### 3 Non-unique solutions

The key to obtaining a unique solution of the system $AX=B$ is to find ${A}^{-1}$ . What happens when ${A}^{-1}$ does not exist?

Consider the system

$\phantom{\rule{2em}{0ex}}2{x}_{1}+3{x}_{2}=5$ (1)

$\phantom{\rule{2em}{0ex}}4{x}_{1}+6{x}_{2}=10$ (2)

In matrix form this becomes

$\phantom{\rule{2em}{0ex}}\left[\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill 4\hfill & \hfill 6\hfill \end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 5\\ \hfill 10\end{array}\right].$

Identify the matrix $A$ and hence find ${A}^{-1}$ .

$A=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill 4\hfill & \hfill 6\hfill \end{array}\right]$ and $\left|A\right|=2×6-4×3=0.$ Hence ${A}^{-1}$ does not exist.

Looking at the original system we see that Equation (2) is simply Equation (1) with all coefficients doubled. In effect we have only one equation for the two unknowns ${x}_{1}$ and ${x}_{2}$ . The equations are consistent and have infinitely many solutions .

If we let ${x}_{2}$ assume a particular value, $t$ say, then ${x}_{1}$ must take the value ${x}_{1}=\frac{1}{2}\left(5-3t\right)$ i.e. the solution to the given equations is:

For each value of $t$ there are unique values for ${x}_{1}$ and ${x}_{2}$ which satisfy the original system of equations. For example, if $t=1$ , then ${x}_{2}=1$ , ${x}_{1}=1$ , if $t=-3$ then ${x}_{2}=-3$ , ${x}_{1}=7$ and so on.

Now consider the system

$\phantom{\rule{2em}{0ex}}2{x}_{1}+3{x}_{2}=5$ (3)

$\phantom{\rule{2em}{0ex}}4{x}_{1}+6{x}_{2}=9$ (4)

Since the left-hand sides are the same as those in the previous system then $A$ is the same and again ${A}^{-1}$ does not exist. There is no solution to the Equations (3) and (4).

However, if we double Equation (3) we obtain

$\phantom{\rule{2em}{0ex}}4{x}_{1}+6{x}_{2}=10,$

which conflicts with Equation (4). There are thus no solutions to (3) and (4) and the equations are said to be inconsistent .

What can you conclude about the solutions of the following systems?

1. $\begin{array}{ccccc}\hfill {x}_{1}& \hfill -& \hfill 2{x}_{2}& \hfill =& \hfill 1\\ \hfill 3{x}_{1}& \hfill -& \hfill 6{x}_{2}& \hfill =& \hfill 3\end{array}$
2. $\begin{array}{ccccc}\hfill 3{x}_{1}& \hfill +& \hfill 2{x}_{2}& \hfill =& \hfill 7\\ \hfill -6{x}_{1}& \hfill -& \hfill 4{x}_{2}& \hfill =& \hfill 5\end{array}$

First write the systems in matrix form and find $\left|A\right|$ :

1. $\left[\begin{array}{cc}\hfill 1& \hfill -2\\ \hfill 3& \hfill -6\end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 3\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}|A|=-6+6=0;$
2. $\left[\begin{array}{cc}\hfill 3& \hfill 2\\ \hfill -6& \hfill -4\end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 7\hfill \\ \hfill 5\hfill \end{array}\right]\phantom{\rule{2em}{0ex}}|A|=-12+12=0.$

Now compare the two equations in each system in turn:

1. The second equation is $3$ times the first equation. There are infinitely many solutions of the

form   ${x}_{2}=t,\phantom{\rule{1em}{0ex}}{x}_{1}=1+2t$  where $t$ is arbitrary.

2. If we multiply the first equation by $\left(-2\right)$ we obtain $-6{x}_{1}-4{x}_{2}=-14$ which is in conflict with

the second equation. The equations are inconsistent and have no solution.