# Tag Archives: transformations

I’m teaching Algebra 2 this summer for my school.  In a recent test on quadratic functions, I gave a question I thought would be a little different, but still reachable for those willing to make connections or exert a little creativity.

Write a system of quadratic functions that has exactly one solution:  (1,1).

Their handheld graphing calculators were allowed.  Some students definitely had difficulty with the challenge, some gave a version of the answer I expected, and one adopted a form I knew was possible, but doubted anyone would actually find during a test situation.

I show my students’ solutions below.  But before you read on, can you give your own solution?

WHAT I EXPECTED

We’ve had many discussions in class about the power of the Rule of 4–that math ideas can be expressed numerically, graphically, algebraically, and verbally.  When you get stumped in one representation, being able to shift to a different form is often helpful.  That could mean a different algebraic representation, or a different Rule of 4 representation altogether.

The question is phrased verbally asking for an algebraic answer.  But it asks about a solution to a system of equations.  I hoped my students would recall that the graphical version of a system solution is equivalent to the point(s) where the graphs of the equations intersected.  In my mind, the easiest way to do this is to write quadratic functions with coincident vertices.  And this is most easily done in vertex form.  The cleanest answer I ever got to this question was

A graphical representation verifies the solution.

Another student recognized that if two parabolas shared a vertex, but had different “slopes”, their only possible point of intersection was exactly the one the question required.  Here’s a graphical version of her answer.

From these two, you can see that there is actually an infinite number of correct solutions.  And I was asking them for just one of these!  🙂

WHAT I KNEW, BUT DIDN’T EXPECT

Another way to solve this question makes use of the geometry of quadratic graphs.  If two quadratics have the same leading coefficients, they are the same graph, intersect exactly once, or never intersect.  This is a very non-trivial idea for most students.  While I’m not convinced the author of the following solution had this in mind when he answered the question, his solution works because of that fact.  Here’s what J wrote on last week’s test and its graph.

J used more equations than he needed, but had he restricted himself to just two equations, I’m not sure the lovely pattern would have been so obvious.

This is a very different (and super cool) answer than what I expected my students to produce.  Lesson re-learned:  Challenge your students, give them room to express creativity and individuality, and be prepared to be amazed by them.

NEXT STEPS

J’s answer actually opens the door to other avenues of exploration.

1. Can you generalize the form of all of J’s equations, essentially defining a family of quadratics?  Can you prove that all members of your generalization satisfy the question posed and that no other answers are possible?
2. Can you find forms of other generalized families of quadratic functions whose only solution is (1,1)?
3. Notice that there were two types of solutions above:  A) those with coincident vertices and different lead coefficients and B) those with identical lead coefficients and different vertices.  Are these the only types of quadratics that can answer this question?  That is, is there a system of quadratics with (1,1) as the only solution that have identical vertices and lead coefficients?  Could both be different and (1,1) be the only solution?
4. If I relax the requirement that the quadratics be functions, what other types of quadratics are possible?  [This could be a very nice calculus question!]

For my part, I’m returning to some of these questions this week to stretch and explore my student’s creativity and problem-solving.

I’d love to hear what you or your students discover.

## Clever math

Here are what I think are three clever uses of math by students.

In my last week of classes at my former school in May, 2013, my entire Honors Precalculus class showed up wearing these shirts designed by one of my students, M.  The back listed all of the students and a lovely “We will miss you.”  As much as I liked the use of a polar function, I loved that M opted not for the simplest possible version of the equation ($r=2-2sin(\theta )$), but for a rotation–a perfect use of my transformations theme for the course.

Now for a throwback.  When I was a graduate student and TA at Syracuse from 1989-1990, one of my fellow grad students designed this shirt for all of the math and math ed students.  I don’t remember who designed it, but I’ve always loved this shirt.

## Exploring Sequences and Lines

Here’s another favorite problem that could be used for any middle or high school students who’ve been exposed to both arithmetic sequences and linear equations.

There is a family of lines, $Ax+By+C=0$, for which $\left\{ A, B, C \right\}$ can be any arithmetic sequence.  What do all members of this family have in common?

