# Matrices

## m1 Matrix addition and multiplication:

• Matrices are added and subtracted by adding/subtracting each element. For example:
$\left(\begin{array}{cc}1& 5\\ 4& -3\end{array}\right)+\left(\begin{array}{cc}5& 2\\ -1& 7\end{array}\right)=\left(\begin{array}{cc}6& 7\\ 3& 4\end{array}\right)$
• To multiply a matrix by a scalar (e.g. 5), multiply each element in the matrix by the scalar
• Matrices are multiplied as follows:
$\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\left(\begin{array}{cc}w& x\\ y& z\end{array}\right)=\left(\begin{array}{cc}aw+by& ax+bz\\ cw+dy& cx+dz\end{array}\right)$
• To add or subtract a pair of matrices, they must be the same size
• To multiply two matrices, the number of columns in the first must equal the number of rows of the second - they conform

## m2 Zero and identity matrices:

• A zero matrix contains only zeroes. Multiplying a matrix by a zero matrix results in a zero matrix
• $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$ is the 2x2 identity matrix
$\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$ is the 3×3 identity matrix
An identity matrix must have equal rows and columns. If $\mathbf{X}$ is any similar size matrix and $\mathbf{I}$ is the identity matrix, $\mathbf{I}\mathbf{X}=\mathbf{X}$ and $\mathbf{X}\mathbf{I}=\mathbf{X}$

## Equal matrices:

• For two matrices to be equal, they must have the same dimensions and be corresponding (same elements in the same positions)

## m3 Order of multiplication and addition:

• Matrix addition is both commutative (the result is not changed if the order is changed, e.g. $\mathbf{A}+\mathbf{B}=\mathbf{B}+\mathbf{A}$) and associative (the result is not changed if the order which the calculations are performed in does not change (e.g. $\mathbf{A}+\mathbf{B}+\mathbf{C}$ $=\left(\mathbf{A}+\mathbf{B}\right)+\mathbf{C}$ $=\mathbf{A}+\left(\mathbf{B}+\mathbf{C}\right)\right)$
• Matrix multiplication is therefore associative but not commutative, $\mathbf{A}\mathbf{B}\ne \mathbf{B}\mathbf{A}$ in most cases

## m4 Linear transformations:

• A shear is where one line stays in its original place, and other points move in a specified direction parallel to it
• Other types of transformation include rotations, reflections, enlargements and two-way stretches. These can all be represented with matrices
• Multiplying the transformation matrix by the points will equal the points of the transformed shape. For example, the transformed shape $ABC$ where $A\left({x}_{1},{y}_{1}\right)$, $B\left({x}_{2},{y}_{2}\right)$ and $C\left({x}_{3},{y}_{3}\right)$ with transformation matrix $\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$ has the points $\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)\left(\begin{array}{ccc}{x}_{1}& {x}_{2}& {x}_{3}\\ {y}_{1}& {y}_{2}& {y}_{3}\end{array}\right)$

## Reflection matrices:

• An $x$-axis ($x=0$) reflection has the matrix $\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$
• A $y$-axis ($y=0$) reflection has the matrix $\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)$
• A $y=x$ reflection has the matrix $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and a $y=-x$ reflection has the matrix $\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)$

## Rotation matrices:

• A rotation θ degrees clockwise around the origin has the matrix $\left(\begin{array}{cc}\mathrm{cos}\theta & \mathrm{sin}\theta \\ -\mathrm{sin}\theta & \mathrm{cos}\theta \end{array}\right)$, and a rotation $\theta$° anti-clockwise has the matrix $\left(\begin{array}{cc}\mathrm{cos}\theta & -\mathrm{sin}\theta \\ \mathrm{sin}\theta & \mathrm{cos}\theta \end{array}\right)$

## Enlargement matrices:

• An enlargement of scale factor $n$, centre $\left(0,0\right)$ has the transformation matrix $\left(\begin{array}{cc}n& 0\\ 0& n\end{array}\right)$. Notice that with scale factor 1, the transformation matrix would be the identity matrix since the points remain in the same positions

## m5 Successive transformations:

• If a point $P$ is to be transformed by matrix $\mathbf{A}$ and then matrix $\mathbf{B}$, this transformation is written as $\mathbf{B}\mathbf{A}$ and the final points are calculated with $\left(\mathbf{B}×\mathbf{A}\right)×P$

## m6 Invariant points:

• Points which map to themselves after a transformation are called invariant points
• The origin is always an invariant point in any matrix transformation

## m7 Determinants:

• The determinant of a 2×2 matrix $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is calculated with $ad-bc$
• It is notated as $det\left(\mathbf{M}\right)$, $|\mathbf{M}|$ or $|\begin{array}{cc}a& b\\ c& d\end{array}|$

