Written by Luka Kerr on June 8, 2018
Introduction To Vectors
Lines
Lines In $\mathbb{R_3}$
Vector Equation Of A Line
$\vec{x} = \vec{a} + λ\vec{v}$ where $\vec{a}$ is the position vector of a point on the line, and $\vec{v}$ is a direction vector on the line. λ is a scalar of the direction vector, and allows to find any point on the line.
Let: $\vec{a} = \begin{pmatrix}a_1 \ a_2 \ a_3\end{pmatrix}$ and $\vec{v} = \begin{pmatrix}v_1 \ v_2 \ v_3\end{pmatrix}$ then $\vec{x} = \begin{pmatrix}a_1 \ a_2 \ a_3\end{pmatrix} + λ \begin{pmatrix}v_1 \ v_2 \ v_3\end{pmatrix}$
Parametric Equation Of A Line
From the vector equation of a line, we can find the parametric equations of the line: \(x = a_1 + λv_1, \\ y = a_2 + λv_2, \\ z = a_3 + λv_3\)
Cartesian Form Of A Line
To convert to cartesian form, we solve for λ: \(\dfrac{x - a_1}{v_1} = \dfrac{y - a_2}{v_2} = \dfrac{z - a_3}{v_3} (= λ)\)
Planes
Linear Combinations
To find a linear combination of vectors, substitute the given vectors into the equation and perform the appropriate arithmetic.
A linear combination ($v$) of two vectors $v_1$ and $v_2$ is a sum of scalar multiples of $ v_1$and $v_2$:
\[v = λv_2 + λv_2\]Planes In $\mathbb{R_3}$
Parametric Vector Equation Of A Plane
To form a vector equation of a plane, we need a point ($a$) on the plane, and two non-collinear vectors ($\vec{v}$ and $\vec{w}$). From this, we are able to let $λ_1$ and $λ_2$ be any value, such that any point on the plane can be reached. \(\vec{x} = a + λ_1\vec{v} + λ_2\vec{w}\)
Parametric Form Of A Plane
From the parametric vector equation of a plane, we can find the parametric form of the plane: \(x_1 = a_1 + λv_1 + λw_1, \\ x_2 = a_2 + λv_2 + λw_2, \\ x_3 = a_3 + λv_2 + λw_3\)
Cartesian Form Of A Plane
The cartesian form of a plane can be represented as $ax_1 + bx_2 + cx_3 = d$, where \(\vec{n} = \begin{pmatrix}a\\b\\c\end{pmatrix}\) is the normal to the plane, and $P = (x_1, x_2, x_3)$ is any point on the plane. Use the point $P$ to solve for $d$.
Example:
Let $n = \vec{(1, 2, 3)}$
Let $P = (-1, 2, 0)$
$x_1 + 2x_2 + 3x_3 = d$
$= 1(-1) + 2(2) + 3(0) = d$
$= -1 + 4 + 0 = d$
$d = -3$
$\therefore$ the cartesian equation of the plane $ = x_1 + 2x_2 +3x_3 = -3$
Vector Geometry
Lengths
The length of a vector can be calculated as: \(| \begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} | = \sqrt{(x_1)^2 + (x_2)^2 + (x_3)^2}\)
The Dot Product
Arithmetic Properties Of The Dot Product
Add the multiplication of each row: \(\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} . \begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix} = (x_1 \times y_1) + (x_2 \times y_2) + (x_3 \times y_3)\)
This can also be written as $\vec{a} \dot{} \vec{b} = |\vec{a}||\vec{b}|cos \theta$ where $\theta \in [0, \pi]$ is the angle between vectors $\vec{a}$ and $\vec{b}$.
To solve for $\cos \theta$: \(\cos{\theta} = \dfrac{\vec{a} \dot{} \vec{b}}{|\vec{a}||\vec{b}|}\)
Geometric Interpretation Of The Dot Product
The dot product can be understood as a measure of how well one vector aligns with another vector. If two vectors are perpendicular (orthogonal) then the dot product evaluates to zero.
Useful for:
- Testing/finding perpendicular (orthogonal) vectors
- Finding the angle between two vectors
Applications
Vector Projection
Having a vector $\vec{AB}$ and a vector $\vec{AC}$, we can find the projection of $\vec{AB}$ onto $\vec{AC}$: \(Proj_{\vec{AC}} (\vec{AB})\) We think of this as the projection of $\vec{AB}$ onto $\vec{AC}$. This projection gives us the shortest distance between $\vec{AB}$ and $\vec{AC}$, or the vector that is perpendicular to $\vec{AC}$.
The general formula for a projection of $\vec{a}$ onto $\vec{b}$ is: \(Proj_{\vec{b}} (\vec{a}) = (\dfrac{\vec{a} \dot{} \vec{b}}{|\vec{b}|^2}) \vec{b}\)
Shortest Distance From A Point To A Line In $\mathbb{R_3}$
Given:
- A line $l$ in vector form
- A point $P$ on the plane
- We can find the point $X$ where the perpendicular line from $l$ to $P$ intersects $l$
- Using $X$, we can then find $\vec{PX}$, the vector from $P$ to $X$
- We know that the direction of $l \dot{} \vec{PX} = 0$, so we can solve for λ
- Once we have λ, we can substitute this back into $\vec{PX}$ and find the magnitude, giving the shortest distance
The Cross Product
Arithmetic Properties Of The Cross Product
The cross product gives us a vector that is perpendicular (orthogonal) to both vectors $\vec{AB}$ and $\vec{AC}$.
When using the cross product, think of the ‘right hand rule’.
