Let’s start off this section with the definition of a linear equation.  Here are a couple of examples of linear equations.

In the second equation note the use of the subscripts on the variables.  This is a common notational device that will be used fairly extensively here.  It is especially useful when we get into the general case(s) and we won’t know how many variables (often called unknowns) there are in the equation.

 

So, just what makes these two equations linear?  There are several main points to notice.  First, the unknowns only appear to the first power and there aren’t any unknowns in the denominator of a fraction.  Also notice that there are no products and/or quotients of unknowns.  All of these ideas are required in order for an equation to be a linear equation.  Unknowns only occur in numerators, they are only to the first power and there are no products or quotients of unknowns. 

 

The most general linear equation is,

(1)

where there are n unknowns, , and  are all known numbers.

 

Next we need to take a look at the solution set of a single linear equation.  A solution set (or often just solution) for (1) is a set of numbers  so that if we set , , … ,  then (1) will be satisfied.  By satisfied we mean that if we plug these numbers into the left side of (1) and do the arithmetic we will get b as an answer.

 

The first thing to notice about the solution set to a single linear equation that contains at least two variables with non-zero coefficents is that we will have an infinite number of solutions.  We will also see that while there are infinitely many possible solutions they are all related to each other in some way.

 

Note that if there is one or less variables with non-zero coefficients then there will be a single solution or no solutions depending upon the value of b.

 

Let’s find the solution sets for the two linear equations given at the start of this section.

 

Example 1  Find the solution set for each of the following linear equations. (a)    [Solution] (b)    [Solution]   Solution (a)   The first thing that we’ll do here is solve the equation for one of the two unknowns.  It doesn’t matter which one we solve for, but we’ll usually try to pick the one that will mean the least amount (or at least simpler) work.  In this case it will probably be slightly easier to solve for  so let’s do that.                                                            Now, what this tells us is that if we have a value for  then we can determine a corresponding value for .  Since we have a single linear equation there is nothing to restrict our choice of  and so we we’ll let  be any number.  We will usually write this as , where t is any number.  Note that there is nothing special about the t, this is just the letter that I usually use in these cases.  Others often use s for this letter and, of course, you could choose it to be just about anything as long as it’s not a letter representing one of the unknowns in the equation (x in this case).   Once we’ve “chosen”  we’ll write the general solution set as follows,                                                       So, just what does this tell us as far as actual number solutions go?  We’ll choose any value of t and plug in to get a pair of numbers  and  that will satisfy the equation.  For instance picking a couple of values of t completely at random gives,                                            We can easily check that these are in fact solutions to the equation by plugging them back into the equation.                                                 So, for each case when we plugged in the values we got for  and  we got -1 out of the equation as we were supposed to.   Note that since there an infinite number of choices for t there are in fact an infinite number of possible solutions to this linear equation. [Return to Problems]   (b)   We’ll do this one with a little less detail since it works in essentially the same manner.  The fact that we now have three unknowns will change things slightly but not overly much.  We will first solve the equation for one of the variables and again it won’t matter which one we chose to solve for.                                                              In this case we will need to know values for both x and y in order to get a value for z.  As with the first case, there is nothing in this problem to restrict out choices of x and y.  We can therefore let them be any number(s).  In this case we’ll choose  and .  Note that we chose different letters here since there is no reason to think that both x and y will have exactly the same value (although it is possible for them to have the same value).   The solution set to this linear equation is then,                                              So, if we choose any values for t and s we can get a set of number solutions as follows.                                    As with the first part if we take either set of three numbers we can plug them into the equation to verify that the equation will be satisfied.  We’ll do one of them and leave the other to you to check.                                          [Return to Problems]

 

The variables that we got to choose values for ( in the first example and x and y in the second) are sometimes called free variables.

 

We now need to start talking about the actual topic of this section, systems of linear equations.  A system of linear equations is nothing more than a collection of two or more linear equations.  Here are some examples of systems of linear equations.

 

 

As we can see from these examples systems of equations can have any number of equations and/or unknowns.  The system may have the same number of equations as unknowns, more equations than unknowns, or fewer equations than unknowns.

 

A solution set to a system with n unknowns, , is a set of numbers, , so that if we set , , … ,  then all of the equations in the system will be satisfied.  Or, in other words, the set of numbers  is a solution to each of the individual equations in the system.

 

For example, ,  is a solution to the first system listed above,

(2)

because,

 

However, ,  is not a solution to the system because,

We can see from these calculations that ,  is NOT a solution to the first equation, but it IS a solution to the second equation.  Since this pair of numbers is not a solution to both of the equations in (2) it is not a solution to the system.  The fact that it’s a solution to one of them isn’t material.  In order to be a solution to the system the set of numbers must be a solution to each and every equation in the system.

 

It is completely possible as well that a system will not have a solution at all.  Consider the following system.

