
# 2.5: Solution Sets for Systems of Linear Equations

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Algebra problems can have multiple solutions. For example $$x(x-1)=0$$ has  two solutions: $$0$$ and $$1$$. By contrast, equations of the form $$Ax=b$$ with $$A$$ a linear operator have have the following property.

If $$A$$ is a linear operator and $$b$$ is a known then $$Ax=b$$ has either

[1.]    One solution

[2.]    No solutions

[3.]    Infinitely many solutions

## 2.5.1: The Geometry of Solution Sets: Hyperplanes

Consider the following algebra problems and their solutions

[1.]    $$6x=12$$, one solution: $$2$$

[2a.]  $$0x=12$$, no solution

[2b.]  $$0x=0$$, one solution for each number: $$x$$

In each case the linear operator is a $$1\times 1$$ matrix. In the first case, the linear operator is invertible. In the other two cases it is not. In the first case, the solution set is a point on the number line, in the third case the solution set is the whole number line.

Lets examine similar situations with larger matrices.

[1.]        $$\begin{pmatrix}6 &0 \\0 &2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix}12 \\ 6\end{pmatrix}$$, one solution: $$\begin{pmatrix}2 \\ 3\end{pmatrix}$$

[2b.]      $$\begin{pmatrix}1 &3 \\0 &0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix}4 \\ 1 \end{pmatrix}$$, no solutions

[2bi.]      $$\begin{pmatrix}1 &3 \\0 &0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix}4 \\ 0\end{pmatrix}$$, one solution for each number $$y$$: $$\begin{pmatrix}4-3y \\ y\end{pmatrix}$$

[2bii.]     $$\begin{pmatrix}0 &0 \\0 &0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix}0 \\ 0\end{pmatrix}$$, one solution for each pair of numbers $$x,y$$:$$\begin{pmatrix}x\\ y\end{pmatrix}$$

Again, in the first case the linear operator is invertible while in the other cases it is not. When the operator is not invertible the solution set can be empty, a line in the plane or the plane itself.

For a system of equations with $$r$$ equations and $$k$$ variables, one can have a number of different outcomes.  For example, consider the case of $$r$$ equations in three variables.  Each of these equations is the equation of a plane in three-dimensional space.  To find solutions to the system of equations, we look for the common intersection of the planes (if an intersection exists).  Here we have five different possibilities:

[1.]        $$\textbf{Unique Solution.}$$  The planes have a unique point of intersection.

[2a.]      $$\textbf{No solutions.}$$  Some of the equations are contradictory, so no solutions exist.

[2bi.]     $$\textbf{Line.}$$  The planes intersect in a common line; any point on that line then gives a solution to the system of equations.

[2bii.]    $$\textbf{Plane.}$$  Perhaps you only had one equation to begin with, or else all of the equations coincide geometrically.  In this case, you have a plane of solutions, with two free parameters.

[2biii.]    $$\textbf{All of \(\mathbb{R}^3$$.}\)  If you start with no information, then any point in $$\mathbb{R}^3$$ is a solution. There are three free parameters.

In general, for systems of equations with $$k$$ unknowns, there are $$k+2$$ possible outcomes, corresponding to the possible numbers (i.e. $$0,1,2,\dots,k$$) of free parameters in the solutions set plus the possibility of no solutions.  These types of "solution sets'' are "hyperplanes'', generalizations of planes the behave like planes in $$\mathbb{R}^3$$ in many ways.

## 2.5.2: Particular Solution $$+$$ Homogeneous solutions

In the standard approach, variables corresponding to columns that do not contain a pivot (after going to reduced row echelon form) are $$\textit{free}$$. We called them non-pivot variables. They index elements of the solutions set by acting as coefficients of vectors.

Example 31    (Non-pivot columns determine terms of the solutions)

