2012 in Review

Since all the cool bloggers (#1 and #2) are writing 2012 in review blog posts, I thought I would throw one out too.

A shot of the fam...

... and an assorted list of some of the important moments for me in this year:

• I became a father - again!  Eleanor Gracelyn was born on August 4th - a date that had previously been quite painful for me - my mother past away on this date in 2004, but this year that date has new joy.  Ellie is now 5 months old and full of smiles and giggles, and spit up and crying.

• I graduated! Again.  This year I completed my Master's degree from Cornerstone University. It was in  Instruction and Curriculum development in Secondary education. This blog is in no small part a product of that education, as I'm sure I wouldn't have started it without it. 

• I flipped a class: I began using a new teaching technique, flipping the classroom, in my physics classes. This involves recording lessons that students watch as their "homework" which frees up class time for labs, projects, one-on-one tutoring, practice, ect. 
• Finished a 100+ page project documenting a flipped unit and capstone-ing my Masters Degree 
• Attended the Alive Music Festival with my wife and the youth group kids. Went to numerous concerts and seminars, and played a lot of guitar
• Got new phones-that work! And an iPad. Been using a lot of new technology that has become very familiar to me now -- it's hard to think that it's only been a couple of months with it!
• Watched a planet travel across the sun! And in general, learned a lot more about using telescopes through a good friend at church

• Went sailing! My dad bought a hobie cat sailboat and we enjoyed getting stuck out there many times:

The Constellation Gemini

This post is the one of a series on constellations and posted throughout the year as each constellation comes into prominence.

Gemini is an important constellation to learn for a number of reasons. Known as the twins, it is home to two of the brightest stars in the sky -- Pollux and Castor. Pollux on the left is the 12th and Castor the 16th brightest stars in the Northern hemisphere. These stars are the heads of two twin brothers standing side by side. On a good night, you can see lines of stars outlining the bodies, as shown below:

Gemini
Image from Wikipedia

Gemini is one of the 12 zodiacal constellations, which means it lies along what's called the ecliptic. The ecliptic is a circle of constellations surrounding us which lies along the same plane as the planets and the solar system as a whole. Therefore, any planets you see are always somewhere along this plane. Gemini is therefore useful to know as planets are often located nearby.  For instance, the bright planet Jupiter will be approaching Gemini early 2013 and spend late 2013 and most of 2014 within the region.

Being along the ecliptic means that the moon will pass through the constellation once a month.

Being along the ecliptic also means that the sun will pass through the constellation (rendering it invisible of the brightness of the sun) once a year. The sun passes through Gemini from May 21 to June 20 each year. During the winter months, the sun is in the opposite side of the sky which makes Gemini an easy constellation to see.

Below is a map of the surrounding constellations for the evening hours in December.

2012 December skies ~ 7pm
2013 January skies ~ 6pm
Image by Skymaps

As you can see from this image, Gemini is pretty easy to spot -- being above and to the left of the familiar constellation Orion. Finding the bright head stars should be easy to do, but next time you have a clear night, see if you can identify some of the other stars making up the body. I have often seen them as forming the Greek letter Ω (Omega).

Of course, a description of Gemini would be incomplete without mentioning the Geminids, a meteor shower that is one of the best in the year. During the nights around December 13 and 14 each year, many bright meteors (sometimes around 100 per hour!) can be visible. They emanate from Gemini - which means they will go outwards away from Gemini. They don't always begin in that constellation, but their paths will tend to point away from Gemini more often than not. Most meteor showers are best observed in the wee morning hours.

A good binoculars object, M35 is a nice cluster of stars that will be my goal to find in the next month. It has the width of the moon, and contains well over 200 stars, but most are invisible to the naked eye. Castor is supposedly juggling it on his foot like a soccer ball.

Probabilities of Phase 10 continued

Yesterday I wrote about some of the probabilities associated with the game phase 10. Today I'll investigate a few others. What are the probabilities of being dealt one of the given phases right away?

