Algebra for Secondary Teachers: Engaging Students to Develop Symbolic Meaning

Continuing our scenario from the introduction. Suppose that a supplier provides four different products with prices, \(p_1, p_2, p_3, p_4 \text{,}\) to three companies. One thing we may want to do is to play with pricing schemes and look at the generated revenue we would get from each of the three companies. Suppose we have orders from the three companies for each of our four products as given in the table below.

Notice that where the product and sum are combined in part (c) we have the scalar values (or prices) of each Product ($12, $9, $15, and $7) being multipled by all entries in their respective vectors where the number of orders of each Product is placed in the vector in order by the company. When these vectors are then added, we get the total revenue for each company organized as an ordered triple (since there are three companies).

Here we are purposefully using an example where the number of entries in each set of values of interest (prices and revenue) are different to highlight that although we can have the same number of entries in each set be the same, it is not necessary. Eventually we will express the prices as a vector consisting of four entries and the revenue as a vector consisting of three entries where we can think of the process of going from prices to revenue as a transformation between worlds so to speak. In this case we will have four entries in one set of vectors/world (prices of different Products) meaning that these vectors exist in \(\mathbb{R}^4\) such as \(\begin{bmatrix} p_1\\p_2\\p_3\\p_4 \end{bmatrix}\) and three entries in the other set of vectors/world (revenue from each of the three companies) existing in \(\mathbb{R}^3\) like \(\begin{bmatrix} r_1\\r_2\\r_3\end{bmatrix} \text{.}\)

In the previous activity, you may have noticed behaviors of the arithmetic when working with the vectors and scalars that are familiar. From your basic experiences in elementary school, you will recall some properties for operating on numbers that seem to also be happening in your work with vectors. Let’s explore these briefly by using the properties of vector addition and scalar multiplication we have just defined.

Since we have seen these properties before from our work with real numbers, we can use them with vectors and scalars so that we can manipulate these combinations of operations. To this end, we can define a "world" where objects like vectors and scalars behave in a way that we expect and can use.

As we have seen, we can use vectors to represent any entity that might have multiple components. This allows us to manipulate values of common components by adding or multiplying them while keeping the components separated in the process. This use of linear combinations of vectors is useful, but the real desire of the previous scenario is to examine how manipulating one set of values (say, prices of products) affects another set of values (revenue from each company). Just as we explored functional relationships in algebra and calculus, we also might want to explore functional relationships with objects that have not just one value as an input and one corresponding value as an output, but rather inputs that have many component values and a corresponding output that also has many component values.

Recall that one of our first experiences with functions dealt with simple linear relationships. For example, Hooke’s Law states that there is a simple linear relationship between force on a stretched or compressed spring and the distance of the associated stretch or compression, \(F=kx\text{,}\) where \(k\) is a constant of proportionality. If we use function notation to express that force, \(F\text{,}\) is a function of stretch, \(x\text{,}\) we indicate the independent variable by placing it in parentheses getting, \(F\left(x\right)=kx\text{.}\)

Now that we have defined what it means to multiply a matrix and a vector, we can go back to the concept that motivated it, namely, the desire to move between an input vector and an output vector. In a way, we are looking to transform one type of vector into another type of vector. When we think of our example of Hooke’s Law as transforming stretch or compression into force, we note that this happens in our daily lives and we are not often even aware it is there.

Consider a trip to the grocery store. In the produce aisle, we take zucchini and place them on a scale. In turn, the scale returns a weight (force) in pounds (if you are in the USA) or mass in kilograms (practically anywhere else in the world). Inside the scale, there is a spring that is being compressed or stretched. This value is then transformed into a value of weight or mass using Hooke’s Law. In fact, it is more likely the case that you are more familiar with a transformation in form of cost instead of weight or mass as most digital scales today allow us to scan the scale’s output barcode registering a cost in our "shop and scan" app. Below is a video illustrating this process.

Recall from your Algebra 1 days, a function like \(F=kx\) is the most basic of linear functions. Therefore, it should come as no surprise that these two aspects of behavior (preservation of scalar multiplication and vector addition) by this type of function are the defining properties of what we call linear transformations.

Lemma 2.4.15.

If \(T\) is a linear transformation, then it must map the zero vector from the domain to the zero vector in the range. In other words, \(T\left(\mathbf{0}_D\right)=\mathbf{0}_R\text{.}\)

Proof.

