Finish writing up notes on matrices lecture 1

mechanical/mmme1026_maths_for_engineering.md

@@ -124,7 +124,7 @@ always hold true as there are many solutions.
 
 - The exponential function $f(x) = \exp x$ may be wirtten as an infinite series:
 
-  $$\exp x = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... $$
+  $$\exp x = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$
 
 - The function $f(x) = e^{-x}$ is just $\frac 1 {e^x}$
 - Note the important properties:
@@ -391,7 +391,7 @@ $$z^3 = 8i$$
 
 </details>
 
-# Matrices (and Simultaneous Equations)
+# Systems of Equations (Simultaneous Equations)
 
 ## Gaussian Elimination
 
@@ -421,7 +421,7 @@ a = 0, b \ne 0 &\rightarrow \text{no solution for $x$} \\
 a = 0, b = 0 &\rightarrow \text{infinite solutions for $x$}
 \end{align*}
 
-### 2x2 System
+### 2x2 Systems
 
 A 2x2 system is one with 2 equations and 2 unknown variables.
 
@@ -457,7 +457,7 @@ You can check the values for $x_1$ and $x_2$ are correct by substituting them in
 
 </details>
 
-### 3x3 System
+### 3x3 Systems
 
 A 3x3 system is one with 3 equations and 3 unknown variables.
 
@@ -496,3 +496,290 @@ These values can be back-substituted into any of the first 3 equations to find o
 \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>