This sort of a question requires you to calculate how many different combinations of cards are possible for each hand, and divide that by how many different combinations of hands there are.  The fact that there are wild cards, and that there are two blue 12's and two green 12's etc is a fact that complicates things beyond what I know how to handle, so I have ignored that in these calculations. I hope that hasn't distorted things too horrendously, and perhaps I'll return and clean this post up if I ever figure out how to handle it. If any of you readers want to instruct me, please leave a comment below.

The first calculation I will find is how many different hands of phase 10 there are. The mathematical function that does this is the choose function. (TotalToChooseFrom choose NumberNeeded) Since the deck has 108 cards, and I need to calculate (108 choose 10). Wolfram Alpha gives us this number: 38,722,819,230,810 different hands.

(Parenthetical note, indicated redundantly by the fact that this paragraph is in parentheses... The actual number of unique hands is less than this because of the aforementioned fact that there are multiple skip cards, multiple wilds and multiple blue12's, red12's, etc.  This is what I don't know how to account for. This also suggests that each of my probabilities calculated below will be too low - except that they all have issues of their own because I didn't account for wild cards or repetitions in many of their calculations either).

So, how to calculate probabilities? Let's begin with my favorite phase -- getting seven cards of all the same color. First, choose which color you'd like to use out of the four.  That's (4c1).  Then, not counting wilds, there are 24 cards of that color, from which you must choose 7, or (24c7).  Then choose 3 cards out of all the cards that are left -- that is 101c7.  Multiply these probabilites together and you get:
Ways of choosing at least 7 of same color with no wilds
   (4c1)(24c7)(101c3) = 230,712,926,400

To calculate proability of being dealt this then, divide this number into the total number of hands:
   .2trillion / 38 trillion = .005 or 0.5%.

Including the possiblity of wilds increases the odds a little -- and for this phase doesn't complicate calculations much.  There are still 4 colors to choose from, but now you have 32 possible cards to draw from instead of just 24:

Ways of choosing at least 7 of same color with wilds

   (4c1)(32c7)(101c3) = 2,243,679,609,600 which is close to 6% probability

What follows is the number of combinations for most of the other phases. Being dealt them exactly is almost always less than 1 percent so I have not calculated the probabilities, but will leave them for you if you want.

Ways of choosing a run of 7 (no wilds):
   (6c1)(8c1)(8c1)(8c1)(8c1)(8c1)(8c1)(101c3) = 262,117,785,600

Ways of choosing a run of 8 (no wilds):
   (5c1)(8c1)(8c1)(8c1)(8c1)(8c1)(8c1)(8c1)(100c2) = 51,904,512,000

Ways of choosing a run of 9 (no wilds):
   (4c1)(8c1)(8c1)(8c1)(8c1)(8c1)(8c1)(8c1)(8c1)(99c1) =  6,643,777,536

Ways of choosing a set of 3 and another set of 3 (no wilds):

   (12c2)(8c3)(8c3)(102c4) = 879,560,035,200 

Ways of choosing a set of 4 and another set of 4 (no wilds):

   (12c2)(8c4)(8c4)(100c2) = 1,600,830,000

Ways of choosing a set of 5 and another set of 2 (no wilds):

   (12c2)(8c5)(8c2)(101c3) = 17,246,275,200

Ways of choosing a set of 5 and another set of 3 (no wilds):

   (12c2)(8c5)(8c3)(100c2) = 1,024,531,200

Ways of choosing a set of 3 and another run of 4 (no wilds):

   (12c1)(8c3)(9c1)(8c1)(8c1)(8c1)(101c3) =  516,044,390,400

Ways of choosing a set of 4 and another run of 4 (no wilds):

   (12c1)(8c4)(9c1)(8c1)(8c1)(8c1)(100c2) = 19,160,064,000

Probabilities of Phase 10

We were playing a card game the other day and I couldn't resist making some back of the napkin calculations of probabilities.

The game was Phase 10 -- a stupid game that I can never win. The point of the game is to collect certain combinations of cards faster than your opponents.

There are 108 cards in a Phase 10 deck:

•   96 numbered cards numbered 1-12, 2 of each in four different colors
•   4 skip cards
•   8 wild cards

To begin each hand, 10 cards are dealt to each player. After dealing and several turns have been played, in one round I have come to have these cards:

What are my chances of drawing what I need?

