Hello.
I am in a Linear Algebra class and we have recently covered matrix arithmetic and multiplication. The multiplication part was terribly counterintuitive, so I tried to find out how it was derived. A textbook in the library explained it by using the following algorithm for arithmetic of a sequence. For example:
1,2,3,4 * 5,6,7,8 = 15+26+37+48 = 5+12+21+32 = 70
How on Earth is this true? How was it derived? This seems utterly arbitrary to me, and the arbitrary has no place in mathematics.
Sure it does. I’ll start with an arbitrary value for x. 
I’m not sure which textbook you’re reading, or the context
in which you found the algorithm, but I’m guessing you found
the matrix addition more intuitive. Wouldn’t the addition of
1,2,3,4 plus 5,6,7,8 be equal to 6,8,10,12 ? In other words,
addition term by term?
If you start with a system of equations like
7 = 3x + 5y
6 = 2x + 4y
then you can express it in matrix notation like
B = AX where B is the 2x1 matrix of the left hand
side, and A is the 2x2 matrix of the coefficients
of the right hand side, and X is the 2x1 matrix
consisting of x and y. But only if you use matrix
multiplication as it is defined.
Yeah, matrix algebra gave me hiccups too, especially with all the esoterica about eigenvalues/eigenvectors and determinants and the whacked formulae that people found relating to them.
I suspect that matrix algebra came about as a direct result of linear equations. This is pretty easy to see… given:
a1 x + a2 y + a3 z = c1
a4 x + b5 y + a6 z = c2
a7 x + a8 y + a9 z = c3
where the a’s and c’s are constants and x, y, and z are unknown variables, you can solve for the unknowns x, y, and z. Manipulating entire equations like this on paper is a pain in the neck. But there is a compact way of representing it:
A = [ a1, a2, a3, a4, a5, a6, a7, a8, a9 ]
U = [ x ]
[ y ]
[ z ]
C = [ c1, c2, c3 ]
So the above set of formulae can be compactly represented as:
AU=C
So matrix multiplication is a compact way of representing the constants-and-unknowns representation of linear algebraic systems. Because we have problems which involve manipulating dozens or hundreds or even thousands of equations of this nature, matrix representation is the only sane way to do it.
As it turns out, you can solve for the unknowns both directly with the equations above, or via the matrix representation. All you have to do is find the matrix inverse of A (a loose correspondence to the inverse of a scalar 1/x):
A^-1 AU = A^-1 C
U=A^-1 C
where I’m using ^ to represent ‘to the power of’.
I won’t go into computing a matrix inverse here (you’ll get into it in your class).
Anyway, it’s not arbitrary how matrix multiplication works. It is defined as being the way it is because of how it helps you combine variables and constants (and even entire formulae) in a linear fashion.
It could have been defined as simply element-by-element multiplication, but then it wouldn’t be remotely as useful.
All this is just more explicitly what RM Mentock was talking about.
Uh, the U matrix was meant to be a vertical matrix (1 column, 3 rows), but my ASCIIart skills here evidently need work.
A should’ve been a 3 x 3 matrix as well.
ARGH.
I’ve been doing this all week. Let’s try again!
A =
[ a1, a2, a3 ]
[ a4, a5, a6 ]
[ a7, a8, a9 ]
U =
[ x ]
[ y ]
[ z ]
C =
[ c1 ]
[ c2 ]
[ c3 ]
Would everybody agree with this?
Sorry for the confusion.
That looks right.
Also, that “sequence arithmatic” you refer to is the dot product of two vectors, which is a very useful operation in physics (and other fields). If I have two vectors, then the dot product of them is the product of the magnitude of the first times the magnitude of the second times the cosine of the angle between them. If you have your vectors in component form, you can also find the dot product using the term-by-term method in the OP. For instance, suppose I have the vectors (4,2,6) and (1,5,9). That is to say, the first vector extends for a distance of 4 in the x direction, 2 in the y direction, and 6 in the z direction, and similarly, the second vector extends 1 in the x direction, 5 in the y, and 9 in the z. Now, suppose I want to find the dot product between these two vectors. I could find the magnitudes of both, and somehow figure out the angle between them, and proceed from there, but that would be an awful lot of work. Or, I could work with components: (4,2,6)[sup]./sup = 41 + 25 + 6*9, or 68.
In fact, if I did want the angle between those two vectors, the way I would proceed is to first find the dot product, then divide by the magnitudes to get the cosine of the angle. In this case, for instance, the first vector has magnitude sqrt(4[sup]2][/sup] + 2[sup]2][/sup] + 6[sup]2][/sup]) = 7.483, and the second vector has magnitude 10.344. So the cosine of the angle is (68/77.408) = 0.8785, and the angle between those two two vectors is 28.543 degrees.