For each row of the two vectors, cross subtract the multiplication of two rows below crossed: \(\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} \times \begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix} = \begin{pmatrix}(x_2 \times y_3) - (y_2 \times x_3)\\(x_3 \times y_1) - (y_3 \times x_1)\\(x_1 \times y_2) - (y_1 \times x_2)\end{pmatrix}\)
\[e.g. \begin{pmatrix}2\\2\\-2\end{pmatrix} \times \begin{pmatrix}2\\1\\1\end{pmatrix} = \begin{pmatrix}4\\-6\\-2\end{pmatrix}\]Geometric Interpretation Of The Cross Product
As stated above, the cross product gives us a vector that is perpendicular (orthogonal) to two vectors.
Useful for:
- Finding a vector perpendicular to two other vectors
- Finding the area of a parallelogram by taking the magnitude of the cross product of two adjacent sides of a parallelogram
Area Of $\triangle$ In $\mathbb{R_2}$ With Vectors
The area of a triangle $\triangle ABC$ is equal to half the area of the parallelogram formed by sides $\vec{AB}$ and$\vec{AC}$: \(= \dfrac{1}{2}|\vec{AB} \times \vec{AC}|\) Where $|\vec{AB} \times \vec{AC}|$ is the length of the cross product of the vectors $\vec{AB}$ and $\vec{AC}$
Complex Numbers
Introduction To Complex Numbers
The number $\sqrt{-1}$ can be represented as $i$. Thus, $i \times i$ or $i^2$ $= 1$.
A complex number is said to be in Cartesian form when it is written in the form $a + bi$, where $a$ is the real part, and $b$ is the imaginary part. Some examples are $3 + 4i$ and $cos \dfrac{\pi}{3} + i sin \dfrac{\pi}{3}$.
Rules Of Arithmetic For Complex Numbers
Let $z = a + bi$ and $w = c + di$
Addition
$z + w = (a + c) + (b + d)i$
Subtraction
$z - w = (a - c) + (b - d)i$
Multiplication
$(a + bi)(c + di) = ac + bci + adi + (bi)(di) = ac + (bc + ad)i + (bd)i^2$
Since $i^2 = -1$ this can be simplified to $(ac - bd) + (bc + ad)i$
Division
Where $w \ne 0$: \(\dfrac{z}{w} = \dfrac{ac + bd}{c^2 + d^2} + \dfrac{bc - ad}{c^2 + d^2}i\)
Propositions
- Uniqueness of Zero. There is one and only one zero in $\mathbb{C}$
- Cancellation Property. If $z, v, w \in \mathbb{C}$ satisfy $z + v = z + w$ then $v = w$.
- Cancellation Property. If $z, v, w \in \mathbb{C}$ satisfy $zv = zw$ and $z \ne 0$ then $v = w$.
- $0z = 0$ for all complez numbers $z$.
- $(-1)z = -z$ for all complex numbers $z$.
- If $z, w \in \mathbb{C}$ satisft $zw = 0$ then either $z = 0$ or $w = 0$, or both.
Real Parts, Imaginary Parts & Complex Conjugates
Real Parts
The real part of an equation $z = a + bi$ is written $Re(z)$, where $a, b \in \mathbb{R}$ is given by $Re(z) = a$.
Imaginary Parts
The imaginary part of an equation $z = a + bi$ is written $Im(z)$, where $a, b \in \mathbb{R}$ is given by $Im(z) = b$.
Complex Conjugates
If $z = a + bi$ where $a, b \in \mathbb{R}$ then the complex conjugate of $z$ is $\bar{z} = a - bi$.
Properties:
- $\bar{\bar{z}} = z$
- $\overline{z + w} = \bar{z} + \bar{w}$ and $\overline{z - w} = \bar{z} = \bar{w}$
- $\overline{zw} = \overline{z}\overline{w}$ and $\overline{\dfrac{z}{w}} = \dfrac{\overline{z}}{\overline{w}}$
- $Re(z) = \dfrac{1}{2}(z + \overline{z})$ and $Im(z) = \dfrac{1}{2i}(z - \overline{z})$
- If $z = a + bi$, then $z\overline{z} = a^2 + b^2$, so $z\overline{z} \in \mathbb{R}$ and $z\overline{z} \ge 0$
Argand Diagram
The argand diagram is a geometric way to represent complex numbers in the form $z = a + bi$. This complex number can be represented by the coordinates $(a, b)$, and where the $y$ axis represents the imaginary axis, and the $x$ axis represents the real axis. For example, $z = 3 - 2i$ would have coordinates $(3, -2)$.
The polar form of a complex number can also be represented on the Argand diagram, where $\theta$ is the angle measured from the positive $x$-axis and $r$ is the distance of a point from the origin.
Polar Form, Modulus & Argument
Polar Form
Polar form of a complex number can be obtained using plane polar corrdinates $r$ and $\theta$ instead of $x$ and $y$.
The relationship between the real and imaginary parts of a complex number $z = x + yi$ and the polar coordinates $r$ and $\theta$ are:
$Re(z) = x = r cos \theta$ and $Im(z) = y = r sin \theta$
Hence a complex number $z \ne 0$ can be written using the polar coordinates $r$ and $\theta$ as:
$z = r(cos \theta + isin\theta)$
It is important to note that $\theta$ for any complex number $z = x + yi$ is not uniquely defined, since adding or subtracting $2\pi$ produces the exact same values for $x$ and $y$, and hence the same complex number $z$.
Modulus
For $z = x + yi$ where $x, y \in \mathbb{R}$, we define the modulus of $z$ to be $|z| = \sqrt{x^2 + y^2}$. This is also known as the magnitude or absolute value of $z$. When written in polar form, the modulus is defined as $r$, or the distance from the point $z$ from the origin as in the Argand diagram.
Argument
The polar coordinate $\theta$ of a complex number $z$ is called the argument of the complex number, and is written as $arg(z)$. This angle can be increased or decreased by $2\pi$ without changing the corresponding complex number.
To find the principle argument of $z$ we choose a value of $\theta$ such that $-\pi \lt \theta \leq \pi$. This is written as $Arg(z)$.