## m8 Uses of determinants:

• The determinant of an enlargement matrix gives the scale factor
• If the determinant is equal to zero, the matrix is known as singular and if not, it's known as non-singular
• If a matrix is singular, it either has infinite solutions (two of the same line) or no solutions at all (parallel lines)

## m9, m10 Inverses:

• If a matrix $\mathbf{M}$ is multiplied by the inverse of the matrix (or ${\mathbf{M}}^{-1}$), it will equal the identity matrix for that order

## Finding the inverse of a non-singular 2x2 matrix:

• ${\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)}^{-1}=$ $\frac{1}{det\mathbf{M}}$$\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$

## m11 Inverse matrices - the product rule:

• $\left(\mathbf{A}\mathbf{B}{\right)}^{-1}={\mathbf{B}}^{-1}{\mathbf{A}}^{-1}$
• This shows that to undo two transformations, you must undo the second before the first
• Proof:
- $\mathbf{A}$ and $\mathbf{B}$ are non-singular matrices, $\mathbf{X}$ is a matrix which multiplies by $\mathbf{A}\mathbf{B}$ to reach $\mathbf{I}$ (identity matrix); $\mathbf{X}$($\mathbf{A}\mathbf{B}$) = $\mathbf{I}$
- $\mathbf{X}\mathbf{A}\mathbf{B}\mathbf{B}$${}^{-1}$ = $\mathbf{I}\mathbf{B}$${}^{-1}$ (multiply by ${\mathbf{B}}^{-1}$)
- $\mathbf{X}\mathbf{A}$ = $\mathbf{B}$${}^{-1}$ (simplify)
- $\mathbf{X}\mathbf{A}\mathbf{A}$${}^{-1}$ = $\mathbf{B}$${}^{-1}$$\mathbf{A}$${}^{-1}$ (multiply by $\mathbf{A}$${}^{-1}$)
- $\mathbf{X}$ = $\mathbf{B}$${}^{-1}$$\mathbf{A}$${}^{-1}$ (simplify)
- ($\mathbf{A}\mathbf{B}$)${}^{-1}$ = $\mathbf{B}$${}^{-1}$$\mathbf{A}$${}^{-1}$ (substitute $\mathbf{X}$ for ($\mathbf{A}\mathbf{B}$)${}^{-1}$)

## m12, m13 Solving linear equations with matrices:

• A pair of simultaneous linear equations can be written as a matrix:
- e.g:
- $3x+2y=9$ and $4x+5y=5$
- Can be written as:
$\left(\begin{array}{cc}3& 2\\ 4& 5\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}9\\ 5\end{array}\right)$
- The solution can be found my finding the inverse of the 2x2 matrix and multiplying it by the third matrix above
• The equations can only be solved if the number of equations is equal to the number of unknowns
• If the two equations never intersect when plotted on a graph, the inverse cannot be found because $det\left(\mathbf{M}\right)=0$ - the matrix is singular
• The matrix may also be singular if both equations are coincident (the same line)

# Complex Numbers

## j1, j2 Solving quadratics and complex number notation:

$x=$ $\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
• The imaginary number, $j$ (sometimes written as $i$) is equal to $\sqrt{-1}$. Therefore, ${j}^{2}=-1$
• A complex number contains both a real and imaginary part, e.g. $5+3j$. The real part is always written first
• The discriminant of a quadratic shows the number of real roots (not complex). It is calculated with ${b}^{2}-4ac$
- If it is positive, there are 2 real roots and 2 $x$-intercepts
- If it $=0$, there is one real root (repeated twice) and one $x$-intercept
- If it is negative, there are no real roots, 2 complex roots and no $x$-intercepts

## j3 Simple calculations:

• To add two complex numbers, add the real parts and imaginary parts separately
• To subtract, subtract them separately
• To multiply, multiply out the brackets and simplify (${j}^{2}=-1$)
• $x-yj$ is the conjugate of $x+yj$. The symbol for a conjugate is: *
• To divide, write the division as a fraction and multiply both numerator and denominator by the conjugate of the denominator

## j4 Zero:

• A complex number is zero only if both the real and imaginary parts are zero

## j5 Complex roots:

• With integer coefficients, a real polynomial has roots in conjugate pairs. It always has the exact number of roots as the order, so ${x}^{3}$ has one real root and one conjugate pair
• Complex numbers are used to solve quadratic equations where the discriminant (see j1) is negative

