Matrices
… A zero matrix has elements which are all zero.
… An identity matrix is a square matrix with the diagonal elements from top left = 1
… Matrices can only be added or subtracted if they are the same size (same number of rows and columns).
… Matrices can be multiplied together if the number of rows in the first matrix = number of columns in the second matrix (‘multiplicatively conformable’).
… The order of multiplication of matrices matters so AB ≠ BA.
However, brackets do not matter so (AB)C = A(BC)
… To find the determinant of a 2×2 matrix M:
a b
c d
det M = ad – bc
If det M = 0, then the matrix is singular and does not have an inverse matrix.
If det M ≠ 0, then M is a non-singular matrix and has an inverse.
… To find the determinant of a 3×3 matrix M:
a b c
d e f
g h i
Find the minor of each element a, b and c. Select each element and cross out its row and column, so for element ‘a’, the minor is the determinant of the 2×2 matrix:
e f
h i
det M = a (det of minors of a) – b ( det of minors of b) + c (det of minors of c)
det M = a(ei – fh) – b(di – fg) + c(dh – eg)
… The inverse of a matrix M can be denoted M-1 (-1 in superscript).
This means that:
M M-1 = M-1 M = I
where I is the identity matrix.
… To find the inverse of a 2×2 matrix M:
a b
c d
M-1 = (1/detM) N
where matrix N can be found by swapping the diagonal ad and changing the sign of b and c:
d -b
-c a
… To calculate the inverse of a 3×3 matrix, A:
- Find detA
- Find the matrix of minors, M
- Find the matrix of cofactors, C, by multiplying by the + – etc. pattern.
- Find the transpose of C (swap rows and columns)
- The inverse of A = (1/detA)(transpose of C)
… Simultaneous equations can be solved using matrices, for example if we have three variables x, y and z in a matrix with one column:
A(x,y,z) = v
then
(x,y,z) = A−1 v
and if A is non-singular (so detA ≠ 0), a unique solution for x, y and z can be found.
… A system of linear equations is ‘consistent’ if there is at least one set of values that satisfies all the equations simultaneously. Otherwise, it is inconsistent.
- a matrix of a set of linear equations which is non-singular will be consistent – i.e. there is a one unique solution.
- if the matrix is singular, then the system is either consistent with infinite solutions, or inconsistent with no solutions.
These cases can be visualised by considering how 3 planes can intersect with each other:
- If they form a point = one unique solution non-singular, consistent
- If they form a sheaf = infinite solutions (singular, consistent)
- if the form a prism = no solutions (singular, inconsistent)
- 2 or more planes are parallel = no solutions (singular, inconsistent)
- All 3 equations represent the same plane = infinite solutions (singular, consistent)
… If one row of a matrix of linear equations is a linear multiple of another row, then these two rows will represent parallel planes.
… If a matrix of linear equations is singular, then to find out the nature of how the planes intersect, eliminate one of the variables, e.g. z, in two ways to form two simultaneous equations in x and y:
(1) 3x – y – 6z = 1
(2) x + 3y + 3z = 2
(3) -3x – y +3z = -2
so using (1) + 2(2) we get 5x + 5y = 5
and using (2) – (3) we get 4x + 4y = 4
- if the two equations are consistent (i.e. infinite number of solutions), then the planes meet as a sheaf.
- if the two equations are inconsistent (i.e. no solutions), then the planes form a prism.
Linear Transformations in 2 Dimensions
… A linear transformation only involves LINEAR terms in x AND y. This means that linear transformations always map the origin onto itself.
… All linear transformations are built by combining simple geometric processes such as rotation, stretching, shrinking, shearing and projection. However, translation is NOT a linear transformation.
For example:
(The following vectors should be in column form):
T:(x, y) → (ax+by, cx+dy)
So:
S:(x, y) → (x+4, y-1) … is not a linear transformation because (0,0) → (4,-1) which is not allowed (it is a translation).
But
T:(x, y) → (2x-y, x+y) … IS a linear transformation
… The linear transformation (the following vectors should be in column form):
T:(x, y) → (ax+by, cx+dy)
can be represented by the matrix:
M = a b
c d
so that a position vector P can be transformed linearly by applying:
M P
… If T is a linear transformation and u and v are vectors, and c is a scalar, then the following is true:
T(u + v) = T(u) + T(v)
T(cv) = cT(v)
… Points which do not move under a linear transformation are called invariant points. Lines which do not move under a linear transformation are called invariant lines.
All the points on an invariant line will be invariant points.
… A reflection in the line y = x is represented by the matrix:
0 1
1 0
What is the invariant line?
… A reflection in the line y = -x is represented by the matrix:
0 -1
-1 0
What is the invariant line?