Properties & Application Of The Polar Form
De Moivre’s Theorem
For any real number $\theta$ and integer $n$ \((cos(\theta) + i sin(\theta))^n = cos(n \theta) + i sin(n \theta)\)
Euler’s Formula
For a real $\theta$ we define \(e^{i\theta} = cos(\theta) + i sin(\theta)\)
The Arithmetic Of Polar Forms
Using Euler’s formula, we can rewrite the compex number $z = r(cos\theta + i sin \theta)$ in an alternative and more useful form. This is called the polar form of the non zero complex number $z = re^{i\theta}$ where $r = |z|$ and $\theta = Arg(z)$.
Four important cases are:
- $1 = e^0$
- $i = e^{i\pi/2}$
- $-1 = e^{i\pi}$
- $-i = e^{-i \pi/2}$
Multiplication
$z_1 z_2 = r_1e^{i\theta_1} r_2e^{1\theta_2} = r_1 r_2e^{i(\theta_1 \theta_2)}$
Division
$\dfrac{z_1}{z_2} = \dfrac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}} = \dfrac{r1}{r2}e^{i(\theta_1-\theta_2)}$
Powers Of Complex Numbers
If $z = re^{i\theta}$ then the properties of exponentials give $z^n = r^n e^{in\theta}$.
Roots Of Complex Numbers
A complex number $z$ is an nth root of a number $z_0$ if $z_0$ is the nth power of $z$, that is $z$ is the nth root of $z_0$ if $z^n = z_0$.
Trigonometric Applications Of Complex Numbers
Binomial Theorem
If $a, b \in \mathbb{C}$ and $n \in \mathbb{N}$ then \((a + b)^n = \sum_{k=0}^{n} \begin{pmatrix}n\\k\\\end{pmatrix} a^{n-k}b^k\) where the numbers \(\begin{pmatrix}n\\k\end{pmatrix} = \dfrac{n!}{k!(n-k)!}\) are the binomial coefficients.
From this, we can express $cos(n \theta)$ or $sin(n\theta)$ in terms of powers of $cos(\theta)$ or $sin(\theta)$ by using De Moivres’s Theorem. For example: \(cos(4\theta) + isin(4\theta) = (cos(\theta) + isin(\theta))^4\) can then be solved using the minomial theorem as it is in the form of $(a + b)^n$.
Complex Polynomials
Roots & Factors Of Polynomials
- A number $\alpha$ is a root or (zero) of a polynomial $p$ if $p(\alpha) = 0$.
- Let $p$ be a polynomial. Then, if there exists polynomials $p_1$ and $p_2$ such that $p(z) = p_1(z)p_2(z)$ for all complex $z$, then $p_1$ and $p_2$ are called factors of $p$.
Remainder Theorem
The remainder $r$ which results when $p(z)$ is divided by $z - \alpha$ is given by $r = p(\alpha)$.
Factor Theorem
A number $\alpha$ is a root of $p$ if and only if $z - \alpha$ is a factor of $p(z)$.
The Fundamental Theorem Of Algebra
A polynomial of degree $n \ge 1$ has at least one root in the complex numbers.
Factorisation Theorem
Every polynomial of degree $n \ge 1$ has a factorisation into $n$ linear factors of the form
\[p(z) = a(z - \alpha_1)(z - \alpha_2)...(z - \alpha_n)\]where the $n$ complex numbers $\alpha_1, \alpha_2…\alpha_n$ are roots of $p$ and where $a$ is the coefficient of $z^n$.
Linear Equations & Matrices
Linear Equations & Matrix Notation
We can express the following linear equations as an augmented matrix: \(x_1 + 2x_2 - 3x_3 = 5 \\ 4x_2 - 5_2 + 4x_2 = 9\) \({ \left( \begin{array}{ccc|c} 1 & 2 & -3 & 5 \\ 4 & -5 & 4 & 9 \\ \end{array} \right) }\)
An augmented matrix contains the coefficients of each unknown, and the right hand side of the linear equation(s). This can be expressed in an equation: $Ax = b$, where $A$ is the coefficient matrix and $x$ is the unkonwn vector. $Ax = b$ can be visualised as: \(A = { \left( \begin{array}{ccc} 1 & 2 & -3 \\ 4 & -5 & 4 \\ \end{array} \right) } ,\quad x =\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} , \quad b = \begin{pmatrix}5\\9\end{pmatrix}\)
Solving Systems Of Equations
Row-Echelon Form
A matrix is in row-echelon form if:
- In every leading row, the leading entry is further to the right than the leading entry in any row higher up in the matrix
For example, the following matrix is in row-echelon form: \(\begin{pmatrix} 2 & 3 & 4 & 5 \\ 0 & 6 & 7 & 8 \\ 0 & 0 & 9 & 10 \end{pmatrix}\)
Reduced Row-Echelon Form
A matrix is in reduced row-echelon form if:
- every leading entry is $1$
- every leading entry is the only non-zero entry in its column
For example, the following matrix is in reduced row-echelon form: \(\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 3 \\ \end{pmatrix}\)
Converting To Row-Echelon Form
The process to convert to row-echelon form is known as Gaussian Elimination. Three row operations can be performed on each row to get the matrix into row-echelon form:
- Swapping rows: $R_n \Leftrightarrow R_m$
- Multiplying a row by itself: $R_n = \lambda R_n$, where $\lambda \ne 0$
- Addition with another row: $R_n = R_n + \lambda R_m$
Converting To Reduced Row-Echelon Form
To convert a matrix to row-echelon form we can perform two types of elementary row operations:
- Multiplication of a row by a constant ($R_n = 3R_n$)
- Adding a multiple of one row to another row ($R_n = R_n + 5R_m$)
The procedure is as follows:
- Start with the lowest row which is not all zeros
- Multipy it by a constant to make its leading entry $1$
- Add multiples of this row to higher rows to get all zeros in the column above the leading entry of this row
- Goto step 1 for the next lowest row
Deducing Solubility From Row-Echelon Form
We can determine the number of solutions to the linear system by examining the position of the leading entries in a row-echelon reduced matrix.
