notes/mechanical/mmme1026_maths_for_engineering.md

---
author: Alvie Rahman
date: \today
title: MMME1026 // Mathematics for Engineering
tags: [ uni, nottingham, mmme1026, maths, complex_numbers ]
---

# Complex Numbers

## What is a Complex Number?

- $i$ is the unit imaginary number, which is defined by:

  $$ i^2 = -1 $$

- An arbritary complex number is written in the form

  $$z = x + iy$$

  Where:

  - $x$ is the real part of $z$ (which you may seen written as $\Re(z) = x$ or Re$(z) = x$)
  - $y$ is the imaginary part of $z$ (which you may seen written as $\Im(z) = y$ or Im$(z) = y$)

- Two complex numbers are equal if both their real and imaginary parts are equal

  e.g. $$(3 + 4i) + (1 -2i) = (3+1) + i(4-2) = 2 + 2i$$

### The Complex Conjugate

Given complex number $z$:

$$z = z + iy$$

The complex conjugate of z, $\bar z$ is:

$$\bar{z} = z -iy$$

### Division of Complex Numbers

- Multiply numerator and denominator by the conjugate of the denominator

<details>
<summary>

#### Example

</summary>

> \begin{align*}
  z_1 &= 5 + i \\
  z_2 &= 1 -i \\
  \\
  \frac{z_1}{z_2} &= \frac{5+i}{1-i} \cdot \frac{\bar{z_2}}{\bar{z_2}} = \frac{(5+i)(1+i)}{(1-i)(1+i)} \\
    &= \frac{5 + i + 5i -1}{1 + 1} \\
    &= \frac{4 + 6i}{2} = 2 + 3i
> \end{align*}

</details>

### Algebra and Conjugation

When taking complex conjugate of an algebraic expresion, we can replace $i$ by $-i$ before or after
doing the algebraic operations:

\begin{align*}
\overline{z_1+z_2} &= \bar{z_1}+\bar{z_2} \\
\overline{z_1z_2} &= \bar{z_1}\bar{z_2} \\
\overline{z_1/z_2} &= \frac{\bar{z_1}}{\bar{z_2}}
\end{align*}

The conjugate of a real number is the same as that number.

#### Application

If $z$ is a root of the polynomial equation

$$0 = az^2 + bz + c$$

with **real** coefficients $a$, $b$, and $c$, then $\bar{z}$ is also a root because

\begin{align*}
0 &= \overline{az^2 + bz + c} \\
  &= \bar{a}\bar{z}^2 + \bar{b}\bar{z} + \bar{c} \\
  &= a\bar{z}^2 + b\bar{z} + c
\end{align*}

### The Argand Diagram

A general complex number $z = x + iy$ has two components so it can can be represented as a point in
the plane with Cartesion coordinates $(x, y)$.

\begin{align*}
4-2i &\leftrightarrow (4, -2) \\
-i &\leftrightarrow (0, -1) \\
z &\leftrightarrow (x, y) \\
\bar z &\leftrightarrow (x, -y)
\end{align*}

### Plotting on a Polar Graph

We can also describe points in the complex plain with polar coordinates $(r, \theta)$:

\begin{align*}
z &= r(\cos\theta + i\sin\theta) &\text{ polar form of $z$} \\
r &= \sqrt{x^2+y^2} &\text{(modulus)}\\
\theta &= \arg z,\text{ where} \tan \theta = \frac y x &\text{(argument)} \\
x &= r\cos \theta \\
y &= r\sin \theta
\end{align*}

Be careful when turning $(x, y)$ into $(r, \theta)$ form as $\tan^{-1} \frac y x = \theta$ does not
always hold true as there are many solutions.

#### Choosing $\theta$ Correctly

1. Determine which quadrant the point is in (draw a picture).
2. Find a value of $\theta$ such that $\tan \theta = \frac y x$ and check that it is consistent.
   If it puts you in the wrong quadrant, add or subtract $\pi$.