Suppose \(T:\mathbb{R}^n \rightarrow \mathbb{R}^m\) is a linear transformation and \(\mathbf{v} \in \mathbb{R}^n\text{.}\) Well since \(-\mathbf{v} \in \mathbb{R}^n\text{,}\) we know that \(T\left(\mathbf{0}_n\right)=T\left(\mathbf{v}+\mathbf{-v}\right)\) where \(\mathbf{0}_n\) here is in \(\mathbb{R}^n\text{.}\) Since vector addition is preserved, we get that this is equal to \(T\left(\mathbf{v}\right)+T\left(\mathbf{-v}\right)\text{.}\) Again, as a linear transformation, we know that scalar multiplication is also preserved and so \(T\left(\mathbf{-v}\right)=-T\left(\mathbf{v}\right)\) and thus \(T\left(\mathbf{v}\right)+T\left(\mathbf{-v}\right)=T\left(\mathbf{v}\right)+ -T\left(\mathbf{v}\right)=\mathbf{0}_m\) where \(\mathbf{0}_m\) here is in \(\mathbb{R}^m\text{.}\)

We began this section looking at how we can represent practical data and relationships using matrices and vectors along with operations allowing us to transform one type of vector into another. In the next few sections, we will extend this view to consider geometric representations that have far reaching applications into areas such as digital file compression and image detection. Eveytime you access your phone using facial recognition, you are using linear algebra. Vectors and matrices play a key role in your everyday life and you may not even be aware of it!

As we ended section 2.1, we wanted to establish whether or not a set of vectors was linealry independent. To do this, we needed to determine whether or not we could find a set of scalar values that would give us a linear combination for other vectors in the set. In addition, we sought to determine the if a vector was in the Span of a set of vectors. For both of these questions, we ultimately want to solve for specific scalars that would meet our desired requirements.

While this process of finding scalars may seem new, it is actually not a new idea. Recall that in high school we solved systems of equations using a variety of techniques including substitution, elimination, and graphing. As the number of variables and equations increased, the techniques of substitution and graphing became very challenging. The process of elimination was generally easier as the number of variables increased and so this process is the one we will refine here.

You may recall that there were certain operations we could perform on equations so that their solution sets did not change. These included multiplying an equation by a nonzero constant, adding equals to equals, and rearranging the equations. For example, since, say, \(5=5\text{,}\) we can also say that \(2\cdot5=2\cdot5\text{.}\) Similarly for an equation like \(x+2y=5\) (for which a solution is clearly \(x=1\) and \(y=2\)), then if we multiply both sides by, say 3, we get \(3\left(x+2y=5\right) \Rightarrow 3x+6y=3\cdot5\Rightarrow 3x+6y=15\text{.}\) Just as \(x=1\) and \(y=2\) is a solution to the original equation, due to the fact that multiplying both sides of an equation by the same value gives a new equation that is also true, \(x=1\) and \(y=2\) is also a solution to \(3x+6y=15\) (i.e. \(3\cdot1+6\cdot2=3+12=15\)).

In a similar fashion, if we start with \(x+2y=5\) and add the same amount to both sides, we get \(x+2y+7=5+7 \Rightarrow x+2y+7=12\text{.}\) If we plug in \(x=1\) and \(y=2\) we get \(1+2\cdot2+7=1+4+7=12\text{.}\) In both cases, if we start with an equation with a certain solution, adding the same constant to both sides or multiplying both sides by the same constant will give a new equation for which the solution to the original equation will be a solution to the new equation.

So now suppose we did not know that the solution to our earlier set of equations, \(x+2y=5\) and \(4x+y=6\text{,}\) was \(x=1\) and \(y=2\text{.}\) How do these manipulations that yield new equations with the same solution help us? As you may recall from your earlier mathematics experiences, we can take the approach of trying to "eliminate" a variable from the system so that we can more easily solve for one of the variables. In this case, consider what would happen if we multipled the top equation by -4. This would give us a new equation, \(-4x-8y=-20\text{,}\) and we can use it in place of the original one. Why multiply it by -4? As we saw, if two equations share the same solution(s), we can add them and the resulting equation will also have the same solution(s). The -4 was selected since it will give a coefficient of \(x\) that is the opposite of the coefficient of \(x\) in the second equation so that when added they will cancel each other to leave us only the variable \(y\text{.}\)

Now that we have worked through a process to reduce a system of equations to fewer and fewer variables, we can summarize the operations that changed the look of the equations without changing the solutions. We essentially have three types of "moves" we can do that will create equivalent systems of equations that maintain the same solutions as the original system.

Scaling Swapping Replacing

As you may have noticed in both of these examples for solving a system with Gaussian elimination, to make it easier to manipulate and perform operations on the equations, we made it a practice to keep the variables aligned with each other and then performed the multiplication by scalars and addition of equations using the three "moves" described above. This helped us make sure we were not accidently adding the coefficients of different variables together. For the most part, this practice just makes "bookkeeping" easier. This should sound kind of familiar. We did the same thing in Chapter 1 when we introduced vectors and matrices. We viewed these arrays of elements as an easier way to keep track of corresponding components that represented values for the same type of quantities (e.g. orders of the same product, price of the same product, revenue from the same company) without needing to label them. So the natural question is, do we need to write the variable name if we already know which variable is associated with which column in our "stacked" representation of the equations? The answer is a resounding, no! So let’s consider what would happen if we were to forego the writing of the variable names.

When we look at the general structure we have been using for our equations in the Guassian elimination process, we wanted our equations to be of the form, \(5x_1+3x_2+2x_3=-7\text{,}\) where the variables are all on the left side of the equation and the right side contains only a numerical value. If we assume this structure for our equations, we can define them in the following way.

Recall from Subsection 2.4.1, we began by looking at systems of equations from the standpoint of transformations.