0 solutions if the right hand column is a leading column: \({ \left( \begin{array}{cc|c} 1 & 2 & 3 \\ 0 & 0 & 4 \\ \end{array} \right) }\)
1 unique solution if every variable is a leading variable: \({ \left( \begin{array}{cc|c} 1 & 0 & 2 \\ 0 & 3 & 4 \\ \end{array} \right) }\)
$\infty$ many solutions if there is at least one non-leading variable: \({ \left( \begin{array}{cc|c} 1 & 2 & 3 \\ 0 & 0 & 0 \\ \end{array} \right) }\)
Solving For An Indeterminate $b$
When dealing with an unknown right hand side, also known as an indeterminate $b$ we perform similar steps as above:
- Convert the system of equations into an augmented matrix
- Reduce this matrix into row-echelon form
- Substitute variables back
Matrices
Matrix Arithmetic And Algebra
An $m \times n$ matrix is an array of $m$ rows by $n$ columns. The notation $[A]_{ij}$ denotes a matrix $A$ and the element at row $i$, column $j$.
Equality, Addition & Multiplication Of A Scalar
Two matrices $A$ and $B$ are equal if:
- The number of rows and columns in $A$ is the same as $B$
- $[A]_{ij} = [B]_{ij}$ for all $i$ and $j$
If $A$ and $B$ are $m \times n$ matrices, then the sum $C = A + B$ is the $m \times n$ matrix whose entries are: \([C]_{ij} = [A]_{ij} + [B]_{ij}\)
For any matrix $A \in M_{mn}$ the negative of $A$ is the $m \times n$ matrix $-A$.
If $A$ and $B$ are $m \times n$ matrices, then the subtraction $C = A - B$ is the $m \times n$ matrix whose entries are: \([C]_{ij} = [A]_{ij} - [B]_{ij}\)
If $A$ is an $m \times n$ matrix and $\lambda$ is a scalar, then the scalar multiple $B = \lambda A$ of $A$ is the $m \times n$ matrix whose entries are: \([B]_{ij} = \lambda[A]_{ij}\)
Matrix Multiplication
If $A$ is an $m \times n$ matrix and $X$ is an $n \times p$ matrix, then the product $AX$ is the $m \times p$ matrix whose entries are given by the formula: \([AX]_{ij} = \sum_{k = 1}^{n} [A]_{ik} [X]_{kj}\) An identity matrix ($I$) is a square matrix with 1’s on the diagonal, and 0’s off the diagonal. For example: \({ \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) } \quad or \quad { \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) }\) Properties of Matrix Multiplication:
- If the product $AB$ exists, then $A(\lambda B) = \lambda(AB) = (\lambda A)B$
- Associative Law of Matrix Multiplication: If the produts $AB$ and $BC$ exist, then $A(BC) = (AB)C$
- $AI = A$ and $IA = A$ where $I$ represents identity matrices of the appropriate sizes
- Right Distributive Law: If $A + B$ and $AC$ exist, then $(A + B)C = AC + BC$
- Left Distributive Law: If $B + C$ and $AB$ exist, then $A(B + C) = AB + AC$
Using these laws allow us to simplify expressions in unknown matrices in almost the same way as simplifying expressions in algebra.
The Transpose Of A Matrix
The transpose of an $m \times n$ matrix $A$ is the $n \times m$ matrix $A^T$ ($A$ transpose) with entries given by: \([A^T]_{ij} = [A]_{ji}\) Properties:
-
A transpose is obtained by changing the order of the subscripts on each entry and hence \([(A^T)^T]_{ij} = [A^T]_{ji} = [A]_{ij}\)
-
If $A, B \in M_{mn}$ and $\lambda, \mu \in \mathbb{R}$ then $(\lambda A + \mu B)^T = \lambda A^T + \mu B^T$
-
If $AB$ exists, the number of columns in $A$ must be equal to the number of rows in $B$
-
A matrix is symmetric if $A = A^T$
The Inverse Of A Matrix
A matrix $X$ is said to be an inverse of a matrix $A$ if both $AX = I$ and $XA = I$, where $I$ is an identity matrix of the appropriate size.
A matrix $X$ is said to be a right inverse of $A$ is $A$ if $r \times c$, $X$ is $c \times r$ and $AX = I_r$.
A matrix $Y$ is said to be a left inverse of $A$ if $A$ is $r \times c$, $Y$ is $c \times r$ and $YA I_c$.
Properties:
- If the matrix $A$ has both a left inverse $Y$ and a right inverse $X$, then $Y = X$
- $A^{-1}$ is the inverse of $A$
- If $A$ is invertible, then $A^{-1}$is also invertible. $(A^{-1})^{-1} = A$
- If $A$ and $B$ are invetible matrices and the product $AB$ exists, then $AB$ is also invertible and $(AB)^{-1} = B^{-1} A^{-1}$
Calculating The Inverse Of A Matrix
Given $A$, an invertible $n \times n$ matrix, we can write the columns of $A^{-1}$ as $x_1, x_2 … x_n$, then $A A^{-1} = I$ can be written as: \(A A^{-1} = A(x_1|x_2|...|x_n) = (e_1|e_2|...|e_n)\) where ${e_1, e_2 … e_n}$ is the standard basis for $\mathbb{R}^n$, and $x_i$ is the unique solution of $Ax = e_i$. If $Ax = e_i$ doesn’t have a solution, then $A$ is not invertible.
