OK, Determinants are Easy, but

What are they good for?

Oops, sorry English Folks, I meant “For what are they good?”

Sometimes it may seem to students that the only value of the thing they have just learned in math is to be able to learn the next thing and although I am a big proponent of the “Math for Math’s sake” school, I occasionally feel a swelling up of my old Engineering background urging me to provide an application of the learning. In just such a mood, I offer the following

APPLICATIONS OF DETERMINANTS

We begin by asking you to recall that moment not so long ago when you were in an algebra or geometry class, and a problem might begin when we gave you the coordinates of three points in the coordinate plane, A=(2,1); B=(5,6) and C= (9,-1). Among the myriad things we might have asked you to do with this given information would be two that are easily answered by the methods of determinants; “Find the area of the triangle formed?” and “Prove that the three points are, or are not, collinear (lie on the same line).”

Finding the area of a triangle If you had not had the benefit of an introduction to determinants, you might have to resort to something like the following to find area. Taking the points in pairs we could find the length of each segment, ([pic], [pic], and [pic]) and then employing the well known formula of Heron (you do remember Heron’s Formula, don’t you?) to find the area. We would arrive at an area of 20.5 square units. But now that you know how to evaluate the determinant of a matrix, we can instead just enter the x values of each point as one column of the matrix, enter the y values as a second column, and add a third column of all ones (someday I will explain about cross products of vectors and this third column will seem more clear). We get the matrix shown at right. And if we evaluate the determinant of the matrix we get –41, as the screen capture shows. Two things may concern you; but if you are at all clever, you realize that the area 20.5 is ½ the determinant of –41 except for the sign. At this point I remind you that the sign of the determinant depends on the order we enter the values, and if we switch the 1st and 2nd rows, we get a positive determinant. So if we can suspend the concerns and accept that the area of the triangle formed is ½ the absolute value of the determinant, we have hit upon a easy, efficient, way to find the area of a triangle using technology and our new friend the determinant. (Now is when you say “Wow, way cool!”) But what about the other problem of proving that the three points lie in a straight line. Well, if they were in a straight line, the area of the triangle would be zero, and the determinant would be zero, and that would tell us. If the points are NOT in a straight line, the determinant will NOT be zero because any three points in a plane which are NOT collinear, form a triangle with a non-zero area. So the same determinant answers both questions. Postscript: After we learn a little about algebraic vectors we will do this same thing with a 2x2 matrix. (I know, you can hardly wait, but be patient my children.)

Extension and generalization

“But Mr. Ballew, we live in three-space and you have shown us something we can use on questions about a two-space plane. Isn’t there any application of a determinant that would apply to REAL life three-space?” you ask. Being only your humble servant, I respond with a three-space analogy to what you have already learned above.

Volume of a Tetrahedron and Co-planer points

If we expand all our points to spatial XYZ coordinates then we can extend the axiom that any three non-collinear points determine a plane and thus a triangle (you remember, geometry) to another that says it takes four points to determine a spatial object. The simple space generalization of a triangle is called a tetrahedron, (from the Greek roots tetra for four, and hedron for surface); but you may have learned it as a triangular based pyramid. You will notice that it has four vertices, and forms four triangles in space out of each set of three points. If we want to find the volume of the tetrahedron we could find the area of one of the triangles, then we would need to find the equation of the plane containing the triangle so that we could find the equation of the line perpendicular to it that passed through the opposite vertex. Next we would need to find the point where this perpendicular intersected the plane of the triangle, and then we could find the distance from this point to the opposite vertex. With all that done we could finally find the volume of the tetrahedron by using the formula V= 1/3 Bh where the B represents the area of the base triangle. Or we could simply assign the four space points into a four by four matrix and adding the column of ones at the right as we did in two-space. Because we are working in three space, we will have to divide the determinant by 3! =6 this time to get the true volume.[Remember that the formula for a pyramid is 1/3 the base area times the height, and the base area is a triangle = ½ b h, so we have 1/3 of ½ or 1/6 of the rectangular box described]. Of course if we make all four points lie on the same plane, the determinant would be zero, so this determinant also tests if four points are co-planer.

Here is an example using the four 3-D points A=(2, 5, 11); B= (8, 3, 6); C=(-3, 12, 2) and D= (3, -1, 4). The screen shot at the right shows theresult before we divide by six to get the volume of 107.5 cubic units.

May we have some practice problems, please?

1) For the given set of points, find the area of the triangles named below

A=(2,3) B=(4, 7, ) C=( 12, 3) D=(-3, -5) E=(5, -3)

(ABC= _____ (ACE= _____ (ACD= _____ (ABC= _____ (BCD= _____

Use the equation y=3x-1 and find three points on the line. Prove the three points are collinear by showing that the triangle they form has an area of zero.

Find the area of the tetrahedron formed by the four spatial points below
A=(-3, 1, 7) B=(6,12,5) C=( 9, -2, 8) D=(1, 4,3)