As with most great math problems, the problem is simply stated and can be approached from many different perspectives.  In the end, this one isn’t that difficult to crack, but the solution (at least to me) is not intuitively obvious from the problem statement. If you’ve not seen this before, please try it before reading further.

WARNING:  Problem Solution Follows

MOST COMMON STUDENT APPROACH:  Given the problem’s phrasing, most begin by writing out a few possible equations using different arithmetic sequences, typically with positive common differences.  After a few tries, most get a little frustrated as no obvious pattern emerges from the equations.

It is this point that is typically the most difficult for me as a teacher.  I want to help anyone who struggles, but “help” often means allowing others the room to struggle and to find ways of solving problems on their own.  Only when a student or group gets really frustrated do I sometimes ask, “Are there any other ways you can look at this problem or at your work?”

Eventually, most decide to graph their equations to see if anything pops out.  The following graph was submitted this past semester by one of my precalculus students using the free online Desmos calculator.

Two lines intersecting in a point is common.  Three or more in a single point almost always indicates something interesting.  Seven lines through a single point is screaming for attention!  From this graph, all lines in this family apparently contain the point (1,-2).  That seems a bit random until you investigate further, but pattern recognition is just half of the battle for a mathematician.  Now that something interesting has been discovered, a mathematician wants to know if this is a fluke or something inherent to all members of the family.

NOTE:  All graphs my students have produced over the years have always contained specific equations.  I don’t know that it’s any more enlightening, but I did create on Desmos a slider version of the graphs of this family with one slider for the initial term (A) and a second for its common difference (d).

UNIQUE SOLUTION METHODS FROM STUDENTS:

All successful solutions eventually rewrite the sequence $\left\{ A,B,C \right\}$ to $\left\{ A,A+d,A+2d \right\}$ where d is the common difference for a generic arithmetic sequence with initial term A.

Method I: After discovering the common point of intersection, most plug it into the left side of the equation and simplify to get

$Ax+By+C=A\cdot 1+\left( A+d\right)\cdot -2+\left( A+2d \right)=0$.

Because the left side reduces to zero for all generic arithmetic sequences, $\left\{ A,A+d,A+2d \right\}$, (1,-2) must be common to all members of this family.

A few students aren’t comfortable dealing with 0=0, so these tend to plug in $x=1$ and solve for y to get $y=-2$, proving that the y-coordinate for $x=1$ for all members of this family is always -2.

Method II:  A few students every year work algebraically from properties of arithmetic sequences.  For any arithmetic sequence, $\left\{ A,B,C \right\}$, $\frac{A+C}{2}=B$.  This rewrites to $1\cdot A-2\cdot B+C=0$, so whenever $\left( x,y \right)=\left(1,-2 \right)$, then $Ax+By+C=0$ is a fundamental property of all arithmetic sequences.

Personally, I think this method gets closest to explaining why the point (1,-2) is the common characteristic of this family.

Method III:  This year, I had a student take an approach I’d never seen before.  She defined one arithmetic sequence as $\left\{ a,a+d,a+2d \right\}$ and another as $\left\{ m,m+n,m+2n \right\}$ for any real values of a,d,m, and n.  This leads to a system of equations: $a\cdot x+(a+d)\cdot y+(a+2d)=0$ and $m\cdot x+(m+n)\cdot y+(m+2n)=0$ .  If you have some younger students or if all the variables make you nervous, the solution is available via Wolfram Alpha.

Still, this system is pretty easy to manipulate.  If you multiply the first equation by m and the second by a, the x-terms will eliminate with subtraction, giving

$m\cdot((a+d)\cdot y+(a+2d))-a\cdot((m+n)\cdot y+(m+2n))=0$.

Solving for y shows that all of the coefficients simplify surprisingly easily.

$((ma+md)-(am+an))\cdot y=-(ma+2md)+(am+2an)$
$(md-an)\cdot y = -2\cdot (md-an)\Longrightarrow y=-2$

From here, determining $x=1$ is easy, proving the relationship.