## j6 Solving higher-degree equations:

• An $n$th degree polynomial has exactly $n$ roots. This includes complex roots and repeated roots
• Example question: $1+2j$ is a root of $4{z}^{3}-11{z}^{2}+26z-15=0$. Find the other roots
- A second root is $1-2j$ (conjugate of $1+2j$), because roots are always in conjugate pairs
- These can be multiplied: $\left(z-1-2j\right)\left(z-1+2j\right)=\left(z-\left(1+2j\right)\right)\left(z-\left(1-2j\right)\right)$
- These brackets expand to ${z}^{2}-z+2zj-z-2zj+1+5={z}^{2}-2z+5$
- Therefore $4{z}^{3}-11{z}^{2}+26z-15=\left({z}^{2}-2z+5\right)\left(ax+b\right)$
- By comparing the dominant term and constant term ($4{z}^{3}$ and $-15$), we can see that $a=4,b=-3$
- The roots are $1+2j,1-2j$ and $4x-3$

## j7 Argand diagrams:

• An Argand diagram is a graph with real numbers horizontally (the $x$-axis on a traditional Cartesian graph) and imaginary numbers vertically ($y$)
• These axis are usually labelled $\mathrm{Re}$ and $\mathrm{Im}$

## j8 Sum and difference on an Argand diagram:

• By representing complex numbers as a vector $\left(x,y\right)$, a line from the origin, vector addition can be used to add:
- A parallelogram can be created by drawing lines for both complex numbers to add. The diagonal of the parallelogram from the origin is therefore the sum
- The other diagonal is equal to to the difference between the two complex numbers
- Alternatively, if the second complex number is drawn from the end of the first line rather than the origin, the sum is the hypotenuse connecting the two lines

## j9 Modulus-argument form:

• The distance from the origin to the complex number is equal to the modulus ($|z|$). This can be calculated using Pythagoras' theorem
• The angle $\theta$, usually measured in radians ($180$° $=\pi$ radians), is measured from the positive real axis. This is the argument, or $\mathrm{arg}\left(\right)$
• However, this would mean that a multiple of $2\pi$ would have two arguments, so it is defined from $-\pi <\theta \le \pi$ (so a $-\pi$ radians argument is not possible and is instead written as 0 radians)
• $\mathrm{arg}\left(0\right)$ is undefined
• Modulus argument form uses these values, in the form $z=r\left(\mathrm{cos}\theta +j\mathrm{sin}\theta \right)$, where $z$ is the complex number, $r$ is the modulus and $\theta$ is the argument
• Therefore, $\mathrm{Im}=r×\mathrm{cos}\theta$ and $\mathrm{Re}=r×\mathrm{sin}\theta$
• $\mathrm{tan}\theta =\mathrm{Im}÷\mathrm{Re}$
- However, $\theta ={\mathrm{tan}}^{-1}\left(\mathrm{Re}/\mathrm{Im}\right)$ only in the first/fourth quadrants. In the second and third, $\pi$ radians must be added
- (the first quadrant is upper-right, second is upper-left, third is bottom-left and fourth is bottom-right)
• If $r$, or $j\mathrm{sin}\theta$ are not positive, the following must be used to rearrange it into modulus-argument form:
- $\mathrm{cos}\left(\pi -a\right)=-\mathrm{cos}\left(a\right)$
- $\mathrm{cos}\left(a-\pi \right)=-\mathrm{cos}\left(a\right)$
- $\mathrm{cos}\left(-a\right)=\mathrm{cos}\left(a\right)$
- $\mathrm{sin}\left(\pi -a\right)=\mathrm{sin}\left(a\right)$
- $\mathrm{sin}\left(a-\pi \right)=-\mathrm{sin}\left(a\right)$
- $\mathrm{sin}\left(-a\right)=-\mathrm{sin}\left(a\right)$
• $\mathrm{arg}\left({z}_{1}-{z}_{2}\right)$ is the angle between the line connecting ${z}_{1}$ and ${z}_{2}$ and a line parallel to the $\mathrm{Re}$ axis

## 10 Loci:

• $|{z}_{2}-{z}_{1}|$ is the distance between points ${z}_{2}$ and ${z}_{1}$ on an Argand diagram
• If $|z-3-4j|=5$, this can be rewritten as $|z-\left(3+4j\right)|=5$ with factorisation. This expression shows that the distance from point $z$ to $3+4j$ is equal to 5. Therefore, the loci is a solid circle centre $\left(3,4\right)$ and with radius 5
• Inequality symbols may also be used
• Take $|z+2|=|z-j|$:
- $|z+2|$ represents the distance from $\left(-2,0\right)$ to a point (because $|z+2|=|z-\left(-2+0j\right)|$)
- $|z-j|$ is the distance from $\left(0,1\right)$ to a point
- Therefore, $|z+2|=|z-j|$ means that the distance from $z$ to $\left(-2,0\right)$ must be same as the distance from $z$ to $\left(0,1\right)$
- The loci is consequently the perpendicular bisector of the line connecting $\left(-2,0\right)$ and $\left(0,1\right)$