(3)

 

It is clear (hopefully) that this system of equations can’t possibly have a solution.  A solution to this system would have to be a pair of numbers x and y so that if we plugged them into each equation it will be a solution to each equation.  However, since the left side is identical this would mean that we’d need an x and a y so that  is both 10 and -3 for the exact same pair of numbers.  This clearly can’t happen and so (3) does not have a solution.

 

Likewise, it is possible for a system to have more than one solution, although we do need to be careful here as we’ll see.  Let’s take a look at the following system.

(4)

 

We’ll leave it to you to verify that all of the following are four of the infinitely many solutions to the first equation in this system.

Recall from our work above that there will be infinitely many solutions to a single linear equation.

 

We’ll also leave it to you to verify that these four solutions are also four of the infinitely many solutions to the second equation in (4).

 

Let’s investigate this a little more.  Let’s just find the solution to the first equation (we’ll worry about the second equation in a second).  Following the work we did in Example 1 we can see that the infinitely many solutions to the first equation in (4) are

 

Now, if we also find just the solutions to the second equation in (4) we get

 

These are exactly the same!  So, this means that if we have an actual numeric solution (found by choosing t above…) to the first equation it will be guaranteed to also be a solution to the second equation and so will be a solution to the system (4).  This means that we in fact have infinitely many solutions to (4).

 

Let’s take a look at the three systems we’ve been working with above in a little more detail.  This will allow us to see a couple of nice facts about systems. 

 

Since each of the equations in (2),(3), and (4) are linear in two unknowns (x and y) the graph of each of these equations is that of a line.  Let’s graph the pair of equations from each system on the same graph and see what we get.

 

 

 

 

From the graph of the equations for system (2) we can see that the two lines intersect at the point  and notice that, as a point, this is the solution to the system as well.  In other words, in this case the solution to the system of two linear equations and two unknowns is simply the intersection point of the two lines.

 

Note that this idea is validated in the solution to systems (3) and (4).  System (3) has no solution and we can see from the graph of these equations that the two lines are parallel and hence will never intersect.  In system (4) we had infinitely many solutions and the graph of these equations shows us that they are in fact the same line, or in some ways they “intersect” at an infinite number of points.

 

Now, to this point we’ve been looking at systems of two equations with two unknowns but some of the ideas we saw above can be extended to general systems of n equations with m unknowns.

 

First, there is a nice geometric interpretation to the solution of systems with equations in two or three unknowns.  Note that the number of equations that we’ve got won’t matter the interpretation will be the same. 

 

If we’ve got a system of linear equations in two unknowns then the solution to the system represents the point(s) where all (not some but ALL) of the lines will intersect.  If there is no solution then the lines given by the equations in the system will not intersect at a single point.  Note in the no solution case if there are more than two equations it may be that any two of the equations will intersect, but there won’t be a single point were all of the lines will intersect.

 

If we’ve got a system of linear equations in three unknowns then the graphs of the equations will be planes in 3D-space and the solution to the system will represent the point(s) where all the planes will intersect.  If there is no solution then there are no point(s) where all the planes given by the equations of the system will intersect.  As with lines, it may be in this case that any two of the planes will intersect, but there won’t be any point where all of the planes intersect at that point.

 

On a side note we should point out that lines can intersect at a single point or if the equations give the same line we can think of them as intersecting at infinitely many points.  Planes can intersect at a point or on a line (and so will have infinitely many intersection points) and if the equations give the same plane we can think of the planes as intersecting at infinitely many places. 

 

We need to be a little careful about the infinitely many intersection points case.  When we’re dealing with equations in two unknowns and there are infinitely many solutions it means that the equations in the system all give the same line.  However, when dealing with equations in three unknowns and we’ve got infinitely many solutions we can have one of two cases.  Either we’ve got planes that intersect along a line, or the equations will give the same plane.

 

For systems of equations in more than three variables we can’t graph them so we can’t talk about a “geometric” interpretation, but we can still say that a solution to such a system will represent the point(s) where all the equations will “intersect” even if we can’t visualize such an intersection point.

 

From the geometric interpretation of the solution to two equations in two unknowns we know that we have one of three possible solutions.  We will have either no solution (the lines are parallel), one solution (the lines intersect at a single point) or infinitely many solutions (the equations are the same line).  There is simply no other possible number of solutions since two lines that intersect will either intersect exactly once or will be the same line.  It turns out that this is in fact the case for a general system.

 

Theorem 1 Given a system of n equations and m unknowns there will be one of three possibilities for solutions to the system. There will be no solution. There will be exactly one solution. There will be infinitely many solutions.   If there is no solution to the system we call the system inconsistent and if there is at least one solution to the system we call it consistent.

 

Now that we’ve got some of the basic ideas about systems taken care of we need to start thinking about how to use linear algebra to solve them.  Actually that’s not quite true.  We’re not going to do any solving until the next section.  In this section we just want to get some of the basic notation and ideas involved in the solving process out of the way before we actually start trying to solve them.

 

We’re going to start off with a simplified way of writing the system of equations.  For this we will need the following general system of n equations and m unknowns.

