Matrices are hard to set up in iWeb, largely because the TAB function does not work the same on all browsers, so I cannot set things up in Word and copy them across. Please read adjacent lines with brackets as if the brackets not only line up but are actually the same bracket. I will eventually find a way of displaying this better.
Principle of matrix multiplication:
(a b) (x) = (ax+by) (a b) (e f) = (ae+bg af+bh)
(c d ) (y) (cx+dy) (c d ) (g h) (ce+dg cf+dh)
What people often say is to mentally pick up the first row (a b) and to slide it vertically next to the first column of the second matrix so that the pairings ax & by or ae & bg become visible (mentally); then to add the multiplied pairs. Whatever, it is a mental contortion that stretches top Y8/9, second set Y9/Y10 and anyone who has not seen it before. What seems to happen is that it provides just enough complexity of arithmetic to cause errors to be revealed, especially in the handling of negatives.
Try these:
(1 2) (5 6) = (19 22) (1 2) (1 0) = (1 2)
(3 4) (7 8) (43 50) (3 4) (0 1) (3 4)
(1 -2) (5 -6) = (-9 -22) (1 0) (11 40) = ( 11 40)
(-3 4) (7 8) (13 50) (0 1) (101 56) = (101 56)
(5 6) (1 2) = (23 34)
(7 8) (3 4) (31 46)
(8 -2) (1 2) = ( 2 8)
(-3 5) (3 4) (12 12)
(8 -6) (5 6) = ( -2 0)
(-7 5) (7 8) ( 0 -2)
(1 2) (4 -2) = (1 0)
(3 4) (-3 1) (0 1)
You might see that some particular manipulations produce the 1 0 0 1 result and that this matrix has no effect in multiplication. You might also see that AB≠BA is generally true but just occasionally AB=BA is true - special cases, then.
(a b) (d -b) = (ad-bc ab-ba) = (ad-bc 0 ) = ad-bc ( 1 0 )
(c d) (-c a) (cd-dc -cb+da) ( 0 ad-bc) ( 0 1 )
The expression ad-bc is called the determinant and is constructed by subtracting the minor diagonal product (that’s bc) from the major diagonal product, ad. I used ad-bc to indicate a bracket that is not a matrix. In some texts this determinant is written as a capital delta, Δ.
(1 0) = 𝑰 is the identity matrix; for any matrix, A, then A𝑰=𝑰A = A, which is equivalent to
(0 1) multiplying by one in ordinary arithmetic [x.1 = 1.x = x]. Think of it as “One”.
You might also see that the special case introduced first allows the following to happen:
Here is a pair of simultaneous equations: 3x + 2y = 7 and 4x + 3y = 9. this can be written as
(3 2) (x) = (7) Now pre-multiply both sides as follows (3 -2) (3 2) (x) = (3 -2) (7)
(4 3) (y) (9) (-4 3) (4 3) (y) (-4 3) (9)
The left hand side reduces (amazingly) and the right hand side gives you the answer
(1 0) (x) = (3) which is to say, x=3 and y=1
( 0 1) (y) (1)
What is special about this is that you can see the pattern of the magic ingredient that ‘undoes’ the problem. I picked an example with a determinant of one, BUT the process works equally quickly however foul the numbers are, so try these:
(3 2) (x) = (17) x= 13 (3 11) (x) = (7) x= -16
(4 3) (y) (19) y=-11 (1 4) (y) = (4) y= 5
When the determinant is not one, I recommend letting it surface, like this,
starting with 3x + 2y = 7 and 7x + 6y = 9 ...
(3 2) (x) = (7) Pre-multiply both sides as before (6 -2) (3 2) (x) = (6 -2) (7) = (24)
(7 6) (y) (9) (-7 3 ) (7 6) (y) (-7 3) (9) (-22)
The two sides separately come to make (4 0) (x) = ( 24) so divide both sides by four and get
(0 4) (y) (-22) x=6, y = -5.5 which checks as correct.
You can find some more of these at the end of the revision exercises. The thing is, this process works with any numbers in the equations - until you make the determinant zero accidentally, when it wouldn’t work because if you sketched the lines, they’d be parallel, so they don’t intersect. In my mind this matrix method takes the same time no matter how stupidly hard the numbers are, so I like it as a method because it is predictably quick.
Try these, where I’ve picked the numbers quite randomly:
2x - 5y = 11 Δ = 34 x=-71/17 3x + 5y = 12 Δ = -26 x=89/26 11x + 12y = 23 Δ = -367 x=775/367
4x + 7y = 13 y= 9/17 4x - 2y = 13 y = 9/26 15x - 17y = 32 y= -7/367
Doing all three took me 200 seconds - and longer still to type in the answers.
This material was developed by me (!) as a Year Nine or Year Ten in 1967 or 1968 — probably the earlier year, given the context: I was set a long homework of some 50 pairs of simultaneous equations to solve. In the language developed above, there was a long run of these with a determinant of one, enough so that I saw the pattern. Quite by chance, the first few non-unity determinants were instead all two, so I could see the correction to make. My father came past and (probably immediately, being always a much more able mathematician than me) recognised what I was doing and told me to use round brackets not the square ones I had written. He later told me to call the correction factor a determinant. My maths master, Mr Boulding, went spare (ape, hyperbolic, mad) at me doing his ‘long’ homework with apparently so little working out. I didn’t see what the fuss was about until I started teaching, a whole lifetime later.
DJS 20130416
