View Full Version : Determinant Demonstraition

bergerud

06-23-2013, 02:45 PM

Ok, math lesson time! The determinant of a 2 X 2 matrix (ad - bc) gives the area of the parallelogram spanned by the two 2D vectors (a,b) and (c,d) which are the rows of the matrix. This is proved with algebra and the students have no idea why it is geometrically true. For years, I have told my students there must be a way draw triangles and see that it is true. I had never done it but I new it must be possible because it is true! Well I finally put some time into it and discovered a most pleasing geometric proof. I have never, in my 25 years of teaching seen this proof before. I is so simple, I cannot understand why it is not all over the internet.

So here is my demo carved out of walnut with arbutus triangles. Can you see why the area of the parallelogram is equal to the determinant (difference of the area of the large square and the small square) ?

cestout

06-23-2013, 03:46 PM

Beautiful work, but I would understand the explanation better if it weren't written in Greek. That is why I stopped at an associates degree.

SteveNelson46

06-23-2013, 04:24 PM

Clever indeed!

badbert

06-23-2013, 04:31 PM

I understood that completely! They were words typed on a page. LOLI have an old friend from school who is a math professor, want me to show it to him?

bergerud

06-23-2013, 05:12 PM

I understood that completely! They were words typed on a page. LOLI have an old friend from school who is a math professor, want me to show it to him?

Sure, ask him is he has ever see it before. It is so simple, I cannot believe it is not in textbooks.

aokweld101

06-23-2013, 08:38 PM

I can see what you mean but I didn't finish high school I got my ged years later and it was from what I learned over the years as being a welder I could not comprehend ..y equaled to z and I don't know y...lol:???:

dltccf

06-23-2013, 09:09 PM

Eureka...Although that was volume not area, but still worthy of one.

dave

gashawk

06-23-2013, 09:55 PM

Bergerud that's very good, now do you think you could show us a Beal conjecture with the CW?

bergerud

06-24-2013, 09:38 AM

Bergerud that's very good, now do you think you could show us a Beal conjecture with the CW?

I do not think so. That is a pure number theory thing with no connection to geometry (since the exponents > 2) that I can see.

lawrence

06-24-2013, 09:49 AM

that thing hurts my head ;)

But looks great. Real nice job on the fit.

Lawrence

bergerud

06-29-2013, 11:54 AM

Ok Lawrence, how is this? I routed out the walnut middle piece and replaced it with arbutus to match the removable pieces. It should now be more obvious that the areas are equal.

(Now the grain does not match so I will have to make another one!)

bergerud,

Elegant visual proof! Having a touch of a math background myself, the only thing missing is that the relationship of a,b,c & d to the parallelogram itself has to be established. But I think it's doable... For instance, the vertex (0,0) and (a,b) of the parallelogram are apparent, but what's not established is the x axis segment 0 to a. This might simply be proved by making the triangle (0,0)-(a,0)-(a,b) movable into the space occupied by the rectangle...

R, Jon

bergerud

07-18-2013, 12:30 PM

bergerud,

Elegant visual proof! Having a touch of a math background myself, the only thing missing is that the relationship of a,b,c & d to the parallelogram itself has to be established. But I think it's doable... For instance, the vertex (0,0) and (a,b) of the parallelogram are apparent, but what's not established is the x axis segment 0 to a. This might simply be proved by making the triangle (0,0)-(a,0)-(a,b) movable into the space occupied by the rectangle...

R, Jon

Sorry for replying so late but I was on a trip. To see where a is you just have to look up to the top. (Remember, vectors can be moved around!) Each vector appears twice in the diagram. There are four vectors. Each vector makes one of the a, b, c, or, d obvious.

I understand where you're coming from, but let me take a counter argument to illustrate my point... What's to say that the vector (0,0)-(c,d) hasn't been fudged (and therefore is not parallel to vector (a,b)-(a+c,b+d)) in order to make the shapes within the triangular areas of the parallelogram fit within the rectangular area of a x d? Since this is a "visual proof", it seems to me that as part of the proof, you have to, for instance, lift the movable triangles c & b, and align them with the large yellow c & b in order to establish equality, thereby completing the "visual proof"...

R, Jon

bergerud

07-18-2013, 10:09 PM

I understand what you are saying also, but I made this for linear algebra students who are familiar with the parallelogram spanned by two vectors. They would take it as a given that the figure is a parallelogram. The vectors are parallel by definition! What is poor about this demo is that the large rectangle is a square. I will make a better one which has more random looking rectangles.

Also, it should be noted that this is not a proof for even more serious mathematical reasons. As the angle between the vectors gets smaller, more and more cuts need to be made.

badbert

07-18-2013, 10:49 PM

I sent the pics to my friend George Chackman. He is a math professor at Old Dominion University, in Norfolk, Va. Here is his response. Of course he dumbed it down, for me, LOL.

Hi Elbert,

What the equation claims is that the area of thelarge diamond is equal to the area of the large rectangle minus the area of thesmall rectangle. The 4 light colored blocks fit in the gaps of thediamond. I wanted to try to prove it, but it would take some time todo. At least it seems true.

Your friend,

George

bergerud

07-19-2013, 12:29 AM

Interesting. Typical safe mathematical response: "At least it seems true".

An engineer, a physicist, and a mathematician are traveling by train through the British countryside. They spot a black sheep grazing on a hillside. The engineer says, "There are black sheep in England". The physicist says, "There is at least one black sheep in England". The mathematician says," There is at least one black sheep in England which is black on at least one side."

ronboley

07-31-2013, 06:36 PM

Interesting. Typical safe mathematical response: "At least it seems true".

An engineer, a physicist, and a mathematician are traveling by train through the British countryside. They spot a black sheep grazing on a hillside. The engineer says, "There are black sheep in England". The physicist says, "There is at least one black sheep in England". The mathematician says," There is at least one black sheep in England which is black on at least one side." Then.....

the conductor pointed out it was a goat.

JDPratt

08-01-2013, 03:40 PM

Mind....trying....to....grasp....mathmatical.....e quation,...Large....rush...of....air...passing.... over....head.

Metallus

08-01-2013, 05:24 PM

May be applicable... (click to view)

63846

mtylerfl

08-06-2013, 09:36 AM

May be applicable... (click to view)

That would be me...;)

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