Download Advanced Higher Mathematics Unit 1

Advanced Higher Mathematics Unit 1
Further Notes on Linear Equations
We have already studied basic Gaussian Elimination, which is quite straightforward, and so I am
assuming that you can perform the algorithm efficiently.
In the examples we have already done things have worked out perfectly and we have obtained a single,
unique, solution to the problem.
But this is not always the case, as will be shown below.
Consider first a simple 2  2 system:
2x  3y  6
4 x  6 y  12
You will see immediately that the second equation is just double the first, and so contributes no new
information to the system.
If we attempt to solve this system by the Gaussian Elimination method we end up with 0  0, which gets
us nowhere!
But consider the Geometric situation:
A linear equation in two variables represents a straight line, and in the above system the second equation
represents the same straight line as the first.
Hence the above system of equations has an infinite number of solutions, each solution being a point
lying on the straight line.
The solutions may be represented parametrically as follows:
Let y   , say. Then from the first equation we get 2 x  3  6  x  3  32 .
Hence the parametric solution is x  3  32  , y  .
Every real number  gives a solution and every solution corresponds to a value of .
Consider now the system
2x  3y  6
4 x  6 y  15
You will see immediately that these are inconsistent equations, for if 2 x  3 y  6, it follows at once that
4 x  6 y  12.
If we attempt to solve this system by the Gaussian Elimination method we end up with 0  3, which gets
us nowhere!
Geometrically here we have a pair of parallel straight lines, which explains why there is no solution to
this system of equations.
Thus when solving a pair of simultaneous linear equations we may end up with a unique solution (as at
“C” Level in S4), an infinite number of solutions (as in the first example, above), or no solution at all (as
in the second example above).
Gaussian Elimination for Systems in Three Variables
The examples we have done in class so far have been “well behaved”, with a unique solution for the three
variables x, y and z.
But this need not be the case, as explained below.
Geometrically, an equation such as x  2 y  3z  1 represents a plane in three-dimensional space, and if
we have a system of three such equations then we have three planes, one for each equation.
There are several possibilities, listed below:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
All three planes meet in one point.
All three planes meet in one line.
The three planes meet, in pairs, in three straight lines.
The three planes form a pair of parallel planes with a transversal.
The three planes are parallel.
Two of the planes are coincident, cut by a third distinct plane.
All three planes are coincident.
[The case where a third plane is parallel to two coincident planes is just a special case of (v)].
Here are illustrations of two of these situations. (remember that a plane is really infinite in extent, so that
the diagrams are showing just part of each plane).
Situations (ii) and (iii) are shown in the above two sketches. You should draw the others for yourself.
Now consider the solution of the following system by Gaussian Elimination:
x  y z 2
x  2 y  3z  1
3x  4 y  5 z  5
The working is shown overleaf:
1
1
3
1
0
0
1
0
0
1
2
4
1
1
1
1
1
0
1
3
5
1
2
2
1
2
0
2
1
5
2
-1
-1
2
-1
0
R2  R1
R3  3R1
R6  R5
We can see that equations 5 and 6 are identical: y  2 z  1.
If we proceed from there, we end up with 0 x  0 y  0 z  0, which is less than helpful!
But if we put z   , say, in the equation y  2 z  1, we can obtain a parametric solution if we back
substitute in the normal way:
z    y  2  1  y  1  2.
Now go back to the first equation: x  y  z  2  x  1  2    2  x  3  
So the system has the parametric solutions
x  3 
y  1  2
z
And the geometric situation is that the three planes meet along the straight line given by the above
parametric equations.
Consider now the system
x yz 2
x  2 y  3z  1
2x  2 y  2z  3
Performing Gaussian Elimination, we get:
1
1
2
1
0
0
1
2
2
1
1
0
1
3
2
1
2
0
2
1
3
2
-1
-1
R2  R1
R3  2 R1
And the last equation in the tableau, 0 x  0 y  0 z  1, has no solutions whatsoever.
The problem here is that the first and third equations represent parallel planes, and the second equation
represents a plane which cuts the first and third planes.
Here are two more examples where it is reasonably obvious what is going on:
Three parallel planes:
x  2 y  3 z  2
x  2 y  3z  0
x  2 y  3z  1
Three coincident planes:
x  2 y  3 z  2
2 x  4 y  6 z  4
3 x  6 y  9 z  6
For those among you who have the wisdom to choose to study Advanced Higher Mathematics Unit 3, we
will be looking in much more detail at Three-Dimensional Coordinate Geometry (using Vector methods),
and also at the Theory of Matrices. These topics will give us more information about the problems
involved when solving simultaneous linear equations.
Those of you doing Numerical Analysis Unit 2 will learn more still.
The most likely questions on Gaussian Elimination in Advanced Higher Unit 1 are:
(i)
(ii)
An “idiot level” Gaussian Elimination question, with a unique solution, or
The case of three planes intersecting in a line, and obtaining the solutions in parametric form, [i.e.
just like the first example on the previous page].
Finally, in this section;
Ill-Conditioning
Consider the 2  2 system:
x y 2
x  0.99 y  1.99
This system has exact solution x  1, y  1.
Now make a small change to the y-coefficient in the second equation:
x y 2
x  1.01 y  1.99
If you solve this system you ill obtain the exact solution x  3, y  1.
This is catastrophic! A change of about one per-cent in a single coefficient has produced changes of
several hundred per-cent in the (exact) computed solution.
This is known as ill-conditioning; where a small change in the input data produces a much larger change
in the computed solution.
Ill-conditioning is a major problem, especially when the coefficients are rounded. More on this in
Advanced Higher Applied Mathematics.
Geometrically speaking the two lines here are nearly coincident, and, in fact, if the coefficient of y in the
second equation becomes 1 exactly, then we have parallel lines and no solutions at all.
If you are doing the Advanced Higher Applied Mathematics Course you will study ill-conditioning in
much more detail, especially in 3  3 systems, but in Mathematics Unit 1 you only have to deal with
simple 2  2 cases such as the one above.