NOTES:

• $AB$ is Matrix $A$ times every column of $B$: $$AB = A \left( \begin{smallmatrix} \mathbf{b_1} \, \mathbf{b_2} \, \mathbf{b_3} \end{smallmatrix} \right) = \left( \begin{smallmatrix} A\mathbf{b_1} \, A \mathbf{b_2} \, A \mathbf{b_3} \end{smallmatrix} \right) $$
• $AB$ is every row of $A$ times Matrix $B$: $$ AB = \left( \begin{smallmatrix} \mathbf{a'_1} \\ \mathbf{a'_2} \\ \mathbf{a'_3} \end{smallmatrix} \right) B = \left( \begin{smallmatrix} \mathbf{a'_1}B \\ \mathbf{a'_2}B \\ \mathbf{a'_3}B \end{smallmatrix} \right) $$
• $AB$ is a summation of $n$ matrices, each of which is attributed to column $i$ of Matrix $A$ times row $i$ of Matrix $B$, where $i$ ranges from $1$ to $n$. $$ AB = \left( \begin{smallmatrix} \mathbf{a_1} \, \mathbf{a_2} \, \mathbf{a_3} \end{smallmatrix} \right) \left( \begin{smallmatrix} \mathbf{b'_1} \\ \mathbf{b'_2} \\ \mathbf{b'_3} \end{smallmatrix} \right) = \mathbf{a_1} \mathbf{b'_1} + \mathbf{a_2} \mathbf{b'_2} + \mathbf{a_3} \mathbf{b'_3} $$
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• $AB_{ij}$, the entry in row $i$ and column $j $ of $AB$ is row $i$ of $A$ times column $j$ of $B$. $$ \left( \begin{matrix} & * & \\ * \, * & AB_{ij} & * \, * \\ & * & \\ & * & \end{matrix} \right) = \left( \begin{matrix} * \\ a'_i \\ * \\ * \end{matrix} \right) \left( \begin{matrix} * \, * \, b_j \, * \end{matrix} \right) $$
• Block elimination: Given a block Matrix $ E = \left( \begin{smallmatrix} A \, B \\ C \, D \end{smallmatrix} \right) $, $A$ is the pivot block in row $1$. The elimination would be: $$ \left( \begin{matrix} I & \mathbf{0} \\ -CA^{-1} & I \end{matrix} \right) \left( \begin{matrix} A & B \\ C & D \end{matrix} \right) = \left( \begin{matrix} A & B \\ \mathbf{0} & D - CA^{-1}B \end{matrix} \right) $$
• Laws of Matrix Addition:
  • Commutative Law: $A + B = B + A$
  • Distributive Law: $c ( A + B) = cA + cB$
  • Associative Law: $ A+ (B + C) = (A + B) + C$
• Laws of Matrix Multiplication:
  • Distributive Law from the left: $A(B + C) = AB + AC $
  • Distributive Law from the right: $(A + B) C = AC + BC$
  • Associative Law for $ABC$: $A(BC) = (AB)C$
  • $AB \ne BA$: the Commutative Law is usually broken