\begin{theorem}[Matrix Properties]\ \\
  Let $A, B, C$ be $m \times p, p \times q, q \times n$ matrices
  \begin{itemize}
    \item Associative Law: $A(BC) = (AB)C$
    \item Distributive Law: $A(B_1 + B_2) = AB_1 + AB_2$
    \item Distributive Law: $(B_1 + B_2)A = B_1A + B_2A$
    \item $c(AB) = (cA)B = A(cB)$
    \item $A\textbf{0}_{p \times n} = \textbf{0}_{m \times n}$
    \item $A\textbf{I}_{n} = \textbf{I}_{n}A = A$
  \end{itemize}
\end{theorem}


\begin{defn}[Powers of Square Matricss]\ \\
  Let $A$ be a $m \times n$. 

  $AA$ is well defined $\iff m = n \iff A$ is square.

  \textbf{Definition.} Let $A$ be square matrix of order $n$. Then Powers of a are
  $$
  A^k = \begin{cases}
    I_n & \text{if } k = 0 \\
    AA...A & \text{if } k \geq 1.
  \end{cases}
  $$

  \textbf{Properties.}
  \begin{itemize}
    \item $A^mA^n = A^{m+n}, (A^m)^n = A^{mn}$
    \item $(AB)^2 = (AB)(AB) \neq A^2B^2 = (AA)(BB)$
  \end{itemize}
\end{defn}

Matrix Multiplication Example: 

\begin{itemize}
  \item Let $A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ 2 & 3 \\ -1 & -2 \end{pmatrix}$
  \item Let $a_1 = \begin{pmatrix}1 & 2 & 3 \end{pmatrix}, a_2 = \begin{pmatrix}4 & 5 & 6 \end{pmatrix}$
  \item $AB = \begin{pmatrix} a_1 & a_2 \end{pmatrix}B = \begin{pmatrix} a_1B \\ a_2B \end{pmatrix}$.
  \item $\begin{pmatrix}
      \begin{pmatrix}1 & 2 & 3 \end{pmatrix} & \begin{pmatrix} 1 & 1 \\ 2 & 3 \\ -1 & -2 \end{pmatrix} \\
      \begin{pmatrix}4 & 5 & 6 \end{pmatrix} & \begin{pmatrix} 1 & 1 \\ 2 & 3 \\ -1 & -2 \end{pmatrix}
      \end{pmatrix} = \begin{pmatrix}
      \begin{pmatrix}2 & 1\end{pmatrix} \\
      \begin{pmatrix}8 & 7\end{pmatrix} \\
    \end{pmatrix}
    $
\end{itemize}

\begin{note}[Representation of Linear System] \ \\
  \begin{itemize}
    \item $\begin{cases}
        a_{11}x_1 + a_{12}x_2 + ... + a_{1n}x_n & = b_1 \\
        a_{21}x_1 + a_{22}x_2 + ... + a_{2n}x_n & = b_2 \\
        \vdots & \vdots \\
        a_{m1}x_1 + a_{m2}x_2 + ... + a_{mn}x_n & = b_m \\
      \end{cases}$

    \item A = $\begin{pmatrix}
        a_{11} & a_{12} & ... & a_{1n} \\
        a_{21} & a_{22} & ... & a_{2n} \\
        \vdots & \vdots & & \vdots \\
        a_{m1} & a_{m2} & ... & a_{mn} \\
      \end{pmatrix}$, Coefficient Matrix, $A_{m\times n}$
    \item $x = \begin{pmatrix}
        x_{1}  \\
        \vdots \\
        x_{n}  \\
      \end{pmatrix}$, Variable Matrix, $x_{n \times 1}$
    \item $b = \begin{pmatrix}
        b_{1}  \\
        \vdots \\
        b_{m}  \\
      \end{pmatrix}$, Constant Matrix, $b_{m \times 1}$. Then $Ax = b$
    \item $A = (a_{ij})_{m\times n} $
    \item $m$ linear equations in $n$ variables, $x_1, ..., x_n$
    \item $a_{ij}$ are coefficients, $b_i$ are the constants
    \item Let $u = \begin{pmatrix} u_1 \\ \vdots \\ u_n \end{pmatrix}$.
      \subitem $x_1 = u_1, \hdots, x_n = u_n$ is a solution to the system
      \subitem $\iff Au = b \iff u$ is a solution to $Ax = b$
    \item Let $a_j$ denote the $j$th column of $A$. Then
      \subitem $b = Ax = x_1a_1 + ... + x_na_n = \sum^n_{j=1}x_ja_j$
  \end{itemize}
\end{note}

\begin{defn}[Transpose]\ \\
  \begin{itemize}
    \item Let $A = (a_{ij})_{m\times n}$
    \item The transpose of $A$ is $A^T = (a_{ji})_{n \times m}$
    \item $(A^T)^T = A$
    \item A is symmetric $\iff A = A^T$
    \item Let $B$ be $m \times n$, $(A+B)^T = A^T + B^T$
    \item Let $B$ be $n \times p$, $(AB)^T = B^TA^T$
  \end{itemize}
\end{defn}

\begin{defn}[Inverse]\ \\
  \begin{itemize}
    \item Let $A, B$ be matrices of same size
      \subitem $A + X = B \implies X = B - A = B + (-A)$
      \subitem $-A$ is the \textit{additive inverse} of $A$
    \item Let $A_{m\times n}, B_{m\times p}$ matrix.
      \subitem $AX = B \implies X = A^{-1}B$.
  \end{itemize}


  Let A be a \textbf{square matrix} of order $n$. 
  \begin{itemize}
    \item If there exists a square matrix $B$ of order $N$ s.t. $AB = I_{n}$ and $BA = I_{n}$, then $A$ is \textbf{invertible} matrix and $B$ is inverse of $A$.
    \item If $A$ is not invertible, A is called singular.
    \item suppose $A$ is invertible with inverse $B$
    \item Let $C$ be any matrix having the same number of rows as $A$.
      $$\begin{aligned}
        AX = C &\implies B(AX) = BC \\
               &\implies (BA)X = BC \\
               &\implies X = BC.
      \end{aligned}$$
  \end{itemize}

  
\end{defn}
