Summary of Part 3 of the course
Determinants
An n n matrix A has an inverse if and only if det A =
6 0. The determinant of
A is given by a crazy formula: for a 2 2 matrix it is simply
a b
A = c d =) det(A) = ad , bc;
for a 3 3 matrix A, the determinant can be computed by expanding using any
row or any column (pick the one with most zeros in). The expansion should
contain three 2 2-determinants, each multiplied by either +1 or ,1 times an
entry from the matrix.
The area of a triangle with vertices (x1 ; y1 ), (x2 ; y2) and (x3 ; y3 ) is equal to
0
1
x1 y1 1
area = 21 det @x2 y2 1A ;
x3 y3 1
where you choose either +1=2 or ,1=2 so that the result is not a negative
number. If this determinant is zero then the three vertices are collinear.
Finally, determinants can be used to solve a system of n-equations in nunknowns (again!!) using Cramer's rule: for 3 equations in 3 unknowns
0x 1
Ax
det Ay
det Az
A @y A = B =) x = det
det A ; y = det A ; z = det A ;
z
where Ax is obtained by replacing the rst column of A by B , Ay is obtained
by replacing the second column of A by B and Az is obtained by replacing the
third column of A by B . There are similar formulas for any n n matrix. Note
that this method only applies if det A 6= 0.
Inequalities and linear programming
The solution to a system of inequalities (in x and y) is best illustrated by drawing
the region of solutions in the plane. To do this:
1. Replace each inequality sign by an equals sign (e.g. 3x + 4y 2 becomes
3x + 4y = 2). This equation denes a line in the plane. Draw one line
(possibly dotted) for each inequality and decide which side of each line
is the side satisfying the inequality. It may help to draw small arrows
pointing to the correct side of each line.
2. The solutions of the system of inequalities correspond to the region of
points that lie on the correct side of every line.
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If all of the inequalities contain either the or sign (i.e., no < or >) and if
each line in the picture is a straight line, then you can nd the maximum or
minimum value of a linear function z = ax + by subject to the inequalities by
substituting the x and y values of each vertex of the region. The largest is the
max, the smallest is the min. Watch out for either
1. cases where the max (or min) occurs at two points. When this happens,
every point on the line joining the pair of points is a max (or min); or
2. cases where the region is unbounded, so there may not be a nite max (or
min) value.
Exponentials and Logarithms
The exponential function f (x) = ax (a > 0 with a 6= 1) is best understood in
terms of its graph; we typically only draw the graph for the case where a > 1
(what is its domain? range?). This function appears in numerous applications
(see x3.5) and we often want to solve equations involving ax. To do this we
required the inverse function, namely f ,1 (x) = loga (x). The whole point of log
are the relations
loga ax = x and aloga (x) = x;
so loga undoes ax , and vice-versa. Again, remember the graph of loga for a > 1
(what is the domain? range?). Key properties of log are:
loga (x)
loga (u v)
loga (u=v)
loga (un )
=
=
=
=
log10 (x)= log10 (a);
loga (u) + loga (v);
loga (u) , loga (v);
n loga (u):
Can you prove these? Remember, y = loga (u) means that ay = u.
These properties are used frequently when solving equations involving logs
or exponentials. To do this, the strategy is always to isolate a single log or
exponential on one side of an equation, then apply or exponential or log.
Word problems
Lots of examples in x7.5 and x3.5. Take care setting up the inequalities in
optimisation problems. For exponential word problems, notice that exponential
growth problems are very similar to exponental decay problems. Typically there
are two quantities that you don't know (e.g. a and b in y = ae,bt). Use the
information from the question to give you one or both of these quantities, then
answer the question.
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