In the picture above, you can see I have a wide variety of cards, and two wilds to use. I am trying to get a run of 9 consecutive cards, and am almost there. I'm wondering, what is the probability of me drawing the card I need from the deck at random? I have to do this because my opponents are being stingy and purposefully leaving me cards that are not helpful, because they have been observing what I've picked up in the past few turns.

Well, at this point in the game there have been 30 cards dealt out, and an additional 12 cards played, which means there are 108-30-12 or 66 cards remaining. I am in search of 3, an 8, a 9, a 12, or a wild card, as any of them would allow me to complete the run and play my hand. I have not seen any 3's, 8's, 9's, 12's, or wilds played by my opponents, but I wouldn't expect to see them -- and especially not wilds, as those are seldomly discarded.

At worst, my opponents could be secretly collecting these cards -- and have all of them in their hands -- or could they?  There are 8 each of the 3's, 8's, 9's, and 12's, and 6 more wild cards out there, which means there are a total of 38 cards that I could win with that are unaccounted for. Even if my opponents were both incredibly lucky (not likely) and incredibly mean (possible...) that would still leave 18 winning cards available out of the 66 in the deck, for a minimum probability of 27%.

At best, my opponents might have none of these cards -- leaving all 38 as possible cards to draw from in the deck of 66 -- yielding a maximum probability of 58%.

Chances are likely that something in between is true -- perhaps my opponents have a few of these cards, but not all of them. I'll have to assume that my potential winning cards have been evenly dispersed amongst all of the other cards I don't have -- and even though I can't draw from my opponents hands, I should include the cards they have in my probability calculations. That means there are 66 + 20 or 86 cards to "draw from" and 38 possible winning cards, or a probability of 44%.

I figure I can last another three turns or so before one of my opponents goes out first. This isn't as much of an arbitrary guess as you might think -- a lot of times you can tell by how many cards a player still has how many turns you have left.  What then are my chances of winning? I should be guarenteed right? 44*3 is over 100%!  Unfortunately it's not that easy, but must be calculated by subtracting my chance of losing from 1. My chance of losing is my chances of not getting a card I want 3 times in a row. There is a 100-44 or 56 percent chance that the card I draw next will not help me.  Therefore, my chances of losing are (.56)(.56)(.56) or 17%.  This means my chances of winning are pretty good -- or around 83%.

Unfortunately, I did not win, but when you're playing your wife, winning is always best, and sometimes losing is a victory. Occasionally you need to lose a battle to win the war. Er... to avoid war? I mean, I love you honey!

Storage Capabilities of the 8.5x11

A former student of mine had an instagram that inspired a dorky series of tweets between the two of us touting the features of this amazing new technology: the sheet of paper. I was trying to be sarcastic because I hadn't seen a sheet of paper in so long, writing all my essays and lesson plans and everything digitally.

The series of tweets:

Ian Williams ‏@ian_d_williamsThe sad tale of an engineer attempting to write an essay. tp://instagr.am/p/TwRo3sRTpt/ 

Rocky Roer ‏@RockyRoer@ian_d_williams what is that thin white note taking device in front of your laptop? Is that a new Christmas present? Wireless? Bluetooth?
Ian Williams ‏@ian_d_williams@RockyRoer It's the newest in pressed wood pulp. Functions without a processor and get's great battery life!
Rocky Roer ‏@RockyRoer@ian_d_williams what kinda storage? 200Gb?
Ian Williams ‏@ian_d_williams@RockyRoer college ruled, 56 lines, expandable to margins. Also compatible with multi-ring long-term storage devices.
Rocky Roer ‏@RockyRoer@ian_d_williams wow 56 lines! I was only familiar with older 33 line models.