Steps to find $A^{-1}$
- Form the augmented matrix $(A | I)$ with $n$ rows and $2n$ columns
- Use Gaussian elimination to convert $(A | I)$ to row-echelon form $(U | C)$. Then, if all entries in the bottom row of $U$ are zero, stop - in this case $A$ has no inverse
- Otherwise, use further row operations to reduce $(U | C)$ to reduced row-echelon form $(I | B)$. The right hand half $B$ is the inverse
For a $2 \times 2$ matrix, the inverse can be found in a simpler way: \({ \left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)^{-1} } = \dfrac{1}{ad - bc} { \left( \begin{array}{cc} d & -b \\ -c & a \\ \end{array} \right) } , \textrm{ provided } ab - bc \ne 0\)
Inverses & Solution Of $Ax = b$
- If $A$ is a square matrix, then $A$ is invertible if and only if $Ax = 0$ implies $x = 0$
- If $A$ is a square matrix, then $A$ is invertible if and only if the matrix equation $Ax = b$ has a unique solution for all vectors $b \in \mathbb{R}^n$. In this case the unique solution if $x = A^{-1}b$
Determinants
The determinant of a matrix $A$ is written as $det(A)$ or $|A|$. A determinant is only defined for a matrix is that matrix is square.
The determinant of an $n \times n$ matrix $A$ is defined as: \(\sum_{k = 1}^{n} (-1)^{1 + k} a_{1k} |A_{1k}|\) Properties:
- $det(A^T) = det(A)$
- If a matrix $B$ is obtained from the matrix $A$ by interchanging two rows or columns, then $det(B) = -det(A)$
- If a matrix contains a zero row or column, then its determinant is zero
- If a matrix $B$ is obtained from the matrix $A$ by the multiplying a row or column of $A$ by a scalar $\lambda$, then $det(B) = \lambda det(A)$
- If any column of a matrix is a multiple of another column of the matrix or any row is a multiple of another row, then the value of $det(A)$ is zero
- If $A$ and $B$ are square matrices such that the product $AB$ exists, then $det(AB) = det(A) det(B)$
The Efficient Numerical Evaluation Of Determinants
- If $U$ is a square $n \times n$ row-echelon matrix, then $det(U)$ is equal to the product of the diagonal entries of $U$. This can be written as:
- If $A$ is a square matrix and $U$ is an equivalent row-echelon matrix formed from $A$ by Gaussian elimination, then $det(A) = \epsilon det(U)$ where $\epsilon = +1$ if an even number of row interchanges have been made, and $\epsilon = -1$ if an odd number of row interchanges have been made
Determinants & Solutions Of $Ax = b$
Given $A$ is a square $n \times n$ matrix:
- If $det(A) \ne 0$, the equation $Ax = b$ has a solution and the solution is unique for all $b \in \mathbb{R}^n$
- If $det(A) = 0$, the equation $Ax = b$ either has no solution of an infinite number of solutions for a given $b$
- For any square matrix $A$, the homogenous system of equations $Ax = 0$ has a non zero solution if and only if $det(A) = 0$
- A square matrix $A$ is invertible if and only if $det(A) \ne 0$
- If $A$ is an invertible matrix, then $det(A^{-1}) = \dfrac{1}{det(A)}$
Proving 3 Points In 3D Space Are Collinear
Given 3 points $A$, $B$ and $C$, to prove that these 3 points are collinear, you must show that $\vec{AB}$ is parallel to $\vec{AC}$. If $\vec{AC}$ is a scalar multiple of $\vec{AB}$, then the two vectors are parallel, and the 3 points are collinear.
Distance Between Two Points
In 2D Space
For points $(x_1, y_1)$ and $(x_2, y_2)$ the distance can be found: \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
In 3D Space
The above forumla can be extended for points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$: \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\)
Coordinates Where A Line Meets A Plane
Given a line $l$ and a plane $II$, we can find the point $P$ where the line intersects the plane in 3 steps:
- Find the parametric equation of the line in terms of λ and $x$, $y$, $z$
- Substitute these values into the equation of the plane to find λ
- Substitute λ back into $x$, $y$ and $z$
The point $P$ is equal to $(z, y, z)$.
Sets, Inequalities & Functions
Elementary Functions
Elementary functions are functions that can be constructed by combining a finite number of polynomials, exponentials, logarithms, roots and trigonometric functions via function composition, addition, subtraction, multiplication and division.
Examples:
$f(x) = e^{sin(x)} + x^2$
Implicitly Defined Functions
When a function is defined by more than one rule it is said to be implicitly defined. An example is the two functions $f : [-a, a] \to \mathbb{R}$ and $g : (-a, a) \to \mathbb{R}$ defined by the rules: \(\left\{ \begin{array}{l} y = f(x) \\ (x^2 + y^2 -1)^3 -x^2y^3 = 0 \\ y \ge b \end{array} \right.\) and \(\left\{ \begin{array}{l} y = g(x) \\ (x^2 + y^2 -1)^3 -x^2y^3 = 0 \\ y \lt b \end{array} \right.\)
Limits
Limits Of Functions At $\infty$
The Pinching Theorem
On a graph, given a function $h$ that always lies above a function $f$, and given that $h$ and $f$ have the same limit as $x \to \infty$, any function, such as $g$, that always lies between $h$ and $f$ (such that is it ‘pinched’), has the same limit of $x \to \infty$.
Suppose that $f$, $g$ and $h$ are all defined on the interval $(b, \infty)$, where $b \in \mathbb{R}$. If: \(f(x) \le g(x) \le h(x) \ \ \ \ \ \ \ \forall x \in (b, \infty)\) and \(\lim_{x \to \infty} f(x) = \lim_{x \to \infty} h(x) = L\) then \(\lim_{x \to \infty} g(x) = L\)
Limits Of The Form $f(x)/g(x)$
When calculating limits of the form \(\lim_{x \to \infty} \dfrac{f(x)}{g(x)}\) where both $f(x)$ and $g(x)$ tend to infinity as $x \to \infty$, to find the limit we have to divide both $f$ and $g$ by the fastest growing term appearing in the denominator $g$.
