2 \end{array}\right],

with a zero wherever an unknown is missing. Reading the rows back gives $1x + 0y + 2z = 5$ and $0x + 3y - 1z = 2$ , and the triple $(3, 1, 1)$ satisfies both.

Row picture

Definition In the row picture each of the $m$ equations is read as a set in $\mathbb{R}^n$ , the space of the unknowns, namely the set of all points that satisfy that one equation. If the coefficients of the equation are not all zero, this set is a line when $n = 2$ , a plane when $n = 3$ and a hyperplane in general. The solution set of the system is the intersection of these $m$ sets. An equation whose coefficients are all zero reads $0 = b_i$ , which every point satisfies when $b_i = 0$ and no point satisfies when $b_i \ne 0$ .

In plain words Draw each requirement as the set of points that meet it. The solutions are the points that lie on every one of these sets.

Example For $x + y = 3$ and $x - y = 1$ , the first line passes through $(0, 3)$ and $(3, 0)$ , and the second passes through $(1, 0)$ and $(0, -1)$ . The lines cross at $(2, 1)$ . A drawing suggests the point, and algebra confirms it. The second equation gives $x = y + 1$ , so the first becomes $2y + 1 = 3$ . Then $y = 1$ and $x = 2$ , and no other point is possible. With three unknowns the equation $x + y + z = 3$ is a plane in $\mathbb{R}^3$ , and a solution of several such equations is a point that lies on all of their planes.

Column picture

Definition Let $A$ have columns $\mathbf{a}_1, \dots, \mathbf{a}_n$ in $\mathbb{R}^m$ . Since $A\mathbf{x} = x_1\mathbf{a}_1 + \cdots + x_n\mathbf{a}_n$ , the system $A\mathbf{x} = \mathbf{b}$ is the vector equation

x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \cdots + x_n\mathbf{a}_n = \mathbf{b}.

In the column picture the unknowns are weights, and a solution is a choice of weights that combines the columns into the target $\mathbf{b}$ . This picture lives in the output space $\mathbb{R}^m$ , while the row picture lives in the input space $\mathbb{R}^n$ .

In plain words The row picture asks which points meet every constraint. The column picture asks how much of each column you need to reach the target.

Example For the system $x + y = 3$ , $x - y = 1$ , the columns of $A$ are $(1, 1)$ and $(1, -1)$ , and the target is $(3, 1)$ . The weights $x = 2$ and $y = 1$ reach it,

2\begin{bmatrix} 1 \\ 1 \end{bmatrix} + 1\begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}.

In the row picture $(2, 1)$ was the point being searched for. In the column picture $(3, 1)$ is the given target, and the numbers $2$ and $1$ are the weights that reach it.

Consistent and inconsistent systems

Definition A linear system is consistent if it has at least one solution and inconsistent if it has none. In the column picture, $A\mathbf{x} = \mathbf{b}$ is consistent exactly when $\mathbf{b}$ lies in $\operatorname{span}\{\mathbf{a}_1, \dots, \mathbf{a}_n\}$ , the span of the columns of $A$ . Every linear system has no solution, exactly one solution, or infinitely many solutions. Indeed, if $\mathbf{p} \ne \mathbf{q}$ are two solutions, then for every real $t$

\begin{aligned} &A\big(\mathbf{p} + t(\mathbf{q} - \mathbf{p})\big) \\[0.4em] &\quad = A\mathbf{p} + t(A\mathbf{q} - A\mathbf{p}) \\[0.4em] &\quad = \mathbf{b} + t(\mathbf{b} - \mathbf{b}) = \mathbf{b}, \end{aligned}

so every point of the line through $\mathbf{p}$ and $\mathbf{q}$ is a solution. Since $\mathbf{q} - \mathbf{p} \ne \mathbf{0}$ , different values of $t$ give different points, so two solutions already force infinitely many.

In plain words Consistent means the requirements can all be met, by one point or by many. Consistent does not mean unique. Inconsistent means the requirements contradict each other. Two lines in the plane can cross once, be parallel and never meet, or coincide, and the same three outcomes, none, one or infinitely many solutions, are the only ones possible for any linear system.

Example In the system $x + y = 3$ , $2x + 2y = 6$ the second equation is twice the first, so the two lines coincide. Every pair $(t, 3 - t)$ with $t$ real is a solution, for instance $(0, 3)$ , $(2, 1)$ and $(-1, 4)$ . Now change the second constant to $7$ . Any solution would have $2x + 2y = 2(x + y) = 6$ , which cannot equal $7$ , so the lines are parallel and the system is inconsistent. In the column picture both columns equal $(1, 2)$ , and their span is a line through the origin. The target $(3, 6) = 3(1, 2)$ lies on that line, and the target $(3, 7)$ does not.

Check yourself

Write the system $2x - y = 4$ , $x + 3y = -5$ as a matrix equation and as an augmented matrix. Is $(1, -2)$ a solution?

Show the answer

The matrix form is

\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ -5 \end{bmatrix},

and the augmented matrix is

\left[\begin{array}{cc|c} 2 & -1 & 4 \\ 1 & 3 & -5 \end{array}\right].

The pair $(1, -2)$ is a solution, since $2(1) - (-2) = 4$ and $1 + 3(-2) = -5$ . In the column picture the same check reads $1 \cdot (2, 1) + (-2) \cdot (-1, 3) = (4, -5)$ .

Describe all solutions of $x - 2y = 1$ , $-3x + 6y = -3$ . What happens if the second constant $-3$ is changed to $4$ ?

Show the answer

The second equation is $-3$ times the first, constant included, so the two lines coincide. Let $y = t$ . The first equation gives $x = 1 + 2t$ , so the solutions are the pairs $(1 + 2t, t)$ for every real $t$ . Each of them also satisfies the second equation, since $-3(1 + 2t) + 6t = -3$ . Conversely, every solution satisfies the first equation, so it has this form.

With the constant $4$ , any solution would satisfy $-3x + 6y = -3(x - 2y) = -3$ , which is not $4$ . The lines are then parallel and the system is inconsistent.

A student claims that three equations in three unknowns always have exactly one solution. Test the claim on the system $x + y + z = 2$ , $2x + 2y + 2z = 4$ , $x - y = 0$ .

Show the answer

The claim is false. The second equation is twice the first, so it adds no new requirement, and the third gives $y = x$ . Let $x = t$ . Then $y = t$ , and the first equation gives $z = 2 - 2t$ . Every triple $(t, t, 2 - 2t)$ satisfies all three equations, because $t + t + (2 - 2t) = 2$ , twice that sum is $4$ , and $t - t = 0$ . Conversely, every solution has $y = x$ and $z = 2 - 2x$ , so it has this form.

The system is consistent with infinitely many solutions, a line in $\mathbb{R}^3$ . Counting equations does not tell how many independent requirements they impose.

Linear Systems as Equations and Geometry

Key terms

Linear system

Matrix form of a system

Augmented matrix

Row picture

Column picture

Consistent and inconsistent systems

Check yourself

How this lesson connects

Builds on

Used later in