## Exponential Functions

- The exponential function $f(x) = \exp x$ may be wirtten as an infinite series:

  $$\exp x = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$

- The function $f(x) = e^{-x}$ is just $\frac 1 {e^x}$
- Note the important properties:

\begin{align*}
e^{a+b} &= e^a e^b \\
(e^a)^b &= e^{ab} 
\end{align*}

## Euler's Formula

$$e^{i\theta} = \cos\theta + i\sin\theta$$

- Properties of $e^{i\theta}$: For any real angle $\theta$ we have

  $$|e^{i\theta}| = |\cos\theta + i\sin\theta| = \sqrt{\cos^2\theta + \sin^2\theta} = 1$$

  and 

  $$ \arg {e^{i\theta}} = \theta $$

- A complex number in *polar form*, where $r = |z|$, and $\theta = \arg z$, may alternatively be
  written in its *exponential form*:

  $$z = re^{i\theta}$$

  **Note**: $$\bar z = r\cos\theta - ir\sin\theta = re^{-i\theta}$$

<details>
<summary>

### Example 1

Write $z = -1 + i$ in exponential form

</summary>

> $\arg z = \frac {3\pi} 4$
> $|z| = \sqrt 2$
> 
> So $z = \sqrt2e^{i\frac{3\pi} 4}$

</details>

<details>
<summary>

### Example 2

The equations for a mechanical vibration problem are found to have the following mathematical
solution:

$$z(t) = \frac{e^{i\omega t}}{\omega_0^2-\omega^2 + i\gamma}$$

</summary>

where $t$ represents time and $\omega$, $\omega_0$ and $\gamma$ are all positive real physical
constants.
Although $z(t)$
is complex and cannot directly represent a physical solution, it turns out that the real and
imaginary parts $x(t)$ and $y(t)$ in $z(t) = x(t) + iy(t)$ can. Polar notation can be used to extract
this physical information efficiently as follows:

a. Put the denominator in the form

   $$ae^{i\delta}$$

   where you should give explicit expressions for $a$ and $\delta$ in terms of $\gamma$, $\gamma_0$,
   and $\gamma$.


   > \begin{align*}
     a &= \sqrt{\gamma^2 + (\omega_0^2 - \omega^2)^2} \\
     \delta &= \tan^{-1}\frac \delta {\omega_0^2 - \omega^2}
   > \end{align*}

b. Hence find the constants $b$ and $\varphi$ such that

   $$x(t) = b\cos(\omega t + \varphi)$$

   and write a similar expression for $y(t)$.

   > \begin{align*}
     z &= \frac{e^i\omega t}{ae^{i\delta}} = \frac 1 a e^{i (\omega t - \delta)} \\
     x + iy &= \frac 1 a \cos(\omega t - \delta) + \frac 1 a \sin(\omega t - \delta) \\
     \therefore \Re z &= x = \frac 1 a \cos(\omega t - \delta), \\
     \Im z &= y = \frac 1 a \sin(\omega t - \delta) \\
     \\
     b &= \frac 1 a = \frac 1 {\sqrt{\gamma^2 + (\omega_0^2 - \omega^2)^2}} \\
     \varphi &= -\delta = \tan^{-1}\frac \delta {\omega_0^2 - \omega^2}\\
     \\
     y(t) &= \frac 1 a \sin(\omega t - \delta) \\
   > \end{align*}

</details>

## Products of Complex Numbers

Suppose we have 2 complex numbers:

$$z_1 = x_1 + iy_1 = r_1e^{i\theta_1}$$  
$$z_2 = x_2 + iy_2 = r_2e^{i\theta_2}$$

Using $e^a e^b = e^{a+b}$, the product is:

\begin{align*}
z_3 = z_1 z_2 &= (r_1e^{i\theta_1})(r_2e^{i\theta_2}) \\
&= r_1r_2e^{i\theta_1}e^{i\theta_2} \\
&= r_1r_2e^{i(\theta_1+\theta_2)} \\
\\
|z_1z_2| &= |z_1|\times|z_2| \\
\arg z_1z_2 &= \arg z_1 \times \arg z_2
\end{align*}

## de Moivre's Theorem

Let $z = re^{i\theta}$. Consider $z^n$.

