Home page for accesible maths 5 Systems of linear equations 5.2 Row operations 5.4 Systems with parameters

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

5.3. Number of solutions

Given a system of linear equations, how many solutions are there?

We shall see that a system of linear equations has either zero, one, or an infinite number of solutions.

Definition 5.3.1.

Let

\left(\begin{array}[]{rrr|r}a_{11}&\dots&a_{1m}&s_{1}\\ \vdots&\vdots&\vdots&\vdots\\ a_{n1}&\dots&a_{nm}&s_{n}\end{array}\right)

be the augmented matrix of a given system of linear equations, for some integers $n,m\geq 1$ .

(i)

The system is called homogeneous if $s_{1}=\cdots=s_{n}=0$ .
(ii)

We call a solution to a system of equations non-trivial, if at least one of the variables has non-zero value.

Example 5.3.2.

1. (a)
  
  The system of equations
  
  $\begin{array}[]{rcrcr}x&+&y&=&3\\ 2x&-&y&=&6\end{array}$
  
  has a non-trivial solution, namely $(x,y)=(3,0)$ .
2. (b)
  
  The homogeneous system of equations
  
  $\begin{array}[]{rcrcr}x&+&y&=&0\\ 2x&-&y&=&0\end{array}$
  
  has no non-trivial solution. Indeed, $(x,y)=(0,0)$ is the unique solution.

An advantage of systems with two equations is that we can work out a graphical solution, as we did in our first example. Indeed, an equation $ax+by=c$ (in the variables $x$ and $y$ ) is a line in the $(x,y)$ -plane, and so the solutions to a system of equations with two variables are the intersection point(s) of all the lines. In other words, the possible outcomes of a system of two linear equations in two variables are as follows:

(i)

The lines are distinct and parallel. (Zero solutions)
(ii)

There is a unique intersection point. (One solution)
(iii)

The lines are the same line. (Infinitely many solutions)

Example 5.3.3.

We consider four systems of linear equations in two variables $x$ and $y$ , all in fact very similar to each other, but leading to four different configurations of solutions.
- (a)
  
  – Unique trivial solution.
  
  $\begin{array}[]{rcrcr}2x&+&5y&=&0\\ -x&+&2y&=&0\end{array}$
  
  The system of linear equations has the following augmented matrix form, which we row reduce:
  
  $\left(\begin{array}[]{rr|r}2&5&0\\ -1&2&0\end{array}\right)\xrightarrow{\begin{array}[]{l}R_{1}\leftrightarrow R_% {2}\\ R_{2}=r_{2}+2r_{1}\end{array}}\left(\begin{array}[]{rr|r}-1&2&0\\ 0&9&0\end{array}\right)\xrightarrow{\begin{array}[]{l}R_{2}=\frac{1}{9}R_{2}\\ R_{1}=-r_{1}\\ R_{1}=r_{1}+2r_{2}\end{array}}(I_{2}\ |\;0).$
  
  The conclusion is that there is a unique solution, namely $(x,y)=(0,0)$ , which is called the trivial solution.
- (b)
  
  – Unique non-trivial solution.
  
  $\begin{array}[]{rcrcr}2x&+&5y&=&16\\ -x&+&2y&=&1\end{array}$
  
  The system of linear equations has the following augmented matrix form, which we row reduce, using the same operations as in (a):
  
  $\left(\begin{array}[]{rr|r}2&5&16\\ -1&2&1\end{array}\right)\xrightarrow{\begin{array}[]{l}R_{1}\leftrightarrow R_% {2}\\ R_{2}=r_{2}+2r_{1}\end{array}}\left(\begin{array}[]{rr|r}-1&2&1\\ 0&9&18\end{array}\right)\xrightarrow{\begin{array}[]{l}R_{2}=\frac{1}{9}R_{2}% \\ R_{1}=-r_{1}\\ R_{1}=r_{1}+2r_{2}\end{array}}\left(\begin{array}[]{rr|r}1&0&3\\ 0&1&2\end{array}\right).$
  
  Converting the final augmented matrix back into equations, we see $x=3$ and $y=2$ . So there is a unique non-trivial solution.
- (c)
  
  – Infinity of solutions.
  
  $\begin{array}[]{rcrcr}2x&-&4y&=&-10\\ -x&+&2y&=&5\end{array}$
  
  The given system of linear equations in the variables $x$ and $y$ has the following augmented matrix form, which we row reduce:
  
  $\left(\begin{array}[]{rr|r}2&-4&-10\\ -1&2&5\end{array}\right)\xrightarrow{\begin{array}[]{l}R_{1}=\frac{1}{2}r_{1}% \\ R_{2}=r_{2}+r_{1}\end{array}}\left(\begin{array}[]{rr|r}1&-2&-5\\ 0&0&0\end{array}\right).$
  
  This final matrix is in reduced row echelon form, and by Theorem 5.2.4, the original system is equivalent to the single equation $x-2y=-5$ , which has infinitely many solutions. Therefore the original system of equations has infinitely many solutions.
- (d)
  
  – No solutions: inconsistent system of equations.
  