For example, evaluate: \(\lim_{x \to \infty} \dfrac{4x^2 -5}{2x^2 +3x}\) Solution: There are two terms appearing in the denominator - $2x^2$ and $3x$. As $x \to \infty$, the fastest growing term is the one involving $x^2$. So we divide the denominator by $x^2$: \(\dfrac{4x^2 -5}{2x^2 + 3x} = \dfrac{4 - 5/x^2}{2 + 3/x} \\ = \dfrac{4 - 0}{2 + 0}\) as $x \to \infty$. Therefore: \(\lim_{x \to \infty} \dfrac{4x^2 -5}{2x^2 +3x} = 2\)
Limits Of The Form $\sqrt{f(x)} = \sqrt{f(x)}$
When calculating limits of the form \(\sqrt{f(x)} - \sqrt{g(x)}\) since both $f(x)$ and $g(x)$ tend to infinity as $x \to \infty$, we have to multiply both the numerator and denominator by $\sqrt{f(x)} + \sqrt{g(x)}$ and expand the numerator as a difference of squares.
For example: \(\sqrt{x + 5} - \sqrt{x + 2} = \dfrac{(\sqrt{x + 5} - \sqrt{x + 2})(\sqrt{x + 5} + \sqrt{x + 2})}{\sqrt{x + 5} + \sqrt{x + 2}} \\ = \dfrac{(x + 5) - (x + 2)}{\sqrt{x + 5} + \sqrt{x + 2}} \\ = \dfrac{3}{\sqrt{x + 5} + \sqrt{x + 2}} \\ \to 0\) Therefore, $\lim_{x \to \infty} \sqrt{x + 5} - \sqrt{x + 2} = 0$
Indeterminate Forms
Given \(\lim_{x \to \infty} \dfrac{f(x)}{g(x)}\) where $f(x) \to \infty$ and $g(x) \to \infty$ as $x \to \infty$, we recognise that a limit of the form $\frac{\infty}{\infty}$ is in indeterminate form.
The Definition Of $\lim\limits_{x \to \infty} f(x)$
Suppose that $L$ is a real number and $f$ is a real-valued function defined on some interval $(b, \infty)$. We say that $\lim_{x \to \infty} f(x) = L$ if for every positive real number $\epsilon$, there is a real number $M$ such that if $x \gt M$, then $|f(x) - L| < \epsilon$.
Limits Of Functions At A Point
Left-hand, Right-hand & Two-sided Limits
If the left-hand limit $\lim_{x \to a^-} f(x)$ and the right-hand limit $\lim_{x \to a^+} f(x)$ both exist and equal the same real number $L$, then we say that the limit of $f(x)$ as $x \to a$ exists and is equal to $L$, and we write \(\lim_{x \to a} f(x) = L\) If any one of these conditions fails then we say that the limit doesn’t exist.
Limits & Continuous Functions
Suppose that $f$ is defined on some open interval containing the point $a$. If $\lim_{x \to a} f(x) = f(a)$ then we say that $f$ is continuous at $a$; otherwise, we say that $f$ is discontinuous at $a$.
If $f : \mathbb{R} \to \mathbb{R}$ is continuous at every point $a$ in $\mathbb{R}$ then we say that $f$ is continuous everywhere.
Rules:
Suppose that $a \in \mathbb{R}$ and that $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exist and are finite real numbers. Then:
- $\lim_{x \to a} (f(x) + g(x)) = \lim_{x \to a} f(x) = \lim_{x \to a} g(x)$
- $\lim_{x \to a}(f(x) - g(x)) = \lim_{x \to a} f(x) - \lim_{x \to a} g(x)$
- $\lim_{x \to a} f(x) g(x) = \lim_{x \to a} f(x) \lim_{x \to a} g(x)$
- $\lim_{x \to a} \dfrac{f(x)}{g(x)} = \dfrac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}$, provided that $\lim_{x \to a} g(x) \ne 0$
Properties Of Continuous Functions
Combining Continuous Functions
Propositions:
- If the functions $f$ and $g$ are continuous at a point $a$. Then $f + g$, $f - g$ and $fg$ are continuous at $a$. If $g(a) \ne 0$ then $f/g$ is also continuous at $a$.
- If $f$ is continuous at $a$ and $g$ is continuous at $f(a)$, then $g \ o \ f$ is continuous at $a$
Continuity On Intervals
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Given $f$ being a real valued function defined on an open interval $(a, b)$, we say that $f$ is continuous on $(a, b)$ if $f$ is continuous on every point in the interval $(a, b)$
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Given $f$ being a real valued function defined on a closed interval $[a, b]$, we say that:
- $f$ is continuous at the endpoint $a$ if $\lim_{x \to a^+} f(x) = f(a)$
- $f$ is continuous at the endpoint $b$ is $\lim_{x \to b^-} f(x) = f(b)$
- $f$ is continuous on the closed interval $[a, b]$ is $f$ is continuous on the open interval $(a, b)$ and at each of the endpoints $a$ and $b$
The Intermediate Value Theorem
Theorem:
Suppose that $f$ is continuous on the closed interval $[a, b]$. If $z$ lies between $f(a)$ and $f(b)$ then there is at least one real number $c$ in $[a, b]$ such that $f(c) = z$.
Assumptions:
- There may be more than one number $c$ in $[a, b]$ that satisfies the intermediate value theorem
- $f$ must be continuous
- $f$ must be defined over $\mathbb{R}$
The Maximum-Minimum Theorem
Definition:
Given $f$ being defined on a closed interval $[a, b]$:
- A point $c$ in $[a, b]$ is an absolute minimum point for $f$ on $[a, b]$ if $f(c) \le f(x) \forall x$ in $[a, b]$. The corresponding value $f(c)$ is called the absolute minimum value of $f$ on $[a, b]$. If $f$ has an absolute minimum point on $[a, b]$, then we say that $f$ attains its minimum on $[a, b]$.