… The matrix of a rotation about the origin is given by:
[ cosθ -sinθ ]
[ sinθ cosθ ]
The only invariant point is…?
How could you prove this is a rotation matrix?
… A transformation matrix represented by:
[ a 0 ]
[ 0 b ]
defines a stretch of scale factor a parallel to the x-axis, and a stretch of scale factor b parallel to the y-axis
When a = b, this represents an enlargement with scale factor a.
… Which lines are invariant in a stretch parallel to the x-axis? a stretch parallel to the y-axis?
If you have stretches in both directions, what are the invariant lines or points?
… The area of the image in reflections and rotations of 2D shapes is the same as the original shape.
However, if the shape is stretched, then the determinant of a transformation matrix M tells us the ‘area scale factor’ for the image. If detM is negative, then the shape has also been reflected.
… Successive transformations, for example PQ, apply from right to left. So PQ represents the application of transformation matrix Q, followed by the transformation matrix P.
The matrix product T = PQ represents the combined effect of both transformations. We can investigate the single transformation T by applying it to the unit vectors (1,0) and (0,1).
… If matrix M represents an enlargement scale factor k, followed by a rotation of angle θ anticlockwise about the origin, then we can break M down into two transformation matrices (note the order of application from right to left):
M = [ cosθ -sinθ ] [ k 0 ]
[ sinθ cosθ ] [ 0 k ]
… Linear Transformations in 3 Dimensions
We can explore a 3×3 transformation matrix by examining its effect on the unit vectors (1,0,0), (0,1,0) and (0,0,1)
If M = [ a b c ]
[ d e f ]
[ g h i ]
Then each of the unit vectors will map onto each column of the transformation matrix from left to right. For example, (1,0,0) will map to (a,d,g)
… Note that a plane can be defined using only one variable, for example z = 0 is a plane parallel to the x and y axes.
… What is the 3D transformation matrix representing a reflection in the plane z = 0 ? Here, we note that only the unit vector (0,0,1) is affected by the reflection and maps to (0,0,-1), so the transformation matrix would be:
[ 1 0 0 ]
[ 0 1 0 ]
[ 0 0 -1 ]
… A 3D rotation about the x-axis by angle θ is represented by the matrix:
[ 1 0 0 ]
[ cosθ -sinθ 0 ]
[ sinθ cosθ 0 ]
Note that points on the x-axis (y=0, z=0) are invariant
… A rotation about the y-axis θ is represented by the matrix:
[ cosθ 0 sinθ ]
[ 0 1 0 ]
[ -sinθ 0 cosθ ]
Note that points on the y-axis (x=0, z=0) are invariant, and watch out for the sneaky sign change on the sin terms!
… A rotation about the z-axis θ is represented by the matrix:
[ cosθ -sinθ 0 ]
[ sinθ cosθ 0 ]
[ 0 0 1 ]
Note that points on the z-axis (x=0, y=0) are invariant
… The inverse of a linear transformation can be found by finding the inverse transformation matrix. Note that reflections are ‘self-inverse’.
For example, if a transformation T is represented by matrix A:
[ 2 4 ]
[ -2 5 ]
and maps a point with coordinates (x,y) onto (6,10), then:
[ 2 4 ] [ x ] = [ 6 ]
[ -2 5 ] [ y ] [ 10 ]
and so
[ x ] = [inverse of A] [ 6 ]
[ y ] [ 10 ]
Complex numbers
Complex numbers help us to analyse and work with waves and any oscillating quantities which have ‘phase difference’, in particular:
- Electromagnetic waves
- Electric circuits (alternating currents)
- Fluid flow
- Signal analysis (e.g. sound processing)
- Control systems (e.g. robotics, aeronautics)
… Euler’s formula gives us a way to express waves with different phases as a power of e:
e^(ix) = cos(x) + isin(x)
Here’s useful video which expands on this elegant formula:
https://youtu.be/_h49ilnTmW4
… To divide complex numbers, multiply the numerator and denominator by the complex conjugate.
… If one root of a quadratic function f(x) = 0 (with real coefficients) is complex, then the other root will be the complex conjugate.
… The roots of a quadratic function are often written as α and β, giving:
(z – α)(z – β) = 0
… For a cubic function, f(x) = 0 (with real coefficients), then you will either have:
All three roots are real, or
One root is real and the other two roots form a complex conjugate pair.
… For a quartic function f(x) = 0 (with real coefficients), then you will either have:
All four roots are real, or
Two are real and the other two form a complex conjugate pair, or
Two form a complex conjugate pair and the other two form a complex conjugate pair.