Since $z = r(\cos\theta + i\sin\theta)$,
\begin{align*}
z^n &= r^n(\cos\theta + i\sin\theta)^n &\text{(1)} \\
\end{align*}

But also

\begin{align*}
z^n &= (re^{i\theta})^n \\
    &= r^n(e^{i\theta})^n \\
    &= r^ne^{in\theta} \\
    &= r^n(\cos{n\theta} + i\sin{n\theta}) &\text{(2)} \\
\end{align*}

By equating (1) and (2), we find de Moivre's theorem:

\begin{align*}
r^n(\cos\theta +i\sin\theta)^n &= r^n(\cos{n\theta} + i\sin{n\theta}) \\
(\cos\theta +i\sin\theta)^n &= (\cos{n\theta} + i\sin{n\theta})
\end{align*}

<details>
<summary>

### Example 1

Write $1+i$ in polar form and use de Moivre's theorem to calculate $(1+i)^{15}$.

</summary>

> \begin{align*}
  r &= |1+i| = \sqrt2 \\
  \theta &= \arg{1+i} = \frac \pi 4 \\
  \\
  \text{So } 1 + i &= \sqrt2(\cos{\frac pi 4} + i\sin{\frac \pi 4}) = \sqrt2e^{i\frac \pi 4} \text{ and}\\
  (i+i)^{15} &= (\sqrt2)^{15}\left(\cos{\frac \pi 4} + i\sin{\frac \pi 4}\right)^{15} \\
             &= 2^{\frac 15 2} \left(\cos{\frac{15\pi} 4} + i\sin{\frac{15\pi} 4}\right) \\
             &= 2^{\frac 15 2} \left(\frac 1 {\sqrt2} - \frac i {\sqrt2}\right) \\
             &= 2^7 (1 - i) \\
             &= 128 - 128i
> \end{align*}
</details>

<details>
<summary>

### Example 2

Use de Moivre's theorem to show that

\begin{align*}
\cos{2\theta} &= \cos^2\theta-\sin^2\theta \\
\text{and} \\
\sin{2\theta} &= 2\sin\theta\cos\theta
\end{align*}

</summary>

> Let $n=2$:

> \begin{align*}
  (\cos\theta+i\sin\theta)^2 &= \cos^2\theta + 2i\sin\theta\cos\theta - \sin^2\theta \\
  \text{Real part: } \cos^2\theta - \sin^2\theta &= \cos{2\theta}\\
  \text{Imaginary part: } 2\sin\theta\cos\theta &= \sin{2\theta}
> \end{align*}

</details>

<details>
<summary>

### Example 3

Given that $n \in \mathbb{N}$ and $\omega = -1 + i$, show that
$w^n + \bar{w}^n = 2^{\frac n 2 + 1}\cos{\frac{3n\pi} 4}$ with Euler's formula.

</summary>

> \begin{align*}
  r &= \sqrt{2} \\
  \arg \omega = \theta &= \frac 3 4 \pi \\
  \\
  \omega^n &= r^n(cos{n\theta} + i\sin{n\theta}) \\
  \bar\omega^n &= r^n(cos{n\theta} - i\sin{n\theta}) \\
  \omega^n + \bar\omega^n &= r^n(2\cos{n\theta}) \\
                          &= 2^{\frac n 2 + 1}\cos{\frac {3n\pi} 4}
> \end{align*}

</details>

## Complex Roots of Polynomials

<details>
<summary>

### Example

Find which complex numbers $z$ satisfy

$$z^3 = 8i$$

</summary>

> 1. Write $8i$ in exponential form,
> 
>    $|8i| = 8$ and $\arg{8i} = \frac \pi 2$
> 
>    $\therefore 8i = 8e^{i\frac \pi 2}$
> 
> 
> 2. Let the solution be $r = re^{i\theta}$.
> 
>    Then $z^3 = r^3e^{3i\theta}$.
> 
> 3. $z^3 = r^3e^{3i\theta} = 8e^{i\frac \pi 2}$
> 
>    i. Compare modulus:
> 
>       $r^3 = 8 \rightarrow r = 2$
> 
>    ii. Compare argument:
> 
>        $$3\theta = \frac \pi 2$$
> 
>        is a solution but there are others since 
> 
>        $$e^{i\frac \pi 2} = e^{i \frac \pi 2 + 2n\pi}$$
> 
>        so we get a solution whenever
> 
>        $$3\theta = \frac \pi 2 + 2n\pi$$
> 
>        for any integer `n`
> 
>        - $n = 0 \rightarrow z = \sqrt3 + i$
>        - $n = 1 \rightarrow z = -\sqrt3 + i$
>        - $n = 2 \rightarrow z = -2i$
>        - $n = 3 \rightarrow z = \sqrt3 + i$
>        - $n = 4 \rightarrow z = -\sqrt3 + i$
>        - The solutions start repeating as you can see
> 
>        In general, an $n$-th order polynomial has exactly $n$ complex roots.
>        Some of these complex roots may be real numbers.
> 
> 4. There are three solutions