 

(5)

 

In this system the unknowns are  and the  and  are known numbers.  Note as well how we’ve subscripted the coefficients of the unknowns (the  ).  The first subscript, i, denotes the equation that the subscript is in and the second subscript, j, denotes the unknown that it multiplies.  For instance,  would be in the coefficient of  in the third equation.

 

Any system of equations can be written as an augmented matrix.  A matrix is just a rectangular array of numbers and we’ll be looking at these in great detail in this course so don’t worry too much at this point about what a matrix is.  Here is the augmented matrix for the general system in (5).

 

Each row of the augmented matrix consists of the coefficients and constant on the right of the equal sign from a given equation in the system.  The first row is for the first equation, the second row is for the second equation etc.  Likewise each of the first m columns of the matrix consists of the coefficients from the unknowns.  The first column contains the coefficients of , the second column contains the coefficients of , etc.  The final column (the m+1^st column) contains all the constants on the right of the equal sign.  Note that the augmented part of the name arises because we tack the  ’s onto the matrix.  If we don’t tack those on and we just have

and we call this the coefficient matrix for the system.

 

Example 2  Write down the augmented matrix for the following system.                                                        Solution There really isn’t too much to do here other than write down the system.                                                  Notice that the second equation did not contain an  and so we consider its coefficient to be zero.

 

Note as well that given an augmented matrix we can always go back to a system of equations.

 

Example 3  For the given augmented matrix write down the corresponding system of equations.                                                               Solution So since we know each row corresponds to an equation we have three equations in the system.  Also, the first two columns represent coefficients of unknowns and so we’ll have two unknowns while the third column consists of the constants to the right of the equal sign.  Here’s the system that corresponds to this augmented matrix.

 

There is one final topic that we need to discuss in this section before we move onto actually solving systems of equation with linear algebra techniques.  In the next section where we will actually be solving systems our main tools will be the three elementary row operations.  Each of these operations will operate on a row (which shouldn’t be too surprising given the name…) in the augmented matrix and since each row in the augmented matrix corresponds to an equation these operations have equivalent operations on equations.

 

Here are the three row operations, their equivalent equation operations as well as the notation that we’ll be using to denote each of them.

 

 

The first two operations are fairly self explanatory.  The third is also a fairly simple operation however there are a couple things that we need to make clear about this operation.  First, in this operation only row (equation) j actually changes.  Even though we are multiplying row (equation) i by c that is done in our heads and the results of this multiplication are added to row (equation) j.  Also, when we say that we add c times a row to another row we really mean that we add corresponding entries of each row.

 

Let’s take a look at some examples of these operations in action.

 

Example 4  Perform each of the indicated row operations on given augmented matrix.                                                           (a)    [Solution] (b)    [Solution] (c)    [Solution] (d)    [Solution] (e)    [Solution]   Solution In each of these we will actually perform both the row and equation operations to illustrate that they are actually the same operation and that the new augmented matrix we get is in fact the correct one.  For reference purposes the system corresponding to the augmented matrix given for this problem is,                                                              Note that at each part we will go back to the original augmented matrix and/or system of equations to perform the operation.  In other words, we won’t be using the results of the previous part as a starting point for the current operation.   (a)   Okay, in this case we’re going to multiply the first row (equation) by -3.  This means that we will multiply each element of the first row by -3 or each of the coefficients of the first equation by -3.  Here is the result of this operation.                   [Return to Problems]     (b)   This is similar to the first one.  We will multiply each element of the second row by one-half or each coefficient of the second equation by one-half.  Here are the results of this operation.                          Do not get excited about the fraction showing up.  Fractions are going to be a fact of life with much of the work that we’re going to be doing so get used to seeing them.   Note that often in cases like this we will say that we divided the second row by 2 instead of multiplied by one-half. [Return to Problems]   (c)   In this case were just going to interchange the first and third row or equation.                           [Return to Problems]   (d)   Okay, we now need to work an example of the third row operation.  In this case we will add 5 times the third row (equation) to the second row (equation).    So, for the row operation, in our heads we will multiply the third row times 5 and then add each entry of the results to the corresponding entry in the second row.   Here are the individual computations for this operation.                                                      For the corresponding equation operation we will multiply the third equation by 5 to get,                                                           then add this to the second equation to get,                                                             Putting all this together and remembering that it’s the second row (equation) that we’re actually changing here gives,                           It is important to remember that when multiplying the third row (equation) by 5 we are doing it in our head and don’t actually change the third row (equation). [Return to Problems]   (e)   In this case we’ll not go into the detail that we did in the previous part.  Most of these types of operations are done almost completely in our head and so we’ll do that here as well so we can start getting used to it.   In this part we are going to subtract 3 times the second row (equation) from the first row (equation).  Here are the results of this operation.                    It is important when doing this work in our heads to be careful of minus signs.  In operations such as this one there are often a lot of them and it easy to lose track of one or more when you get in a hurry. [Return to Problems]

 

Okay, we’ve now got most of the basics down that we’ll need to start solving systems of linear equations using linear algebra techniques so it’s time to move onto the next section.