Anyway, taking it a step too far I'm sure, I decided to investigate the storage capabilities of a simple sheet of paper. To begin, I found a sheet of paper -- though it took a while. I also had to find a stylus that was compatible with the paper -- after a handful of failed attempts, I found one with enough battery left (you can always count on Sharpie models!) to input text:

I counted the number of characters on each line, including spaces and punctuation.  I found 63, 62, and 65 characters on each line, and so I'm going to average them and suggest that a typical line stores 64 bytes. I'm thrilled this is a power of 2 -- it makes me confident that it's technology, because everything else in technology comes in powers of two.  Let me save that number in memory as 2^6.

After counting the number of available rows, I found another convenient power of 2, or 32 lines.  2^5.

Multiplying these numbers together, I get 2^11 bytes, or 2*(2^10).  2^10 is 1 KB, and so one side of one sheet of paper is 2KB. Utilizing both sides of every sheet of paper yields the following storage capabilities for the following standard models:

70 sheet 1 subject sprial notebook:  280 KB

100 sheet looseleaf unbound package: 400 KB

500 sheet ream of paper: 2 MB

Of course, using different fonts and font sizes could significantly increase or decrease storage capabilities.    

A Ton Of Snow

The other day my students were taking their pre-calc exam and due to my writing it too long, many needed to stay after class was over to finish. So many in fact, that I emailed "Wow, I have a ton of students needing to stay after!" to my principal.

After hitting send, I was instantly annoyed that I had used "ton" as an exaggeration -- something I have been known to chide my students about. But I was comforted when I did a quick estimation and realized that I was actually fairly accurate:

As it turned out, I had 14 students in my room, and many of them were the bigger athletes (certainly bigger than 120 lb) so I'm sticking to my original statement as literally true. And is 120 lb an appropriate average weight of a student?  Who knows.

Well, today I was out shoveling snow for the first time this season, and I caught myself again saying "This is a ton of snow!"  Well, was it? Was it really?

To calculate the weight of snow, I need to find the volume of snow on my driveway, and multiply it by the density of snow.  The volume of snow is easy enough to approximate -- I'll assume I have an even rectangular driveway with the same height of snow everywhere, and so volume of this rectangular prism is just length * width * height:

 The density of snow is a little more difficult, as there is heavy snow, light fluffy snow, solid ice, etc.  Wikipedia says snow has a density of anywhere from 8% - 50% of water, depending on many things, but mostly on how compacted it is as it melts, freezes, melts, refreezes and more snow falls on top of it. This was relatively new snow, and so I'll assume it has a density of about 25% of water.

The density of water is 1 kg/L.  This "coincidence" of having such a clean number is actually not a coincidence at all, but was by design -- as 1 kg was originally defined to the weight of 1 liter of water as the metric system was being invented. Since then we have defined the kg more precisely than that using more complicated methods. Though it's no longer exactly 1, this value remains accurate enough for our purposes (I think it's 1.003 or something close?)  Converting kg to pounds and L to cubic inches is a tough exercise:

If my snow was 25% the density of water, than:

The weight of the snow on my driveway is then:

Looks like I wasn't exaggerating after all! I moved over a ton of snow today!

"Random" Thoughts

I hear the word "random" used all the time, and so often incorrectly. It seems like it is some of my high school students favorite words, and expressions. I bet I hear at least once a day, "That's so random!!"

 Here's a few thoughts on the word, and some alternatives that perhaps you ought to consider instead.

Random: If something happens randomly, it means that one option out of many different equally possible options occurred. This was not something someone chose. This event could have occurred a different way if things had happened just a little differently. Imagine a dice rolling and coming up a four. It could just as easily come up a 5. This is what random means. Side note: Random things will sometimes repeat. A dice will sometimes come up with the same side showing twice in a row. In fact we can predict how often that will occur (about 17%).  Your iPod does not play songs randomly -- because if it did you would complain that it wasn't "random enough".  If it really did pick a song at random, you would hear repeats occasionally, and the first music players actually did this. Your iPod probably is using a shuffled play list - which did use random choice to create the list by choosing one of the songs to play first, one of the remaining songs to go second, one of the remaining songs to go next and so on. 

Haphazard:  Lack of a plan, order or direction. A person who doesn't know where they are going is not driving "randomly". They are not flipping a coin to decide when to turn left or right.