SOLUTIONS and APPROACHES NOT YET OFFERED BY STUDENTS:

Approach A:  High school students don’t often think about ways to simplify problem situations, especially at the beginning of problems.  One way I did that for this problem in later class discussions was to recognize that it one of the terms in the arithmetic sequence was 0, you didn’t need to deal with nearly as many terms. For example, if your sequence was ${1,0,-1}$, the linear equation would be $x-1=0$.  Similarly, the sequence $\left\{ 01,2 \right\}$ leads to $y+2=0$.  Obviously, the only thing these two lines have in common is the point (1,-2).  A proof of the property must still be established, but this is one of the fastest ways I’ve seen to identify the central property.

Approach B:  A purely algebraic approach to this problem could redefine the arithmetic sequence as $\left\{ a,a+d,a+2d\right\}$as before, giving:

$a\cdot x+(a+d)\cdot y+(a+2d)=0$

Collecting like terms gives

$(x+y+1)\cdot a+(y+2)\cdot d=0$.

The values of a and d must remain as parameters to include all possible arithmetic sequences.  Because the equation always equals 0, the coefficients of a and d are both 0, making $y=-2$ (for the coefficient of d) and therefore $x=1$.

EXTENSION:

We once had a test question at the end of the unit containing this exercise.  Basically, it reminded students that they had discovered that all lines $Ax+By+C=0$, for which $\left\{ A, B, C \right\}$ can be any arithmetic sequence contained the point (1,-2).  It then asked for an equation of a family of linear functions using the same arithmetic $\left\{ A, B, C \right\}$ that all contained the point (1,2).

The two most common responses we’ve seen involve a reflection or a vertical translation.  (1,-2) can become (1,2) by reflecting over the x-axis, so making the y-values negative would do the trick:  $Ax-By+C=0$.  Similarly, (1,-2) can become (1,2) by translating up 4 units, giving $Ax+B(y-4)+C=0$.

## Exponentials and Transformations

Here’s an old and (maybe) a new way to think about equations of exponential functions.  I suspect you’ve seen the first approach.  If you understand what exponentials functions are, my second approach using transformations is much faster and involves no algebra!

Members of the exponential function family can be written in the form $y=a\cdot b^x$ for real values of a and postive real values of b.  Because there are only two parameters, only two points are required to write an equation of any exponential.

EXAMPLE 1: Find an exponential function through the points (2,5) and (4,20).

METHOD 1:  Plug the points into the generic exponential equation to get a 2×2 system of equations.  It isn’t necessary, but to simplify the next algebra step, I always write the equation with the larger exponent on top.

$\left\{\begin{matrix} 20=a\cdot b^4 \\ 5=a\cdot b^2 \end{matrix}\right.$

If the algebra isn’t the point of the lesson, this system could be solved with a CAS.  Users would need to remember that $b>0$ to limit the CAS solutions to just one possibility.

If you want to see algebra, you could use substitution, but I recommend division.  Students’ prior experience with systems typically involved only linear functions for which they added or subtracted the equations to eliminate variables.  For exponentials, the unknown parameters are multiplied, so division is a better operational choice.  Using the system above, I get $\displaystyle \frac{20}{5}=\frac{a\cdot b^4}{a\cdot b^2}$.  The fractions must be equivalent because their numerators are equal and their denominators are equal.

Simplifying gives $4=b^2\rightarrow b=+2$ (because $b>0$ for exponential functions) and $a=\frac{5}{4}$.

This approach is nice because the a term will always cancel from the first division step, leaving a straightforward constant exponent to undo, a pretty easy step.

METHOD 2:  Think about what an exponential function is and does.  Then use transformations.

Remember that linear functions ($y=m\cdot x+b$) “start” with a y-value of b (when $x=0$) and add m to y every time you add 1 to x.  The only difference between linear and exponential functions is that exponentials multiply while linears add.  Therefore, exponential functions ($y=a\cdot b^x$) “start” with a y-value of a when $x=0$ and multiply by b every time 1 is added to x.

