Matrix Formulas

$\textbf{A}_{n,m}\textbf{B}_{m,p} = \textbf{C}_{n,p}$	Matrix mult result size; also cols A must equal rows B
$\textbf{M}_{i,j}^T = \textbf{M}_{j,i}$	Matrix transpose
$\textbf{M} = [\textbf{a} \textbf{b} \textbf{c}]$	Column vector representation
$\textbf{M} = \begin{bmatrix} X & a & b \\ a & X & c \\ b & c & X \end{bmatrix}$	Symmetric matrix (X is be any value)
$\textbf{M}^T = \textbf{M}$
$\textbf{M} = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$	Antisymmetric matrix or skew-symmetric matrix (diagonals must be 0)
$\textbf{M}^T = -\textbf{M}$
$\textbf{M}\textbf{v} = v_x \textbf{a} + v_y \textbf{b} + v_z \textbf{c}$	Column vector mult notation (a,b,c cols of M)
$\begin{vmatrix}a & b \\ c & d \end{vmatrix} = ad - bc$	2D Matrix determinant
$det(\textbf{M}) = \sum\limits_{i=0}^{n-1} M_{ik}(-1)^{i+k}\vert\textbf{M}_{\bar{ik}}\vert$	Recursive determinant (each $\vert M_{ik}\vert$ is the sub matrix excluding the current col)
	k is chosen as any row; good when row is mostly zeros
$C_{ij}(\textbf{M}) = (-1)^{i+j}\vert\textbf{M}_{\bar{ij}}\vert$	Cofactor matrix definition
$\textbf{M}^{-1} = \frac{1}{det(\textbf{M})} \textbf{C}^T(\textbf{M})$	Inverse; adjugate matrix
$\textbf{A}^{-1} = \frac{1}{A_{00}A_{11} - A_{01}A_{10}} \begin{bmatrix} A_{11} & -A_{01} \\ -A_{10} & A_{00} \end{bmatrix}$	2D Matrix inverse
$\textbf{B}^{-1} = \frac{1}{det(\textbf{B})} \begin{bmatrix} B_{11}B_{22} - B_{... ...{20} & B_{01}B_{20} - B_{00}B_{21} & B_{00}B_{11} - B_{01}B_{10} \end{bmatrix}$	3D Matrix inverse
$\textbf{M}^{-1} = \frac{1}{[\textbf{a},\textbf{b},\textbf{c}]} \begin{bmatrix}... ...\textbf{c} \times \textbf{a} \\ \textbf{a} \times \textbf{b} \\ \end{bmatrix}$	3D matrix inverse (a,b,c cols of M)
$\textbf{M}^{-1} = \frac{1}{\textbf{s} \cdot \textbf{v} + \textbf{t} \cdot \text... ...\times \textbf{c} - z\textbf{s} \vert \textbf{c} \cdot \textbf{s} \end{bmatrix}$	4D Matrix inverse. a,b,c,d = 3D column vectors of M.
	x,y,z,w = last row of M, x = $a \times b$ , t = $c \times d$ , u = , v =