Arbitrary: Determined by chance, whim, or impulse, and not by necessity, reason, or principle. An arbitrary decision is one that could have been lots of things, and the decider just picked one of them because a decision had to be made. Perhaps they could have made the decision randomly, by drawing draws or rolling a dice, but instead just made up their mind.

Spur-of-the-moment: occurring or done without advance preparation or deliberation; extemporaneous; unplanned. I pick a lot of numbers on the spur of the moment over the course of a day, as I make up quick examples for class. I don't pick them randomly, and I know that I favor certain numbers like 2's and 3's. There is usually little meaning behind the choice of numbers, but they were not chosen randomly.

Spontaneous: resulting from internal or natural processes, with no apparent external influence. I often hear of people described as random -- especially those that are really funny and come up with the weirdest things "from out of nowhere". These people aren't dice-rollers. They don't have thousands of thoughts rolling around in their head waiting to fall out their mouths like some sort of lottery. They are spontaneous, and the world is a better, funnier place because of them. Learn the word. Use it.

Assorted: Literally (and don't get me started about "literally" misusages) "not"-sorted. This rant was purposefully misnamed Random thoughts, thought a more appropriate word to use is assorted. I suppose I could have put these words into a shuffling algorithm and randomized them -- or sorted them alphabetically, but I just put them down as they came to me.

Unexpected: If you didn't expect something was going to happen, then describe that outcome as unexpected, not random. The fact that we had a fire alarm during third hour was not random -- in fact, it was probably planned. You just didn't expect it. Surprise! 

Aimless: Sometimes people do things that are pointless, and serve no purpose. Putting a picture of a cute penguin on a blog post about randomness might not serve any purpose, but that doesn't make it random.  

Transition Matrices

This post is a part of a series of guest-posts on the applications of matrix multiplication. These posts were written by my pre-calc students:

Transition Matrices

by Leanna Krueger 

Imagine that you own an apple orchard and you sell 3 different types of apples; Honeycrisp (A), Golden Delicious (B), and Gala (C). Honeycrisp apples sell the most while Golden Delicious is the runner up and Gala sells the least. You planted apple trees for each of these apples, and the cost of replacing each tree with the tree that produces apple A is too expensive until you get more money. In order to predict your earnings and decide when you can replant your orchard trees you need to estimate what people will buy. You know that this week the current market for apples is shown in the "state matrix" S shown below:

	A	B	C
Percentage of People Buying	60%	30%	10%

Suppose you surveyed repeated buyers and found the following:

A.                If people are buying A this week, the probability of them buying A next week is 70%. The probability of them buying B next week is 20% and the probability for C is 10%.

B.                 If people are buying B this week, the probability of them buying A next week is 50%. The probability of them buying B next week is 30% and the probability for C is 20%.

C.                 If people are buying C this week, the probability of them buying A next week is 30%. The probability of them buying B next week is 40% and the probability for purchasing C again is 30%.

You could set this information up in a transition matrix, T, shown below:

	A	B	C
A	.7	.2	.1
B	.5	.3	.2
C	.3	.4	.3

If a family was buying apple B, and you wanted to know the probability of them buying apple A 4 weeks later you would have to multiply the 2 matrices above forming S*T. After multiplying them once, you would have to multiply them a second time, and third time, and a fourth time. In a sense, doing that is the same thing as finding S*T to the fourth power:

	A	B	C
1 week	.6	.25	.15
2 weeks	.59	.255	.155
3 weeks	.587	.2565	.1565
4 weeks	.5861	.25695	.156951

Therefore, the probability of the family buying apple A 4 weeks later is 58.61%.  