Here we are separating out the variable names into a single vector and the coefficients are isolated into a matrix. If you wanted to give a name to this matrix, you might call it...you guessed it, a coefficient matrix.

In the last activity, you took a system of equations and converted it into a coefficient matrix, a vector whose components represented the variables, and another vector whose components represented the values from the right-hand side of the equations. From Chapter 1, we viewed these equations as transformations and in this section we used Gaussian elimination to find solutions for the variables in the equations (see Example 2.4.17 and Activity 2.4.7). So how are these connected? Think back to our use of Gaussian elimination. What did we really manipulate? We kept the variables aligned for convenience and multiplied entire equations by scalar values and from time to time added equations together. However, the only things we really changed in the process were the values of the coefficients (the very things that become isloated in the coefficient matrix in our transformation version of the equations). So the natural question is, why do we waste our time writing down the variable names if we are just going to keep them aligned anyway. Instead, why don’t we simply work with the numerical values of the coefficients?

To enable us to more easily work with the solution process, we can tweak our coefficient matrix just a bit, so that the numerical values from the right-hand side of our equations (which do change as we manipilate) are also included. We can simply augment our coefficient matrix with these values. Consider the system of equations given below. We can use the processes of scaling equations and adding them together like in Activity 2.4.8 to try and simplify the system to find solutions.

In this case we can create the coefficient matrix and then augment it with an additional column containing the values from the right-hand side of the equations. In this case we get

Once we have reduced the equations to an augmented matrix with the coefficients on the left side of the bar and the numerical values from the right side of the equations on the right side of the bar, we can use the process we used earlier to operate on the matrix and reduce it to a simplified form. Recall that before, we wanted to eliminate variables so that we could get an equation with just one variable in it. Here we will do the same, but instead, we wish to manipulate the matrix so that we have entries that are zero in locations (this is like eliminating the variable from that particular equation). Note that the rows represent equations and the columns represent the variables, \(x_1\text{,}\) \(x_2\text{,}\) and \(x_3\text{,}\) respectively.

Suppose we want to eliminate the variable, \(x_1\text{,}\) from the second row (equation). As we did earlier, we can multiply the first row (equation) by \(-4\) and add it to the second row and use the new resulting row (equation) to replace the old version of the second row. This would give us the following.

This can also be done step-by-step on most basic graphing calculators. In this case, the first entry in the \(\mathbf{mRowAdd}\) command is the scalar (-4), followed by the matrix, then the row that is being scaled (row 1), and finally the row that is being added to and thus \(\mathbf{replaced}\) (row 3).

Now we can do the same thing for the third row multiplying Row 1 by \(-2\) and adding it to Row 3 and replacing Row 3 with the result.

At this point, we can begin to back-solve since the last row represents an equation that contains only the variable \(x_3\text{.}\) Once we have this value, we can use it in \(R_2\) since it now has entries representing \(x_2\) and \(x_3\) and inserting our value for \(x_3\) leaves just \(x_2\text{.}\) Finally, we then use the values we have obtained for \(x_2\) and \(x_3\) in the first row to get our value for \(x_1\text{.}\)

From this version of the augmented matrix, we can see that \(-3x_3=-6\) and so \(x_3=2\text{.}\) As a result, we can see that \(-3x_2-12\cdot2=-18\) and so \(x_2+4\cdot2=6 \Rightarrow x_2=6-8=-2\text{.}\) Using the first row with these values gives us \(x_1+2\cdot-2+4\cdot2=5 \Rightarrow x_1-4+8=5\text{.}\) This results in \(x_1=5+4-8=1\text{.}\)

The fact that our matrix has zeros in the lower left that enable us to easily back-solve means that this kind of matrix is special and thus deserves a special name. We refer to this structure of a matrix as upper triangular since the nonzero values of the coefficient matrix portion of the augmented matrix form a triangle.

If we want, we could make our back-solving easier by scaling the last two rows. If we multiply both the second and third rows by \(-\frac{1}{3}\text{,}\) we get

Here the \(\mathbf{mRow}\) command first takes the scalar, followed by the matrix and then the row that is being scaled.

From this version of the augmented matrix, we can more easily see that \(x_3=2\text{.}\) As a result, we can see that \(x_2+4\cdot2=6\) and so \(x_2=6-8=-2\text{.}\) Using the first row with these values gives us \(x_1+2\cdot-2+4\cdot2=5 \Rightarrow x_1-4+8=5\text{.}\) This results in \(x_1=1\text{.}\)

Now that we have converted a system of equations to augmented matrix form and then used the process of Gaussian elimination to reduce the system and find solutions, we might ask the question, will we always be able to reduce the matrix to a nice upper triangular form where back-solving to obtain a unique solution is possible? As you will recall, not all linear combinations of vectors (and thus systems) end with a unique solution. This means that it will not always be able to find a unique solution to a given system of equations and thus get a nice upper triangular matrix. If you have taken a formal linear algebra course, you may recall that we can use the form of a matrix for linear systems to determine whether or not the solution(s) is (are) unique, infinite, or non-existent by simplying looking at a reduced version of the matrix.