- A point $d$ in $[a, b]$ is an absolute maximum point for $f$ on $[a, b]$ if $f(x) \le f(d)$ for all $x$ in $[a, b]$. The corresponding value $f(d)$ is called the absolute maximum value of $f$ on $[a, b]$. If $f$ has an absolute maximum point on $[a, b]$, then we say that $f$ attains its maximum on $[a, b]$.
Theorem:
If $f$ is continuous on a closed interval $[a, b]$, then $f$ attains its minimum and maximum on $[a, b]$. Essentially, there exists points $c$ and $d$ in $[a, b]$ such that $f(c) \le f(x) \le f(d)$ for all $x$ in $[a, b]$.
Differentiable Functions
Gradients Of Tangents & Derivatives
Suppose that $f$ is defined on some open interval containing the point $x$. We say that $f$ is differentiable at $x$ if \(\lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}\) exists. If the limit exists, we denote if by $f^\prime(x)$. This is known as the derivative of $f$ at $x$.
Rules For Differentiation
- $(f + g)^\prime(x) = f^\prime(x) + g^\prime(x)$
- $(C.f)^\prime(x) = C.f^\prime(x)$, where $C$ is a constant
- $(fg)^\prime(x) = f^\prime(x) g(x) + f(x) g^\prime(x)$
- $\bigg(\dfrac{f}{g}\bigg)^\prime (x) = \dfrac{f^\prime(x)g(x) - f(x)g^\prime(x)}{g(x)^2}$, given that $g(x) \ne 0$
Rules 3 and 4 are called the product and quotient rules respectively, and can be expressed as \(\dfrac{d(uv)}{dx} = v \dfrac{du}{dx} + u \dfrac{dv}{dx}\) and \(\dfrac{d}{dx} \bigg(\dfrac{u}{v}\bigg) = \dfrac{v \dfrac{du}{dx} - u \dfrac{dv}{dx}}{v^2}\) Where $u$ and $v$ are both functions of $x$.
The Chain Rule
If $y = f(u)$ and $u = g(x)$, then \(\dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx}\)
Implicit Differentiation
Given that a number $q$ is rational, then \(\dfrac{d}{dx} x^q = q x^{q - 1}\)
Differentiation, Continuity & Split Functions
If $f$ is differentiable at $a$, then $f$ is continuous at $a$
If $a$ is a fixed real number and the function $f$ is defined by
\[f(x) = \left\{ \begin{array}{l} p(x) \ \ \ \ \ \ \ if \ \ x \ge a \\ q(x) \ \ \ \ \ \ \ if \ \ x < a \end{array} \right.\]where $p(x)$ and $q(x)$ are continuous and differentiable in some interval containing $a$, then if $f$ is continuous at $a$ and $p^\prime(a) = q^\prime(a)$, then $f$ is differentiable at $x = a$.
Local Maximum, Local Minimum & Stationary Points
Suppose that $f$ is defined on some interval $I$. We say that a point $c$ in $I$ is a local minimum point if there is a positive number $h$ such that $f(x) \le f(x)$ whenever $x \in (c - h, c + h)$ and $x \in I$.
We say that a point $d$ in $(a, b)$ is a local maximum point if there is a positive number $h$ such that $f(x) \le f(d)$ whenever $x \in (d - h, d + h)$ and $c \in I$.
Theorem
Given $f$ is defined on $(a, b)$ and has a local maximum or minimum point $c$ for some $c$ in $(a, b)$. If $f$ is differentiable at $c$ then $f^\prime(c) = 0$.
Stationary Point Definition
If a function $f$ is differentiable at a point $c$ and $f^\prime(c) = 0$, then $c$ is called a stationary point of $f$.
The Mean Value Theorem & Applications
The Mean Value Theorem
Theorem
Suppose that $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$. Then there is at least one real number $c$ in $(a, b)$ such that \(\dfrac{f(b) - f(a)}{b - a} = f'(c)\) Essentially, given two points and their values, there lies a point $c$ in between, where the tangent to $c$ has the same gradient as the line from $a$ to $b$.
The Sign Of A Derivative
Suppose that a function $f$ is defined on an interval $I$. We say that
- $f$ is increasing on $I$ if for every two points $x_1$ and $x_2$ in $I$, $x_1 < x_2$ implies that $f(x_1) < f(x_2)$
- $f$ is decreasing on $I$ if for every two points $x_1$ and $x_2$ in $I$, $x_1 < x_2$ implies that $f(x_1) > f(x_2)$
Theorem
Suppose that $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$
- If $f^\prime(x) > 0$ for all $x$ in $(a, b)$ then $f$ is increasing on $[a, b]$
- If $f^\prime(x) < 0$ for all $x$ in $(a, b)$ then $f$ is decreasing on $[a, b]$
- If $f^\prime(x) = 0$ for all $x$ in $(a, b)$ then $f$ is constant on $[a, b]$
The Second Derivative & Applications
The Second Derivative Test
Suppose that a function $f$ is twice differentiable on $(a, b)$ and that $c \in (a, b)$
- If $f^\prime(c) = 0$ and $f^{\prime\prime}(c) > 0$ then $c$ is a local minimum point of $f$
- If $f^\prime(c) = 0$ and $f^{\prime\prime}(c) < 0$ then $c$ is a local maximum point of $f$
Note:
- If $f^\prime(c) = 0$ and $f^{\prime\prime}(c) = 0$, then the second derivative test cannot be applied. The point $c$ may be a local maximum, minimum or neither
Critical Points, Maxima & Minima
Suppose that $f$ is defined on $[a, b]$. We say that a point $c$ in $[a, b]$ is a critical point for $f$ on $[a, b]$ if $c$ satisfies one of the following properties:
- $c$ is an entpoint $a$ or $b$ on the interval $[a, b]$
- $f$ is not differentiable at $c$
- $f$ is differentiable at $c$ and $f^\prime(c) = 0$
Theorem
If $f$ is continuous on $[a, b]$, then $f$ has a global maximum and a global minimum on $[a, b]$. Also, the global maximum point and the global minimum point are both critial points for $f$ on $[a, b]$.