</details>

# Systems of Equations (Simultaneous Equations)

## Gaussian Elimination

Gaussian eliminiation can be used when the number of unknown variables you have is equal to the
number of equations you are given.

I'm pretty sure it's the name for the method you use to solve simultaneous equations in school.

For example if you have 1 equation and 1 unknown:

\begin{align*}
2x &= 6 \\
x &= 3
\end{align*}

### Number of Solutions

Let's generalise the example above to

$$ax = b$$

There are 3 possible cases:

\begin{align*}
a \ne 0 &\rightarrow x = \frac b a \\
a = 0, b \ne 0 &\rightarrow \text{no solution for $x$} \\
a = 0, b = 0 &\rightarrow \text{infinite solutions for $x$}
\end{align*}

### 2x2 Systems

A 2x2 system is one with 2 equations and 2 unknown variables.

<details>
<summary>

#### Example 1

\begin{align*}
3x_1 + 4x_2 &= 2 &\text{(1)} \\
x_1 + 2x_2 &= 0 &\text{(2)} \\
\end{align*}

</summary>

\begin{align*}
3\times\text{(2)} = 3x_1 + 6x_2 &= 0 &\text{(3)} \\
\text{(3)} - \text{(1)} = 0x_1 + 2x_2 &= -2 \\
x_2 &= -1
\end{align*}

We've essentially created a 1x1 system for $x_2$ and now that's solved we can back substitute it
into equation (1) (or equation (2), it doesn't matter) to work out the value of $x_1$:

\begin{align*}
3x_1 + 4x_2 &= 2 \\
3x_1 - 1 &= 2 \\
3x_1 &= 6 \\
x_1 &= 2
\end{align*}

You can check the values for $x_1$ and $x_2$ are correct by substituting them into equation (2).

</details>

### 3x3 Systems

A 3x3 system is one with 3 equations and 3 unknown variables.

<details>
<summary>

#### Example 1

\begin{align*}
2x_1 + 3x_2 - x_3 &= 5 &\text{(1)} \\
4x_1 + 4x_2 - 3x_3 &= 5 &\text{(2)} \\
2x_1 - 3x_2 + x_3 &= 5 &\text{(3)} \\
\end{align*}

</summary>

The first step is to eliminate $x_1$ from (2) and (3) using (1):

\begin{align*}
\text{(2)}-2\times\text{(1)} = -2x_2 -x_3 &= -1 &\text{(4)} \\
\text{(3)}-\text{(1)} = -6x_2 + 3x_3 &= -6 &\text{(5)} \\
\end{align*}

This has created a 2x2 system of $x_2$ and $x_3$ which can be solved as any other 2x2 system.
I'm too lazy to type up the working, but it is solved like any other 2x2 system.

\begin{align*}
x_2 &= -2
x_3 &= 5
\end{align*}

These values can be back-substituted into any of the first 3 equations to find out $x_1$:

\begin{align*}
-2x_1 + 3x_2 - x_3 = 2x_1 + 6 - 3 = 5 \rightarrow x_1 = 1
\end{align*}

</details>

<details>
<summary>

#### Example 2

\begin{align*}
x_1 + x_2 - 2x_3 &= 1 &R_1 \\
2x_1 - x_2 - x_3 &= 1 &R_2 \\
x_1 + 4x_2 + 7x_3 &= 2 &R_3 \\
\end{align*}