What makes the given points a bit annoying is that neither is a y-intercept.  No problem.  If you don’t like the way a problem is phrased, CHANGE IT!    (Just remember to change the answer back to the original conditions!)

If you slide the given points left 2 units, you get (0,5) and (2,20).  It would also be nice if the points were 1 x-unit apart, so halving the x-values gives (0,5) and (1,20).  Because the y-intercept is now 5, and the next point multiplies that by 4, an initial equation for the exponential is $y = 5\cdot 4^x$ . To change this back to the original points, undo the transformations at the start of this paragraph:  stretch horizontally by 2 and then move right 2.  This gives $y = 5\cdot 4^\frac{x-2}{2}$.

This is algebraically equivalent to the $y=\frac{5}{4}\cdot 2^x$ found early.  Obviously, my students prove this.

One student asked why we couldn’t make the (4,20) point the y-intercept.  Of course we can!  To move more quickly through the set up, starting at (4,20) and moving to (2,5) means my initial value is 20 and I multiply by $\frac{1}{4}$ if the x-values move left 2 from an initial x-value of 4.  This gives $y = 20\cdot\left( \frac{1}{4} \right) ^\frac{x+4}{-2}$.  Of course, this 3rd equation is algebraically equivalent to the first two.

Here’s one more example to illustrate the speed of the transformations approach, even when the points aren’t convenient.

EXAMPLE 2: Find an exponential function through (-3,7) and (12,13).

Starting at (-3,7) and moving to (12,13) means my initial value is 7, and I multiply by $\frac{13}{7}$ if the x-values move right 15 from an initial x-value of -3.  This gives $y = 7\cdot\left( \frac{13}{7} \right) ^\frac{x-3}{15}$.

Equivalently, starting at (12,13) and moving to (-3,7) means my initial value is 13, and I multiply by $\frac{7}{13}$ if the x-values move left 15 from an initial x-value of 12.  This gives $y = 13\cdot\left( \frac{7}{13} \right) ^\frac{x+3}{-15}$.

If you get transformations, exponential equations require almost no algebraic work, no matter how “ugly” the coordinates.  I hope this helps give a different perspective on exponential function equations and helps enhance the importance of the critical math concept of equivalence.

## Quadratics, Statistics, Symmetry, and Tranformations

A problem I assigned my precalculus class this past Thursday ended up with multiple solutions by the time we finished.  Huzzah for student creativity!

The question:

Find equations for all polynomial functions, $y=f(x)$, of degree $\le 2$ for which $f(0)=f(1)=2$ and $f(3)=0$.

After they had worked on this (along with several variations on the theme), four very different ways of thinking about this problem emerged.  All were valid and even led to a lesson I hadn’t planned–proving that, even though they looked different algebraically, all were equivalent.  I present their approaches (and a few extras) in the order they were offered in our post-solving debriefing.

The commonality among the approaches was their recognition that 3 non-collinear points uniquely define a vertical parabola, so they didn’t need to worry about polynomials of degree 0 or 1.  (They haven’t yet heard about rotated curves that led to my earlier post on rotated quadratics.)

Solution 1–Regression:  Because only 3 points were given, a quadratic regression would derive a perfectly fitting quadratic equation.  Using their TI-Nspire CASs, they started by entering the 3 ordered pairs in a Lists&Spreadsheets window.  Most then went to a Calculator window to compute a quadratic regression.  Below, I show the same approach using a Data&Statistics window instead so I could see simultaneously the curve fit and the given points.

The decimals were easy enough to interpret, so even though they were presented in decimal form, these students reported $y=-\frac{1}{3}x^2+\frac{1}{3}x+2$.

For a couple seconds after this was presented, I honestly felt a little cheated.  I was hoping they would tap the geometric or algebraic properties of quadratics to get their equations.  But I then I remembered that I clearly hadn’t make that part of my instructions.  After my initial knee-jerk reaction, I realized this group of students had actually done exactly what I explicitly have been encouraging them to do: think freely and take advantage of every tool they have to find solutions.  Nothing in the problem statement suggested technology or regressions, so while I had intended a more geometric approach, I realized I actually owed these students some kudos for a very creative, insightful, and technology-based solution.  This and Solution 2 were the most frequently chosen approaches.