… The modulus of a complex number z = x + iy is the length of the vector (x,y):
|z| = √(x² + y²)
… The argument of a complex number z = x + iy is the angle between the real axis and the vector (x,y):
arg(z) = α = arctan(y/x)
where the principle argument lies in the interval -π ≤ α ≤ π
… It’s best to draw out the Argand diagram for the complex number as it’s then easier to work out the argument using the geometry.
… For a complex number z with
|z| = r
and
arg(z) = θ
the modulus-argument form of z can be written:
z = r(cosθ + isinθ)
… For any two complex numbers z1 and z2:
- |z1z2| = |z1||z2|
- arg (z1z2) = arg(z1) + arg(z2)
This means that:
(z1)(z2) = (r1)(r2)( cos(θ1+θ2) + isin(θ1+θ2) ) - |z1/z2| = |z1|/|z2|
- arg(z1/z2) = arg(z1) – arg(z2)
This means that:
z1/z2 = (r1/r2)( cos(θ1-θ2) + isin(θ1-θ2) )
… Always take care that a complex number is given in the correct modulus-argument form. For example:
z = 3(cosθ – isinθ) is not in the correct form of z = r ( cosθ + isinθ)
We’ll need to rewrite z using the fact that:
sin(θ) = -sin(-θ)
and
cos(θ) = cos(-θ)
… For two complex numbers z1 and z2, |z2 – z1| represents the distance between z1 and z2 on an Argand diagram.
This is the same idea as in vectors where AB = OB – OA and |AB| = |OB – OA|
… If z1 = x1 + iy1,
then the locus (path) of point z on an Argand diagram such that
|z – z1| = r
| z – (x1 +iy1) | = r
is a circle with radius r and centre (x1, y1)
… To find the cartesian equation of the circle from the form |z – z1| = r
| (x + iy) – (x1 + iy1) | = r
| (x – x1) + i(y – y1) | = r
√[ (x – x1)² – (y – y1)² ] = r
(x – x1)² – (y – y1)² = r²
… For two complex numbers z1 = x1 + iy1, and z2 = x2 – iy2:
|z – z1| = |z – z2|
represents the perpendicular bisector of the line segment joining z1 and z2.
… For a complex number z1 = x1 + iy1:
arg(z -z1) = θ
represents the locus of points z from the fixed point z1 (but not including this point), which is at an angle θ to the line parallel to the real axis from point z1.
We can find the cartesian equation of this line by substituting z1 = x1 + iy1 and z = x + iy:
z – z1 = x + iy – (x1 – iy1)
z – z1 = (x – x1) – i(y – y1)
and so
arg (z – z1) = arctan( (y-y1)/(x-x1) )
If arg (z – z1) = θ, then
arctan( (y-y1)/(x-x1) ) = θ
(y-y1)/(x-x1) = tanθ
and
y – y1 = tanθ (x – x1) … which is the line with gradient tanθ through (x1, y1)
… Regions on an Argand diagram can be defined in a similar way to normal Cartesian inequalities, for example:
| z – 4 – 2i | ≤ 2 to represent the area within a circle
| z – 4 | < | z – 6 | to represent the area to the left of the perpendicular bisector of the line segment joining (0,4) and (0,6)
0 ≤ arg(z – 2 – 2i) ≤ π/4 to show the area between a half line and the line at an angle π/4.
Use set notation to show where a complex number lies, for example:
{z ∈ C : |z| ≤3 } ∩ {z ∈ C : π/4 ≤ arg(z + 3) ≤ π}
Series
We saw how various summations can be represented by the standard results:
[r=1]to[n] Σ1 = n
[r=1]to[n] Σr = ½n(n+1)
[r=1]to[n] Σr² = ⅙n(n+1)(2n+1)
[r=1]to[n] Σr³ = ¼n²(n+1)²
We also saw how to split up and rearrange sigma notation so that the above results can be used.
Roots of Polynomials
… Finding coefficients of a polynomial f(x) using information given about the roots of f(x) = 0.