Adjacency Matrices

This post is a part of a series of guest-posts on the applications of matrix multiplication. These posts were written by my pre-calc students:

Adjacency Matrices 

By John Taylor 

Let’s say you have to go over to your friend Will’s house, but you have to get some things done first. How can you know how many routes you can take and gets to Will’s? Enter the adjacency matrix. This helpful matrix can tell you how many different routes you can take of the same number of moves to get to the same place. Now the first thing you should do when you are starting to get ready to make your matrix is to look at your map:

 Now figure out what these places are. The house is, well, your house. All the others, except W, are stores you need to go to: The sock shop (S), the book store (B), the library (L), the electronics store (E), the farmer’s market (F) and the drug store (D). The W is Will’s house. The next step is to write a matrix recording these relations. For each road from one place to another record a 1. If there is no connection put a zero. For a one-way street only write a 1 for the direction the arrow faces. If there are two options, say between home and the sock shop, enter a 2. This matrix will look like this.

Now we have this maps matrix. That was the hard part; now it’s quite simple. However many places you want to go to and raise the matrix to that power. (Note: This will give you all ways even if they repeat points. For the sake of this example you are very forgetful and often have to go back because you forgot many things at these stores.)

 Let’s go back to the trip to Will’s. You want to go to six places before you head over. Just raise the whole matrix to 6 and in the cell H,W you will see how many routes there are that reach Will’s in six moves. The resulting matrix will look like this:

Now look in the cell H,W and you will find out your answer. Even though most of the numbers are big, there are only 10 ways to get from your house to Will’s in 6 moves. This is the way adjacency matrices work. I hope this example has helped you understand these helpful matrices more fully.

Solving a systems of Equations with a Matrix

This post is a part of a series of guest-posts on the applications of matrix multiplication. These posts were written by my pre-calc students:

Systems of Equations Matrix Project

by Jake Melton

Matrices have many different applications and can be helpful in many situations. One such situation is solving systems of equations. Solving systems of equations using matrices takes several steps, all of which will be explained in the following blog post. All of the steps will be shown as if you were using a TI-89 calculator (or Mr. Roer's edits show steps for a TI-82, -83, -84). 

The following system of equations will be used as the example throughout the following post:

To begin the process we need to first write all of the equations in standard form (i.e. variables=number.) This will have to be done by hand. After we finish this process we get: 

Once we have written this new system of equations there are two ways we could go. First, we could complete the system of equations by hand. This process would include writing two matrices A and B. A would be a coefficient matrix, that is a matrix written using the number by which the variable is multiplied; this matrix would be 4 by 4. The second matrix, B would be a 4 by 1 matrix written using the answers. Remember, if there is no variable as seen in the second equation where there is no x, you must write a 0. The system of equations is then summarized by the matrix multiplication:

To find the solutions, you can simply multiply the inverse of A by B and the resulting 4 by 1 matrix would be the answers, illustrated here. 

*Note if the determinant of A equals zero the system has either no solution or many solutions. 

However, to solve the system of equations we could follow another much more simple process using our calculators. Using a TI-89 (or TI-82, -83, or-84) first press “on” and scroll up to the data/matrix editor  and press enter (or press 2nd Matrix). Select “new” then “matrix”. Next, enter the name you would like under “variable” and continue by pressing enter. The data should be empty, now simply enter your data into the table, making sure you enter a 4 by 5 matrix. This matrix is augmented [A|B] and will appear as follows. 

As you can see, this matrix is simply both matrices shown on the previous page combined. Now, simply press “enter” then “home”. Next you will need to click the “catalog” button found in the middle of your calculator, then press “2” and scroll to “rref(” and hit enter. (or press 2nd Matrix, over to calc, and down to B: rref ) Now you will need to enter the name of your matrix using “alpha” and the name of your matrix. The rref you just entered stands for “reduced row echelon form of a” or as Mr. Roer calls it, “really ridiculously easy form” and it simplifies the equations for you to give you:

Each row is a simplified equation. The first row simply means that 

1w +0x +0y +0z = 3.5  or, w = 3.5 

So now that you have found this matrix you can conclude that: 

w = 3.5 

x = -4 

y = -1.5 

z = 2.5 

These numbers shown above mean that w, x, y, and z are the only numbers that could be put into the original system of equations and make ALL of them true. And that is how you would go about finding the answer to a system of equations using matrices. Now you know how to solve these systems by hand or using your TI-89 calculator (or TI-82, -83, -84). Now use these steps to solve any difficult system of equations with ease. I hope you leave feeling enlightened and much smarter, or more confused, but hopefully the former.