  $\begin{array}[]{rcrcr}2x&-&4y&=&4\\ -x&+&2y&=&5\end{array}$
  
  We row reduce the associated augmented matrix:
  
  $\left(\begin{array}[]{rr|r}2&-4&4\\ -1&2&5\end{array}\right)\xrightarrow{\begin{array}[]{l}R_{1}=\frac{1}{2}r_{1}% \\ R_{2}=r_{2}+r_{1}\end{array}}\left(\begin{array}[]{rr|r}1&-2&2\\ 0&0&7\end{array}\right).$
  
  Translating this final matrix back into equations, we find that the original system is equivalent to one involving the equation $0x+0y=7$ . Since the left-hand side is 0, and can never equal 7, there are no solutions.
  
  Geometrically, the lines corresponding to each equation do not intersect, because they are parallel and distinct.

We end this section on systems of linear equations with a useful result concerning homogeneous systems of equations. This concept is relevant in linear algebra, in the study of eigenvalues and eigenvectors which is the topic of Section 7.

Lemma 5.3.4.

If $A_{r}\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ is a matrix in reduced row echelon form, then the equation $A_{r}v=0$ , (where $v$ is a column vector in $\operatorname{M}_{{n\times 1}}({{\mathbb{R}}})$ ) has a non-trivial solution if and only if $A_{r}$ is not the identity matrix.

Proof.

If $A_{r}$ is the identity matrix, then clearly the only solution is $v=0$ , the trivial one. If $A_{r}$ is not the identity matrix, then it has at most $n-1$ non-zero rows, and hence at most $n-1$ leading coefficients. Choose a variable $x_{i}$ corresponding to a column not containing a leading coefficient. Whatever value we choose for $x_{i}$ , it is easy to construct a solution for the other variables which satisfy the equations. So we have constructed a non-trivial solution. ∎

Theorem 5.3.5.

Let $A=(a_{ij})\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ , for some integer $n\geq 1$ , and let

v=\begin{pmatrix}x_{1}\\ \vdots\\ x_{n}\end{pmatrix}\quad\in\operatorname{M}_{{n\times 1}}({{\mathbb{R}}}).

Then the homogeneous system

Av=0\quad\in\operatorname{M}_{{n\times 1}}({{\mathbb{R}}})

has a non-trivial solution if and only if $\det A=0$ .

Proof.

By Theorem 5.2.4, the solutions of $Av=0$ are the same as the solutions of $A_{r}v=0$ . By Theorem 4.4.4 together with Lemma 3.3.3, we know that $\det A=0$ if and only if $A_{r}$ is not the identity. So by Lemma 5.3.4, the result follows. ∎

Example 5.3.6.

Consider this Theorem applied to Example 5.3(a): Since $\det\begin{pmatrix}2&5\\ -1&2\end{pmatrix}=9$ , is not zero, the only solution is the trivial one.

Corollary 5.3.7.

Let $A=(a_{ij})\in\operatorname{M}_{{n\times m}}({{\mathbb{R}}})$ and let

v=\begin{pmatrix}x_{1}\\ \vdots\\ x_{m}\end{pmatrix}\in\operatorname{M}_{{m\times 1}}({{\mathbb{R}}}).

Assume that $n<m$ . Then the homogeneous system

Av=0\quad\hbox{in $\operatorname{M}_{{m\times 1}}({{\mathbb{R}}})$ has a non-% trivial solution.}\quad

Proof.

Note that the above system of equations is equivalent to the one given by the augmented matrix

\left(\begin{array}[]{c|r}\hbox{\large{\bf A}}_{n\times m}&0_{n\times 1}\\ \\ \hline\\ 0_{(m-n)\times m}&0_{(m-n)\times 1}\end{array}\right),

where, for sake of clarity, the indices indicate the sizes of the given matrices. In other words, we complete the matrix $A$ into a square $m\times m$ matrix by adding rows of zeros at the bottom, which we can do at least once, because $n<m$ by assumption. It follows that the determinant of the left-hand side $m\times m$ matrix

\left(\begin{array}[]{c}\hbox{\large{\bf A}}_{n\times m}\\ \hline 0_{(m-n)\times m}\end{array}\right)\quad\in\operatorname{M}_{{m}}({{% \mathbb{R}}})

is zero and so, the result follows from Theorem 5.3.5. ∎