$$\begin{pmatrix} 1 & 0 & 1 & -1 \\ 0 & 1 & -1& 1 \\ 0 &0 & 0 & 0 \\ \end{pmatrix} \begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix} = \begin{pmatrix}1\\1\\0\end{pmatrix} \Leftrightarrow \left\{ \begin{array}{lcr} 1x_1 +0x_2+ 1x_3 - 1x_4 & = 1 \\ 0x_1 +1x_2 - 1x_3 + 1x_4 & = 1 \\ 0x_1 +0x_2 + 0x_3 + 0x_4 & = 0 \end{array} \right.$$
Following the standard approach, express the pivot variables in terms of the non-pivot variables and add "freebee equations". Here $$x_3$$ and $$x_4$$ are non-pivot variables.
$$\begin{eqnarray*} \left. \begin{array}{rcl} x_1 & = &1 -x_3+x_4 \\ x_2 & = &1 +x_3-x_4 \\ x_3 & = &\phantom{1+~\,}x_3\\ x_4 & =&\phantom{1+x_3+~}x_4 \end{array} \right\} \Leftrightarrow \begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix} = \begin{pmatrix}1\\1\\0\\0\end{pmatrix} + x_3\begin{pmatrix}-1\\1\\1\\0\end{pmatrix} + x_4\begin{pmatrix}1\\-1\\0\\1\end{pmatrix} \end{eqnarray*}$$
The preferred way to write a solution set is with set notation.  $$S = \left\{\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix} = \begin{pmatrix}1\\1\\ 0\\0\end{pmatrix} + \mu_1 \begin{pmatrix}-1\\1\\1\\0\end{pmatrix} + \mu_2 \begin{pmatrix}1\\-1\\ 0 \\1\end{pmatrix} : \mu_1,\mu_2\in {\mathbb R} \right\}$$
Notice that the first two components of the second two terms come from the non-pivot columns
Another way to write the solution set is
$S= \left\{ X_0 + \mu_1 Y_1 + \mu_2 Y_2 : \mu_1,\mu_2 \in {\mathbb R} \right\}$
where
$X_0= \begin{pmatrix}1\\1\\0 \\0\end{pmatrix}, Y_1=\begin{pmatrix}-1\\1\\1\\0\end{pmatrix} , Y_2= \begin{pmatrix}1\\-1\\0 \\1\end{pmatrix}$

Here $$X_0$$ is called a particular solution while $$Y_1$$ and $$Y_2$$ are called homogeneous solutions.

## 2.5.3: Linearity and these parts

With the previous example in mind, lets say that the matrix equation $$MX=V$$ has  solution set  $$\{ X_0 + \mu_1 Y_1 + \mu_2 Y_2):\mu_1,\mu_2 \in {\mathbb R} \}$$. Recall that matrices are linear operators. Thus
$$M( X_0 + \mu_1 Y_1 + \mu_2 Y_2) = MX_0 + \mu_1MY_1 + \mu_2MY_2 =V$$
for $$\textit{any}$$ $$\mu_1, \mu_2 \in \mathbb{R}$$. Choosing $$\mu_1=\mu_2=0$$, we obtain
$$MX_0=V\, .$$
This is why $$X_0$$ is an example of a  $$\textit{particular solution}$$.

Setting $$\mu_1=1, \mu_2=0$$, and using the particular solution $$MX_0=V$$, we obtain
$$MY_1=0\, .$$
Likewise, setting $$\mu_1=0, \mu_2=1$$, we obtain $$MY_2=0\, .$$
Here $$Y_1$$ and $$Y_2$$ are examples of what are called $$\textit{homogeneous}$$ solutions to the system. They $$\textit {do not}$$ solve the original equation $$MX=V$$, but instead its associated $$\textit {homogeneous equation}$$ $$M Y =0$$.

One of the fundamental lessons of linear algebra: the solution set to $$Ax=b$$ with $$A$$ a linear operator consists of a particular solution plus homogeneous solutions.

$$general ~solution = particular~ solution + homogeneous~ solutions.$$

Example 32

Consider the matrix equation of the previous example. It has solution set
$S = \left\{\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix} = \begin{pmatrix}1\\1\\0 \\0\end{pmatrix} + \mu_1 \begin{pmatrix}-1\\1\\1\\0\end{pmatrix} + \mu_2 \begin{pmatrix}1\\-1\\ 0\\1\end{pmatrix} \right\}$
Then $$MX_0=V$$ says that $$\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix} = \begin{pmatrix}1\\1\\0 \\ 0\end{pmatrix}$$ solves the original matrix equation, which is certainly true, but this is not the only solution.

$$MY_1=0$$ says that $$\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix} = \begin{pmatrix}-1\\1\\1\\ 0\end{pmatrix}$$ solves the homogeneous equation.

$$MY_2=0$$ says that $$\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix} = \begin{pmatrix}1\\-1\\0 \\1\end{pmatrix}$$ solves the homogeneous equation.

Notice how adding any multiple of a homogeneous solution to the particular solution yields another particular solution