Matrix Multiplication Cryptography

This post is a part of a series of guest-posts on the applications of matrix multiplication. These posts were written by my pre-calc students:

Matrix Cryptography
by David Stanley

Cryptography, put simply, is the art of encoding messages.  It serves to answer the simple question of how you get a message to a friend without your enemy being able to read it.  Cryptography has been used for centuries by militaries and intelligence agencies to send important messages, while insuring that the information the messages contain does not fall into enemy hands. 

Although there are many forms of cryptography, one of the simplest yet most effective forms of encryption still utilizes the simple matrix.The message is placed in matrix form, and then multiplied by a random square matrix or encoding matrix. 

 The first step is to write down the message that you wish to send.  I will use this completely true and totally non-brown nosing message as an example:

Mr Roer is the best math teacher ever in the history of humanity.

Secondly, you must create an encoding matrix.  This matrix must be a square matrix. An example of this would be:

In this kind of encryption, letters are assigned numbers for their place in the alphabet.  A would be 1, B would be 2, C would be 3 and so on.  Spaces are assigned the number 27, as their are only 26 letters in the alphabet.  So the message in matrix form would be:

Notice how my encoding matrix has the same amount of columns as the message matrix does rows.  This is required or else they cannot be multiplied.  All that is left to do is to multiply the encryption matrix by the message matrix.  This gives you:

Now if you received this in the mail, you would have no idea at all what it said.  In order to figure this out in a timely manner, you would have to have the decoding matrix.  The decoding matrix is the inverse of the encoding matrix.  This can easily be found on your calculator.  The decoding matrix for this problem is quite long, so I will round to four decimal places. 

The exact elements in the decoding matrix have more digits and would give the exact numbers as the original message. Using this rounded decryption matrix gives numbers that can all be rounded to the original message, though occasionally a letter might be slightly off.  

In summary, the encoding process written in calculator language when 

           [A] is the encoding matrix,

           [B] is the original message,  

    and [C] is the encoded message is [A]x[B]=[C]. 

The process of decoding the message is ([A]^-1)x[C]=[B]. 

And there you have it.  That is matrix cryptography in a nutshell. 

Meaningful Matrix Multiplication

This post is a part of a series of guest-posts on the applications of matrix multiplication. These posts were written by my pre-calc students:

The Christmas Party Competition by Kiera Suywn

Svetlana and Isabelle are frien-emies. Not only that, but they both love Christmas. Every year they each plan a Christmas party. This year their parties happened to be on the same day. So they are trying to beat each other out on who can get the best price for her party supplies. Svetlana needs 1 ½ gallon of eggnog, 6 twelve packs of candy canes, 3 gingerbread house kits, and 7 hot chocolate mixes. Isabelle needs 5 ½ gallons of eggnog, 2 twelve packs of candy canes, 7 gingerbread house making kits, and 4 hot chocolate mixes. They are both comparing prices from Meijer, Target, and Forest Hills Foods. To figure out who would have the best price at what store, they used matrices, something they learned a very long time ago in Pre-Calculus class and thought they would never use again.
Here is a chart showing the prices of the items per store. They put this as matrix [A] which is (stores x prices of food).

	Eggnog	Candy Canes	Gingerbread	Hot Chocolate
Target	2.79	2.49	9.99	1.99
Meijer	2.19	2.00	9.99	1.39
Forest Hills	3.99	1.25	7.99	4.99

            The next matrix they made was one that showed the (food x people) this went in the matrix [B] spot. 

	Svetlana	Isabelle
Eggnog	1	5
Candy Canes	6	2
Gingerbread house kit	3	7
Hot Chocolate	7	4

They multiplied these two matrices together which produced a (stores x people) matrix. This is what they got for their answers. This gave them the total amounts that each of their supplies would be.

	Svetlana	Isabelle
Target	89.49	112.74
Meijer	73.35	101.56
Forest Hills Foods	70.39	98.34

Forest Hills Foods had the best prices for both of them but Svetlana won by $27.95. Maybe next year Isabelle!