Antiderivatives
Suppose that $f$ is continuous on an open interval $I$. A function $F$ is said to be an antiderivative of $f$ on $I$ if $F^\prime(x) = f(x)$ for all $x$ in $I$.
Theorem
Suppose that $f$ is a continuous function on an open interval $I$ and that $F$ and $G$ are two antiderivatives of $f$ on $I$. Then there exists a real constant $C$ such that $G(x) = F(x) + C$ for all $x$ in $I$.
Well Known Antiderivatives
Function | Antiderivative |
---|---|
$x^r$, where $r$ is rational and $r \ne -1$ | $\dfrac{1}{r + 1} x^{r + 1} + C$ |
$sin(x)$ | $-cos(x) + C$ |
$cos(x)$ | $sin(x) + C$ |
$e^{ax}$ | $\dfrac{1}{a}e^{ax} + C$ |
$\dfrac{f^\prime(x)}{f(x)}$ | $ln|f(x)| + C$ |
L’Hopital’s Rule
L’Hopital’s Rule
Suppose that $f$ and $g$ are both differentiable functions and $a$ is a real number. Suppose also that either one of the two following conditions hold:
- $f(x) \to 0$ and $g(x) \to 0$ as $x \to 0$
- $f(x) \to \infty$ and $g(x) \to \infty$ as $x \to 0$
If $\lim_{x \to a} \dfrac{f^\prime(x)}{g^\prime(x)}$ exists, then \(\lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a} \dfrac{f'(x)}{g'(x)}\) This theorem also holds for
- Limits as $x \to \infty$ or $x \to - \infty$
- One sided limits ($x \to a^+$ or $x \to a^-$)
Inverse Functions
One To One Functions
A function is one to one if $f(x_1) = f(x_2)$ implies that $x_1 = x_2$ whenevery $x_1, x_2 \in Dom(f)$.
- Every element of the domain has to map to exactly one element in the co-domain
- There can be elements in the co-domain that aren’t mapped to anything in the domain
Inverse Functions
Suppose that a function $f$ is a one to one function. Then the inverse function of $f$ is the unique function $g$ such that $g(f(x)) = x$ and $f(g(x)) = x$, and $Dom(g) = Range(f), \ Range(g) = Dom(f)$. The inverse function for $f$ is often written as $f^{-1}$
The Inverse Function Theorem
Suppose that $I$ is an open interval, $f : I \to \mathbb{R}$ is differentiable, and $f^\prime(x) \ne $ 0for all $x \in I$. Then:
- $f$ is one to one and has an inverse function $g : Range(f) \to Dom(f)$
- $g$ is differentiable at all points in $Range(f)$
- The derivative of $g$ is given by the formula $g^\prime(x) = \dfrac{1}{f^\prime(g(x))}$ for all $x$ in $Range(f)$.
Curve Sketching
Curves Defined By A Cartesian Equation
Checklist For Sketching Curves
- Identify the domain of $f$
- Identify any symmetries
- $f$ is odd if $f(-x) = -f(x)$
- $f$ is even if $f(-x) = f(x)$
- Find $x$ and $y$ axis intercepts
- Identify vertical asymptotes
- Examine the behaviour of $f(x)$ as $x \to \pm \infty$
- Use calculus to identify stationary points and other features if necessary
Oblique Asymptotes
Suppose that $a$ and $b$ are real numbers and that $a \ne 0$. We say that a straight line given by the equation $y = ax + b$ is an oblique asymptote for a function $f$ if $\lim_{x \to \infty} (f(x) - (ax + b)) = 0$.
Essentially, this is saying that given a line $y = ax + b$ and a function $f$, if $f$ approaches $y$ as $x \to \infty$, the line $y$ is known as an oblique asymptote.
Integration
Areas & The Reimann Integral
The Definition Of Area Under The Graph Of A Function & The Reimann Integral
Suppose that $f$ is a bounded function on $[a, b]$ and that $f(x) \ge 0$ for all $x \in [a, b]$. In this subsection we define what is meant by ‘the area under the graph of $f$ from $a$ to $b$’. This is done by constructing upper and lower Reimann sums with respect to partitions of $[a, b]$.
A finite set $P$ of points in $\mathbb{R}$ is said to be a partition of $[a, b]$ if $P = {a_0, a_1, a_2, …, a_n}$ and $a = a_0 < a_1 < a_2 < … < a_n = b$.
Definition of the area under the graph of a function
Suppose that $f$ is bounded on $[a, b]$ and $f(x) \ge 0$ for all $x \in [a, b]$. If there exists a unique real number $A$ such that $\underline{S}_p(f) \le A \le \bar{S}_p(f)$ for every partition $P$ of $[a, b]$, then we say that $A$ is the area under the graph of $f$ from $a$ to $b$.
Definition of the Reimann integral
Suppose that a function $f$ is bounded on $[a, b]$. If there exists a unique real number $I$ such that $\underline{S}_p(f) \le I \le \bar{S}_p(f)$ for every partition $P$ of $[a, b]$, then we say $f$ is Reimann integrable on the interval $[a, b]$. If $f$ is Reimann integrable, then the unique real number $I$ is called the definite integral of $f$ from $a$ to $b$ and we write $I = \int_{a}^{b} f(x) \ dx$.
The function $f$ is called the integrand of the definite integral, while the points $a$ and $b$ are called the limits of the definite integral.