Solution 2–Systems:  Equations of quadratic functions are typically presented in standard, factored, or vertex form.  Since neither two zeros nor the vertex were explicitly given, the largest portion of the students used the standard form, $y=a\cdot x^2+b\cdot x+c$ to create a 3×3 system of equations.  Some solved this by hand, but most invoked a CAS solution.  Notice the elegance of the solve command they used, working from the generic polynomial equation that kept them from having to write all three equations, keeping their focus on the form of the equation they sought.

This created the same result as Solution 1, $y=-\frac{1}{3}x^2+\frac{1}{3}x+2$.

CAS Aside: No students offered these next two solutions, but I believe when using a CAS, it is important for users to remember that the machine typically does not care what output form you want.  The standard form is the only “algebraically simple” approach when setting up a solution by hand, but the availability of technology makes solving for any form equally accessible.

The next screen shows that the vertex and factored forms are just as easily derived as the standard form my students found in Solution 2.

I was surprised when the last line’s output wasn’t in vertex form, $y=-\frac{1}{3}\cdot \left ( x-\frac{1}{2} \right )^2+\frac{25}{12}$, but the coefficients in its expanded form clearly show the equivalence between this form and the standard forms derived in Solutions 1 and 2–a valuable connection.

Solution 3–Symmetry:  Two students said they noticed that $f(0)=f(1)=2$ guaranteed the vertex of the parabola occurred at $x=\frac{1}{2}$.  Because $f(3)=0$ defined one real root of the unknown quadratic, the parabola’s symmetry guaranteed another at $x=-2$, giving potential equation $y=a\cdot (x-3)(x+2)$.  They substituted the given (0,2) to solve for a, giving final equation $y=-\frac{1}{3}\cdot (x-3)(x+2)$ as confirmed by the CAS approach above.

Solution 4–Transformations:  One of the big lessons I repeat in every class I teach is this:

If you don’t like how a question is posed.  Change it!

Notice that two of the given points have the same y-coordinate.  If that y-coordinate had been 0 (instead of its given value, 2), a factored form would be simple.  Well, why not force them to be x-intercepts by translating all of the given points down 2 units?

The transformed data show x-intercepts at 0 and 1 with another ordered pair at $(3,-2)$.  From here, the factored form is easy:  $y=a\cdot (x-0)(x-1)$.  Substituting $(3,-2)$ gives $a=-\frac{1}{3}$ and the final equation is $y=-\frac{1}{3}\cdot (x-0)(x-1)$ .

Of course, this is an equation for the transformed points.  Sliding the result back up two units, $y=-\frac{1}{3}\cdot (x-0)(x-1)+2$, gives an equation for the given points.  Aside from its lead coefficient, this last equation looked very different from the other forms, but some quick expansion proved its equivalence.

Conclusion:  It would have been nice if someone had used the symmetry noted in Solution 3 to attempt a vertex-form answer via systems.  Given the vertex at $x=\frac{1}{2}$ with an unknown y-coordinate, a potential equation is $y=a\cdot \left ( x-\frac{1}{2} \right )^2+k$.  Substituting $(3,0)$ and either $(0,2)\text{ or }(1,2)$ creates a 2×2 system of linear equations, $\left\{\begin{matrix} 0=a\cdot \left ( 3-\frac{1}{2} \right )^2+k \\ 2=a\cdot \left ( 0-\frac{1}{2} \right )^2+k \end{matrix}\right.$.  From there, a by-hand or CAS solution would have been equally acceptable to me.

That the few alternative approaches I offered above weren’t used didn’t matter in the end.  My students were creative, followed their own instincts to find solutions that aligned with their thinking, and clearly appreciated the alternative ways their classmates used to find answers.  Creativity and individual expression reigned, while everyone broadened their understanding that there’s not just one way to do math.