# Curve Sketching

## C1 - C2 Graphs of rational functions:

• Numbers which can be expressed like $\frac{n}{d}$ with an integer numerator and an integer, non-zero denominator are rational
• A rational function is similar, but with $n$ and $d$ being functions
• An asymptote is a line on a graph which the curve tends to but never touches, as $x$ or $y$ tends to infinity or -infinity
- Typically, asymptotes are drawn with a dashed line
- To find a vertical asymptote, set the denominator to zero and solve for $x$
• Intercepts are where the line(s) cut the axis. They are found by setting $y$ to 0 and solving for $x$, and then setting $x$ to 0 and solving for $y$
• To find the behaviour of the graph each side of the asymptotes as $x$ and $y$ tend to infinity and negative infinity, set $x$ and then $y$ to a positive and then a negative number (e.g. -100 and +100) and then mark the behaviour on the graph
• The final step is to use the points to complete the sketch. A graphical calculator could be used to confirm the shape

## C3 Solving inequalities:

• To solve an inequality such as $\frac{x-5}{\left(x+2\right)\left(x-3\right)}$ $>0$, one method is to plot the graph for $y=f\left(x\right)$ and find all values of $x$ for which $y$ is greater than 0
• If the inequality is in the form $f\left(x\right)$ $g\left(x\right)$, then the solution is the values of $x$ where the graphs intersect or the graph for $f\left(x\right)$ is lower than that for $g\left(x\right)$

# Proof

TO DO: p1 - language; remember divider!

## p2 Identities:

• In an identity, (≡), all possible values will satisfy the identity, whereas in an equation (=), not all values satisfy the equation
• For example, $\left(x+y{\right)}^{2}\equiv {x}^{2}+2xy+{y}^{2}$ is an identity because it is always true for any value of $x$ and $y$
• However, $x=y$ is an equation, because it is only true if $x$ and $y$ have the same value

## p3 Finding unknown constants in an identity:

• Because identities are true for all variable values, one method to find the values of constants (often represented by upper-case letters) is to substitute multiple random values into a variable and solve the set of resulting equations
• Another method is to manipulate each side so that the structure of each side is equal, and then compare the coefficients

## p4 Proof by induction:

• The steps for proof by induction are as follows:
- Prove that the result is true for a starting value (e.g. $n=1$)
- Assume that the result is true for a value of $k$
- Prove that if it is true for $k$, it is also true for $k+1$
- This therefore proves that the proposition is true. A small sentence should be used to show this: So if the result is true when $n=k$, then it is true when $n=k+1$. As it is true for $n=1$, it is true for all $n$ 1 by induction.
• Example: Prove that the sequence $1+2+3+...+n=$ $\frac{n\left(n+1\right)}{2}$
- Step 1: When $n=1$, LHS = 1 and RHS = $1\left(1+1\right)/2=1$. Therefore, the hypothesis is true for $n=1$
- Step 2: Let n = k. After this substitution, LHS = $1+2+3+...+k$ and RHS = $\frac{k\left(k+1\right)}{2}$
- Step 3: Now let $n=k+1$:
- LHS: We know that $1+2+3+...+k=$ $\frac{k\left(k+1\right)}{2}$. Therefore, we must add $\left(k+1\right)$ to get $\frac{k\left(k+1\right)}{2}$ $+\left(k+1\right)$
- Create a common denominator: $\frac{k\left(k+1\right)}{2}$ $+$ $\frac{2\left(k+1\right)}{2}$
- The left-hand side therefore simplifies to $\frac{\left(k+1\right)\left(k+2\right)}{2}$
- RHS: If we replace all instances of $n$ in $\frac{n\left(n+1\right)}{2}$ with $\left(k+1\right)$, we get $\frac{\left(k+1\right)\left(k+1+1\right)}{2}$ $=$ $\frac{\left(k+1\right)\left(k+2\right)}{2}$
- LHS is now equal to RHS. We show that the proof is complete by writing: so if the result is true when $n=k$, then it is true when $n=k+1$. As it is true for $n=1$, it is true for all $n\ge 1$ by induction.

