The concept of a linear equation will now be extended to systems of linear equations.
A system of linear equations is understood to be one or more linear equations with one or more unknowns.
A solution of the system of equations is a list of values of the unknowns that, when substituted into each equation of the system, gives a set of equations that are valid (a true statement). Not every system of equations has a solution: the system \[ \left\{\;\begin{aligned} x + y\;&= 1 \\ x+ y\;&=2\end{aligned} \right.\] for example, has no solution. Such a system is called conflicting.
Solving a system of equations is the determination of all the solutions. The result is also called the solution.
Two systems of linear equations are called equivalent if they have the same solution.
A system of linear equations is often solved by reduction, that is, by replacing it by another system of linear equations that is both simpler than the previous one, and equivalent to it.
In case the system is conflicting, mathematicians speak of the empty set of solutions or simply the empty solution. We will not use this terminology in this course; we will simply say it has no solution.
We usually fix an order of the unknowns and write solutions as a list of values of the variables in this order; but other formats are also widely used.
Note the difference between a solution and the solution: the solution of the linear equation #x=x+y# in the unknowns #x# and #y# is #y=0#, but #x=1 \land y=0# is a solution.
The solution of an equation with the two unknowns #x# and #y# is generally a line in the #x,y# plane. The solutions of two equations with two unknowns are the points that belong to the lines of both equations. In the example given, \[ \left\{\;\begin{aligned} x + y\;&= 1 \\ x+ y\;&=2\end{aligned} \right.\] we are dealing with two parallel lines.
In the reduction, usually elementary operations are being used. These are the following three operations on systems of linear equations:
- The multiplication of both sides of a equation by the same number distinct from zero.
- The addition of a multiple of one equation to one of the other equations.
- The interchange of two equations.
Later we will discuss these operations more extensively.
Consider the system \[\left\{\;\begin{aligned} 2x + 3y\;&= 1 \\ \phantom{2}x+\phantom{3}y\;&=1\end{aligned} \right.\] with unkowns \(x\) and \(y\), which we also write as \[{2 x +3y = 1 \quad \land\quad x +y =1 }\] Here, \(\land\) is the logical "and" operator.
A solution of this system is \[x=2\ \text{ and }\ y=-1\] In order to see that this is a solution, we substitute these values in the equations: \[\left\{\begin{aligned} 2\cdot 2+3\cdot (-1)\;&=1 \\ \phantom{2\cdot } 2+\,\phantom{3\cdot } (-1)\;&=1\end{aligned} \right.\] These equalities hold, so \(x=2\land y=-1\) is a solution. This solution can also be written as \ \[\rv{x,y}=\rv{2,-1}\] or \[x=2\quad\land\quad y=-1\] Solving the system is finding all solutions. In this case, there are no more solutions, as we show below.
If we subtract the second equation twice from the first equation, we get a new linear equation, namely \(y=-1\). The original system of equations is equivalent to \[\left\{\begin{aligned} \phantom{x+}y\;&=-1 \\ x+y\;&=\phantom{-}1\end{aligned} \right.\] Subtracting the first equation of the new system from the second equation, we get the following equivalent system: \[\left\{\begin{aligned} y\;&=-1 \\ x\;&=\phantom{-}2\end{aligned} \right.\] Interchanging the two equations gives the solution in a familiar order of variables.
We discuss how a system of linear equations can be seen as a system of vector equations.
The system of equations \[\left\{\;\begin{aligned} 2x + 3y\;&= 1 \\ \phantom{2}x+\phantom{3}y\;&=1\end{aligned} \right.\] can also be written in vector form as \[x\cdot\cv{2\cr 1\cr}+y\cdot\cv{3\cr 1\cr}= \cv{1\cr 1\cr}\] In this form, solving the system is nothing but finding an expression of the vector \(\cv{1\cr 1\cr}\) as a linear combination of the vectors \(\cv{2\cr 1\cr}\) and \(\cv{3\cr 1\cr}\).
The vectors are written here as column vectors. The column vector \(\cv{3\cr 1}\) corresponds to the row vector #\rv{3,1}#. The system of equations \[\left\{\;\begin{aligned} 2x + 3y\;&= 1 \\ \phantom{2}x+\phantom{3}y\;&=1\end{aligned} \right.\] can be also written in terms of row vectors: \[x\cdot\rv{2, 1}+y\cdot\rv{3, 1}= \rv{1, 1}\]
We show why the system of linear equations is equivalent to the equation in vector form. We have
\[ x\cdot\cv{2\cr 1\cr}+y\cdot\cv{3\cr 1\cr}= \cv{2x\cr x\cr}+\cv{3y\cr y\cr} =\cv{2x+3y\cr x+y\cr}\]
The vector equation can therefore be written as
\[\cv{2x+3y\cr x+y\cr}=\cv{1\cr 1\cr}\]
This amounts to an equality for each coordinate:
\[\left\{\;\begin{aligned} 2x + 3y\;&= 1 \\ \phantom{2}x+\phantom{3}y\;&=1\end{aligned} \right.\]
Thus, we recovered the original system of equations.
We will often see that a linear algebra problem eventually leads to solving a system of linear equations. Therefore we pay considerable attention to finding solutions.
A system of linear equations can have more equations than unknowns. We give an example of two equations with one unknown. By repeating the dynamic example at the bottom of this page to generate repeated below you can see that in this case there are three possibilities:
- There is no solution: the equations may be conflicting. That means that no solution of one equation form the system is also a solution of the other. Also, it may be that one of the equations has no solution.
- There is exactly one solution.
- There is more than one solution: in this case the number of solutions is even infinite, because all the values of the unknown will satisfy both equations.
We have already seen that the system \[ \left\{\;\begin{aligned} x + y\;&= 1 \\ x+ y\;&=2\end{aligned} \right.\] is conflicting.
The system \[ \left\{\;\begin{aligned} x \;&= 1 \\ y\;&=2\end{aligned} \right.\] with unkowns #x# and #y# has exactly one solution.
The system \[ \left\{\;\begin{aligned} x \;&= x \\ y\;&=3\end{aligned} \right.\]
has more than one solution: for each value #a# of #x# , the pair #\rv{x,y} = \rv{a,3}# is a solution.
Solve the following system of equations for #x#: \[\lineqs{4 x + 5 &=& 2 x-1 \cr 8 x -4 &=& 2 x - 22 \cr}\]
#x=-3#
There are two linear equations with the same unknown, namely #x#. We can solve these individually: \[\lineqs{ x &=& -3 \\ x &=&\displaystyle -3 }\] A look at the two separate solutions gives that there is one solution.
Therefore, the answer is #x=-3#.
The notion of a system of linear equations