It was a good day.

## Cubics and CAS

Here’s a question I posed to one of my precalculus classes for homework at the end of last week along with three solutions we developed.

Suppose the graph of a cubic function has an inflection point at (1,3) and passes through (0,-4).

1. Name one other point that MUST be on the curve, and
2. give TWO different cubic equations that would pass through the three points.

SOLUTION ALERT!  Don’t read any further if you want to solve this problem for yourself.

The first question relies on the fact that every cubic function has 180 degree rotational symmetry about its inflection point.  This is equivalent to saying that the graph of a cubic function is its own image when the function’s graph is reflected through its inflection point.

That means the third point is the image of (0,-4) when point-reflected through the inflection point (1,3), which is the point (2,10) as shown graphically below.

From here, my students came up with 2 different solutions to the second question and upon prodding, we created a third.

SOLUTION 1:  Virtually every student assumed $y=a\cdot x^3$ was the parent function of a cubic with unknown leading coefficient whose “center” (inflection point) had been slid to (1,3).  Plugging in the given (0,-4) to $(y-3)=a\cdot (x-1)^3$ gives $a=7$.  Here’s their graph.

SOLUTION 2:  Many students had difficulty coming up with another equation.  A few could sketch in other cubic graphs (curiously, all had positive lead coefficients) that contained the 3 points above, but didn’t know how to find equations.  That’s when Sara pointed out that if the generic expanded form of a cubic was $a\cdot x^3+b\cdot x^2 +c\cdot x+d$ , then any 4 ordered pairs with unique x-coordinates should define a unique cubic.  That is, if we picked any 4th point with x not 0, 1, or 2, then we should get an equation.  That this would create a 4×4 system of equations didn’t bother her at all.  She knew in theory how to solve such a thing, but she was thinking on a much higher plane.  Her CAS technology expeditiously did the grunt work, allowing her brain to keep moving.

A doubtful classmate called out, “OK.  Make it go through (100,100).”  Following is a CAS screen roughly duplicating Sara’s approach and a graph confirming the fit.  The equation was onerous, but with a quick copy-paste, Sara had moved from  idea to computation to ugly equation and graph in just a couple minutes.  The doubter was convinced and the “wow”s from throughout the room conveyed the respect for the power of a properly wielded CAS.

In particular, notice how the TI-Nspire CAS syntax in lines 1 and 3 keep the user’s focus on the type of equation being solved and eliminates the need to actually enter 4 separate equations.  It doesn’t always work, but it’s a particularly lovely piece of scaffolding when it does.

SOLUTION 3:  One of my goals in Algebra II and Precalculus courses is to teach my students that they don’t need to always accept problems as stated.  Sometimes they can change initial conditions to create a much cleaner work environment so long as they transform their “clean” solution back to the state of the initial conditions.

In this case, I asked what would happen if they translated the inflection point using $T_{-1,-3}$ to the origin, making the other given point (-1,-7).  Several immediately called the 3rd point to be (1,7) which “untranslating” — $T_{1,3}(1,7)=(2,10)$ — confirmed to be the earlier finding.

For cases where the cubic had another real root at $x=r$, then symmetry immediately made $x=-r$ another root, and a factored form of the equation becomes $y=a\cdot (x)(x-r)(x+r)$ for some value of a.  Plugging in (-1,-7) gives a in terms of r.

The last line slid the initially translated equation using $T_{1,3}$ to re-position the previous line according to the initial conditions.  While unasked for, notice how the CAS performed some polynomial division on the right-side expression.

I created a GeoGebra document with a slider for the root using the equation from the last line of the CAS image above.  The image below shows one possible position of the retranslated root.  If you want to play with a live version of this, you will need a free copy of GeoGebra to run it, but the file is here.

What’s nice here is how the problem became one of simple factors once the inflection point was translated to the origin.  Notice also that the CAS version of the equation concludes with $+7x-4$, the line containing the original three points.  This is nice for two reasons.  The numerator of the rational term is $-7x(x-2)(x-1)$ which zeros out the fraction at x=0, 1, or 2, putting the cubic exactly on the line $y=7x-4$ at those points.