# Algebra

## a1 Sequences and series:

• A sequence is a set of numbers
• A series is the sum of a sequence

## a2 Summation:

• The notation is to use the sign (uppercase sigma). ${a}_{r}$ where ${a}_{r}$ is a sequence formula, is equal to the sum of all terms added together, to infinity
• To make the series finite, limits can be added. The upper limit is written above the and the lower limit is written below
- The limits are included - if summing ${a}_{r}={r}^{2}$ from $r=1$ to $r=4$, the sum would equal ${1}^{2}+{2}^{2}+{3}^{2}+{4}^{2}$
• The following can be combined and simplified to sum a sequence. The middle two are provided in the exam:
- Linear: $\sum r=\frac{1}{2}n\left(n+1\right)$
- Quadratic: $\sum {r}^{2}=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)$
- Cubic: $\sum {r}^{3}=\frac{1}{4}{n}^{2}\left(n+1{\right)}^{2}$
- For constants: $\sum 1=n$ (so replace any constant terms with $n$
• To sum a finite series:
- Arithmetic sequences: ${S}_{n}=\frac{1}{2}n\left(2a+\left(n-1\right)d\right)$
- Geometric sequences: ${S}_{n}=$ $\frac{a\left(1-{r}^{n}\right)}{1-r}$
• Relationships:
- Where $d$ is the common difference and $a$ is the first term
- Arithmetic: ${a}_{k}=a+\left(k-1\right)d$
- Geometric: ${a}_{k}=a{r}^{k-1}$

## The method of differences:

• The first step to many method of differences questions is to simplify an addition/subtraction of two fractions. This is done by multiplying their denominators to create a common denominator, and then multiplying each fraction by what's required to make it reach that common denominator
• In some sequences, each term can be expressed as the difference between several fractions. For example:
• Find $\frac{1}{r}-\frac{1}{r+1}$
- When $r=1$: $\frac{1}{1}-\frac{1}{2}$
- When $r=2$:
- When $r=3$: etc...
- You can see that the middle section cancels. Therefore, you can find the sum by only calculating the first and last 3 terms
• When finding the sum to $n$ instead of a set limit, the final three terms will be $\left(n-2\right)$, $\left(n-1\right)$ and $n$

## a3 Convergent series:

• A convergent series approaches a number but never reach it. For example, ${a}_{k}=1/k$
• Arithmetic series can therefore never be convergent
TO DO: a4 "Be able to manipulate simple algebraic inequalities, to deduce the solution of suchan inequality." a5 "Appreciate the relationship between the roots and coefficients of quadratic, cubic and quartic equations." a6 "Be able to form a new equation whose roots are related to the roots of a given equation by a linear transformation."

## a5 The relationships between the roots and coefficients of polynomials:

• The sum of the roots ($\sum \alpha$) is equal to $\frac{-b}{a}$ ($a{x}^{2}+bx+c$)
• The product of the roots ($\alpha \beta$) is $\frac{c}{a}$
• Example: The roots of the equation $2{z}^{2}+3z+5=0$ are $\alpha$ and $\beta$. Find an equation with the roots $2\alpha$ and $2\beta$
- Step 1: Find the sum of roots and product of roots:
- $\alpha +\beta =\frac{-b}{a}=\frac{-3}{2}$
- $\alpha \beta =\frac{c}{a}=\frac{5}{2}$
- Step 2: Find the same for the new equation
- $2\alpha +2\beta =2\left(\alpha +\beta \right)=2\left(\frac{-3}{2}\right)=-3$
- $2\alpha ×2\beta =4\alpha \beta =4\left(\frac{5}{2}\right)=10$
- Step 3: Find the new equation
- Since $\frac{-b}{a}$ must equal 3, we can try $a=1$ and find that $b=3$
- Similarly, $\frac{c}{a}=10$ so $c=10$
- Therefore, the new equation is ${z}^{2}+3z+10=0$
- (if there were fractional coefficients here, $a=1$ should be changed to make them integers

## Cubics:

• Sum of roots = $\frac{-b}{a}$
• Sum of the products of roots in pairs ($\alpha \beta +\beta \gamma +\gamma \alpha$ or $\sum \alpha \beta$) = $\frac{c}{a}$
• Product of the roots ($\alpha \beta \gamma$)= $\frac{-d}{a}$

## Quartics:

• Sum of roots = $\frac{-b}{a}$
• Sum of the product of roots in pairs = $\frac{c}{a}$
• Sum of the product of roots in threes ($\alpha \beta \gamma +\beta \gamma \delta +\gamma \delta \alpha +\delta \alpha \beta$ or $\sum \alpha \beta \gamma$) = $\frac{-d}{a}$
• Product of the roots ($\alpha \beta \delta \gamma$) = $\frac{e}{a}$