So
f(x) = k(x – α)(x – β)
f(x) = k [ x² – (α + β)x + αβ ]
Comparing this with a quadratic polynomial ax² + bx + c = 0 leads to…
k = a
The sum of the roots, α + β = -b/a
The product of the roots, αβ = -c/a
So a quadratic can be written as
x² – (α + β)x + αβ = 0
… We can use these results with complex numbers, for example:
The equation z² + (3 + i)z + p = 0 has a root of 2-i. Find the value of p and the other root of the equation:
Since 2-i is a root, then z = 2-i is one solution (root), which we can call α
Therefore
(2-i)² + (3 + i)(2-i) + p = 0
4 -4i -1 + 6 -i + 1 + p = 0
10 – 5i + p = 0
p = 5i – 10
z² – (α + β)z + αβ = 0
The sum of the roots, α + β = -b/a = -(3+i)
α + β = -(3+i)
and as α = 2-i
β = -(3+i) – (2-i)
β = -5
… For a cubic polynomial things get a little more complex, but it’s a similar idea:
If ax³ + bx² + cx + d = 0,
The sum of the roots, α + β + γ = -b/a
The sum of the products of root pairs, αβ + βγ + αγ = c/a
The product of the roots, αβγ = -d/a
… Using these results, we can find integer values for a, b, c and d of a cubic polynomial if we know the roots α, β and γ. However, note that we can have an infinite number of integer values for a, b, c and d which will all result in the same roots for the equation. This is because we could multiply both sides of the cubic equation by any number and the equation would remain unchanged:
ax³ + bx² + cx + d = 0
x³ + (b/a)x² + (c/a)x + (d/a) = 0
Now sub in for b/a, c/a and d/a using the sum of roots etc.
… For a quartic equation:
If ax^4 + bx³ + cx² + dx + e = 0,
The sum of the roots: Σα = α + β + γ + δ = -b/a
The sum of the products of root pairs: Σαβ = αβ + αγ + αδ + βγ + βδ + γδ = c/a
The sum of the products of the root triplets: Σαβγ = αβγ + αβδ + αγδ + βγδ = -d/a
The product of the roots: αβγδ = e/a
Note the sigma notation used to abbreviate these sums.
Proof By Mathematical Induction
… We can prove that a summation formula is true for all positive integers by using mathematical induction:
- The Basis step: Prove that the formula is true when n = 1.
- State the assumption that the formula is true when n = k.
- The Inductive step: Prove that the statement is true when n = k+1. To do this, show that the LHS = RHS for the equation:
f(k+1) = f(k) + [(k+1)th term]. - Conclusion: If the statement is true for n = k, n = k+1 and n = 1, then the statement is true for all positive integers (n ∈ Z+ or n ∈ N, natural numbers).
Note that the tricky part is usually step 3, where we have to rearrange the RHS to be in the same form as the LHS.
… Prove by induction that for any positive integer n,
[r=1 to n]Σr = ½n(n+1)
Step 1: If n = 1, then:
Σ(1) = ½(1)(1+1)
1 = 1
So the formula is true for n = 1
Step 2: Assume the formula is true for n = k, so that:
[r=1 to k]Σr = ½k(k+1)
Step 3: If n = k+1, then:
[r=1 to k+1]Σr = ½(k)(k + 1) + (k+1)th term
[r=1 to k+1]Σr = ½(k)(k + 1) + (k+1)
now we're aiming to show that the RHS can be in the form ½n(n+1) when n = k+1
[r=1 to k+1]Σr = ½(k + 1)(k + 2) ...done!
Step 4:
If the summation formula is true for n = k, then we have shown that it is also true for n = k + 1
As the result is true for n = 1, it is now true for all n = Z+ by mathematical induction.… Proving a divisibility by induction
This is a slightly different method where we show that f(k+1) – f(k) is divisible by a given number. The proof relies upon this stage and the first stage where we prove that f(1) is divisible.
Here’s a good video tutorial to take you through it: https://youtu.be/5JGjELKJYD0
Conics
… A parabola of the form y^2 = 4ax has a focus at (a,0) and a directrix equation x = -a
… The distance from a point on a parabola to the focus will be equal to the shortest distance between the point and the directrix line.
… For a hyperbola, the RATIOS of the distances: Point / focus = Point / directrix … are equal.
… It is worth noting that the rectangular hyperbola with equation xy = c^2 has foci at (±c√2, ±c√2) and directrices x + y = ±c√2.
First Order Differential Equations
… Can be solved using either:
Separating the variables
Integrating factor
1) The separation of variables method
… is used when a differential equation can be written in the form:
dy/dx = f(y)g(x) where f is a function of y only and g is a function of x only.
… Rewrite the problem as y/f(y) = g(x)dx and integrate both sides. Use the initial condition to find the constant of integration.
2) The integrating factor method is used when the differential equation is (or can be rearranged) in the form:
dy/dx + P(x)y = Q(x) where P and Q are functions of x only.
… The integrating factor is e^∫P(x)dx
… Rearrange the differential equation (if needed) into the standard (“exact”) form and find the integrating factor. Multiply through by the integrating factor and rewrite the left hand side using the product rule ‘squish’:
uv’ + u’v = d/dx(uv),
or
f(x)dy/dx + f'(x)y = d/dx (f(x)y)
… Integrating both sides gives the general solution.
… Note that an equation is in its “exact” form if it looks like uv’ + u’v
Take a look at http://www.maths.surrey.ac.uk/explore/vithyaspages/firstorder.html which has some useful notes on these methods.
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