The only r-values are in the denominator, so as $r\rightarrow\infty$, the rational term also becomes zero.  Graphically, you can see this happen as the cubic “unrolls” onto $y=7x-4$ as you drag $|x|\rightarrow\infty$.  Essentially, this shows both graphically and algebraically that $y=7x-4$ is the limiting degenerate curve the cubic function approaches as two of its transformed real roots grow without bound.

## Transformations II and a Pythagorean Surprise

In my last post, I showed how to determine an unknown matrix for most transformations in the xy-plane and suggested that they held even more information.

Given a pre-image set of points which can be connected to enclose one or more areas with either clockwise or counterclockwise orientation.  If a transformation T represented by matrix $[T]= \left[ \begin{array}{cc} A & C \\ B & D \end{array}\right]$ is applied to the pre-image points, then the determinant of $[T]$, $det[T]=AD-BC$, tells you two things about the image points.

1. The area enclosed by similarly connecting the image points is $\left| det[T] \right|$ times the area enclosed by the pre-image points, and
2. The orientation of the image points is identical to that of the pre-image if $det[T]>0$, but is reversed if $det[T]<0$.  If $det[T]=0$, then the image area is 0 by the first property, and any question about orientation is moot.

In other words, $det[T]$ is the area scaling factor from the pre-image to the image (addressing the second half of CCSSM Standard NV-M 12 on page 61 here), and the sign of $det[T]$ indicates whether the pre-image and image have the same or opposite orientation, a property beyond the stated scope of the CCSSM.

Example 1: Interpret $det[T]$ for the matrix representing a reflection over the x-axis, $[T]=\left[ r_{x-axis} \right] =\left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]$.

From here, $det[T]=-1$.  The magnitude of this is 1, indicating that the area of an image of an object reflected over the line $y=x$ is 1 times the area of the pre-image—an obviously true fact because reflections preserve area.

Also, $det \left[ r_{x-axis} \right]<0$ indicating that the orientation of the reflection image is reversed from that of its pre-image.  This, too, must be true because reflections reverse orientation.

Example 2: Interpret $det[T]$ for the matrix representing a scale change that doubles x-coordinates and triples y-coordinates, $[T]=\left[ S_{2,3} \right] =\left[ \begin{array}{cc} 2 & 0 \\ 0 & 3 \end{array} \right]$.

For this matrix, $det[T]=+6$, indicating that the image’s area is 6 times that of its pre-image area, while both the image and pre-image have the same orientation.  Both of these facts seem reasonable if you imagine a rectangle as a pre-image.  Doubling the base and tripling the height create a new rectangle whose area is six times larger.  As no flipping is done, orientation should remain the same.

Example 3 & a Pythagorean Surprise:  What should be true about  $det[T]$ for the transformation matrix representing a generic rotation of $\theta$ units around the origin,  $[T]=\left[ R_\theta \right] = \left[ \begin{array}{cc} cos( \theta ) & -sin( \theta ) \\ sin( \theta ) & cos( \theta ) \end{array} \right]$ ?

Rotations preserve area without reversing orientation, so $det\left[ R_\theta \right]$ should be +1.  Using this fact and computing the determinant gives

$det \left[ R_\theta \right] = cos^2(\theta ) + sin^2(\theta )=+1$ .

In a generic right triangle with hypotenuse C, leg A adjacent to acute angle $\theta$, and another leg B, this equation is equivalent to $\left( \frac{A}{C} \right) ^2 + \left( \frac{B}{C} \right) ^2 = 1$, or $A^2+B^2=C^2$, the Pythagorean Theorem.  There are literally hundreds of proofs of this theorem, and I suspect this proof has been given sometime before, but I think this is a lovely derivation of that mathematical hallmark.

Conclusion:  While it seems that these two properties about the determinants of transformation matrices are indeed true for the examples shown, mathematicians hold out for a higher standard.   I’ll offer a proof of both properties in my next post.