</summary>

1. Eliminate $x_1$ from $R_2$, $R_3$:

   \begin{align*}
   x_1 + x_2 - 2x_3 &= 1 &R_1' = R_1\\
   - 3x_2 - 5x_3 &= -1 &R_2' = R_2 - 2R_1 \\
   3x_2 + 5x_3 &= 1 &R_3' = R_3 - R_1 \\
   \end{align*}

   We've created another 2x2 system of $R_2'$ and $R_3'$

2. Eliminate $x_2$ from $R_3''$

   \begin{align*}
   x_1 + x_2 - 2x_3 &= 1 &R_1'' = R_1' = R_1\\
   - 3x_2 - 5x_3 &= -1 &R_2'' = R_2' = R_2 - 2R_1 \\
   0x_3 &= 0 &R_3'' = R_3 '+ R_2' \\
   \end{align*}

   We can see that $x_3$ can be any number, so there are infinite solutions. Let:

   $$x_3 = t$$

   where $t$ can be any number

3. Substitute $x_3$ into $R_2''$:

   $$R_2'' = -3x_2 - 5t = -1 \rightarrow x_2 = \frac 1 3 - \frac{5t} 3$$

4. Substitute $x_2$ and $x_3$ into $R_1''$:

   $$R_1'' = x_1 + \frac 1 3 - \frac{5t} 3 + 2t = 1 \rightarrow x_1 = \frac 2 3 - \frac t 3$$

</details>

## Systems of Equations and Matrices

Many problems in engineering have a very large number of unknowns and equations to solve
simultaneously.
We can use matrices to solve these efficiently.

Take the following simultaneous equations::

\begin{align*}
3x_1 + 4x_2 &= 2 &\text{(1)} \\
x_1 + 2x_2 &= 0 &\text{(2)}
\end{align*}

They can be represented by the following matrices:

\begin{align*}
A &= \begin{pmatrix} 3 & 4 \\ 1 & 2 \end{pmatrix} \\
\pmb x  &= \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \\
\pmb b  &= \begin{pmatrix} 2 \\ 0 \end{pmatrix} \\
\end{align*}

You can then express the system as:

$$A\pmb x = \pmb b$$

<details>
<summary>

#### A 3x3 System as a Matrix

</summary>

\begin{align*}
2x_1 + 3x_2 - x_3 &= 5 \\
4x_1 + 4x_2 - 3x_3 &= 3 \\
2x_1 - 3x_2 + x_3 &= -1
\end{align*}

Could be expressed in the form $A\pmb x = \pmb b$ where:

\begin{align*}
A &= \begin{pmatrix} 2 & 3 & -1 \\ 4 & 4 & -3 \\ 2 & -3 & -1 \end{pmatrix} \\
\pmb x  &= \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \\
\pmb b  &= \begin{pmatrix} 5 \\ 3 \\ -1 \end{pmatrix} \\
\end{align*}

</details>

<details>
<summary>

#### An $m\times n$ System as a Matrix

</summary>

\begin{align*}
a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n &= b_1 \\
a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n &= b_2 \\
\cdots \\
a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n &= b_m \\
\end{align*}

Could be expressed in the form $A\pmb x = \pmb b$ where:

\begin{align*}
A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\
                a_{21} & a_{22} & \cdots & a_{2n} \\
                \vdots &        &     & \vdots \\
                a_{m1} & a_{m2} & \cdots & a_{mn} 
                \end{pmatrix}, 
\pmb x  = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}, 
\pmb b  = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix}
\end{align*}

</details>

# Matrices

## Order of a Matrix

The order of a matrix is its size e.g. $3\times2$ or $m\times n$

## Column Vectors

- Column vectors are matrices with only one column:

  $$ \begin{pmatrix} 1 \\ 2 \end{pmatrix} \begin{pmatrix} 4 \\ 45 \\ 12 \end{pmatrix} $$

- Column vector variables typed up or printed are expressed in $\pmb{bold}$ and when it is
  handwritten it is \underline{underlined}:

  $$ \pmb x = \begin{pmatrix} -3 \\ 2 \end{pmatrix}$$

## Matrix Algebra

### Equality

Two matrices are the same if:

- Their order is the same
- Their corresponding elements are the same

### Addition and Subtraction

Only possible if their order is the same.
\begin{align*}
A + B&= \begin{pmatrix} a_{11} + b_{11} & a_{12} + b_{12} & \cdots & a_{1n} + b_{1n} \\
                a_{21} + b_{21} & a_{22} + b_{22} & \cdots & a_{2n} + b_{2n} \\
                \vdots &        &     & \vdots \\
                a_{m1} + b_{m1} & a_{m2} + b_{m2} & \cdots & a_{mn} + b_{mn}
                \end{pmatrix} \\
A - B&= \begin{pmatrix} a_{11} - b_{11} & a_{12} - b_{12} & \cdots & a_{1n} - b_{1n} \\
                a_{21} - b_{21} & a_{22} - b_{22} & \cdots & a_{2n} - b_{2n} \\
                \vdots &        &     & \vdots \\
                a_{m1} - b_{m1} & a_{m2} - b_{m2} & \cdots & a_{mn} - b_{mn}
                \end{pmatrix}, 
\end{align*}

### Zero Matrix

This is a matrix whose elements are all zeros.
For any matrix $A$,

$$A + 0 =A$$

We can only add matrices of the same order, therefore 0 must be of the same order as $A$.

### Multiplication

Let

$$
\begin{matrix}
A & m\times n \\
B & p\times q
\end{matrix}
$$

To be able to multiply $A$ by $B$, $n = p$.

If $n \ne p$, then $AB$ does not exist.

$$
\begin{matrix}
A & B & = & C \\
m\times n & p \times q & & m\times q
\end{matrix}
$$

When $C = AB$ exists,

$$C_{ij} = \sum_r\! a_{ir}b_{rj}$$

That is, $C_{ij}$ is the 'product' of the $i$th row of $A$ and $j$th column of $B$.

#### Multiplication of a Matrix by a Scalar

If $\lambda$ is a scalar, we define

$$
\lambda a = \begin{pmatrix} \lambda a_{11} & \lambda a_{12} & \cdots & \lambda a_{1n} \\
                \lambda a_{21} & \lambda a_{22} & \cdots & \lambda a_{2n} \\
                \vdots &        &     & \vdots \\
                \lambda a_{m1} & \lambda a_{m2} & \cdots & \lambda a_{mn} 
                \end{pmatrix}, 
$$

<details>
<summary>

#### Example 1

</summary>

$$
\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix}
\begin{pmatrix} 0 & 1 \\ 3 & 2 \end{pmatrix} = 
\begin{pmatrix} -3 & -1 \\ 3 & 4 \end{pmatrix}
$$
$$
\begin{pmatrix} 0 & 1 \\ 3 & 2 \end{pmatrix}
\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix} =
\begin{pmatrix} 2 & 1 \\ 7 & -1 \end{pmatrix}
$$

</details>

<details>
<summary>

#### Example 2

</summary>

$$
A = \begin{pmatrix} 4 & 1 & 6 \\ 3 & 2 & 1 \end{pmatrix},\,
B = \begin{pmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 0 \end{pmatrix}
$$
$$
AB = \begin{pmatrix} 11 & 6 \\ 6 & 7 \end{pmatrix},\,
BA = \begin{pmatrix} 7 & 3 & 7 \\ 10 & 5 & 8 \\ 4 & 1 & 6 \end{pmatrix}
$$

</details>

### Other Properties of Matrix Algebra

- $(\lambda A)B = \lambda(AB) = A(\lambda B)$
- $A(BC) = (AB)C = ABC$
- $(A+B)C = AC + BC$
- $C(A+B) = CA + CB$
- In general, $AB \ne BA$ even if both exist
- $AB = 0$ does not always mean $A = 0$ or $B = 0$:

  $$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix}3 & 0 \\ 0 & 0 \end{pmatrix} = 
  \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = 0$$

  <details>
  <summary>

  It follows that $AB = AC$ does not imply that $B=C$ as

  $$AB = AC \leftrightarrow A(B + C) = 0$$

  and as $A$ and $(B-C)$ are not necessarily 0, $B$ is not necessarily equal to $C$:

  </summary>

  $$AB = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix} = 
  \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$$

  and

  $$AC = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix}1 & 2 \\ 1 & 0 \end{pmatrix} = 
  \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = AB$$

  but $B \ne C$

  </details>