$$ \text{row} \overbrace{ \left \{ \left[ \begin{matrix} a_{1,1} & \cdots & a_{1,n} \\ \vdots & \ddots & \vdots \\ a_{n,1} & \cdots & a_{n,n} \end{matrix} \right] \right . }^{\text{column}} $$
$$ \begin{aligned} \alpha \times \left[\begin{matrix} a&b \\ c&d \end{matrix}\right]&= \left[\begin{matrix} \alpha a&\alpha b \\ \alpha c&\alpha d \end{matrix}\right] \\ \left[\begin{matrix} a&b \\ c&d \end{matrix}\right]\pm \left[\begin{matrix} \alpha&\beta \\ \gamma&\delta \end{matrix}\right]&= \left[\begin{matrix} \alpha\pm a&\beta\pm b \\ \gamma\pm c&\delta\pm d \end{matrix}\right] \\ \left[\begin{matrix} a&b \\ c&d \end{matrix}\right] \left[\begin{matrix} \alpha \\ \beta \end{matrix}\right] &= \left[\begin{matrix} a&b \\ c&d \end{matrix}\right] \Large{\{} \normalsize \overbrace{\left[\begin{matrix} \alpha a +\beta b \\ \alpha c + \beta d \end{matrix}\right]}^{ \left[\begin{matrix} \alpha \\ \beta \end{matrix}\right]} \end{aligned} $$
$$ \begin{aligned} \text{Identity: }I&= \left[\begin{matrix} 1&0\\ 0&1 \end{matrix}\right] \text{ such that: }I\times x=x \\ \text{Null: }0&= \left[\begin{matrix} 0&0\\ 0&0 \end{matrix}\right] \text{ such that: }0\times x=0 \\ \text{stretch matrix: }S&= \left[\begin{matrix} k&0\\ 0&k \end{matrix}\right] \text{ such that: }S\times x=k\times x \\ \text{x-axis reflection: }r_x&= \left[\begin{matrix} 1&0\\ 0&-1 \end{matrix}\right] \\ \text{y-axis reflection: }r_y&= \begin{bmatrix} -1&0 \\ 0&1 \end{bmatrix} \\ \text{x=y-axis reflection: }r_{xy}&= \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix} \\ \text{rotation: }R(\theta)&= \left[\begin{matrix} \cos(\theta)&-\sin(\theta)\\ \sin(\theta)&\cos(\theta) \end{matrix}\right] \end{aligned} $$
$A,B\,\&\,C$ are matrices and $I$ is the identity matrix:
$$ (AB)C=A(BC) \\A(B+C)=AB+AC \\A\times B \ne B \times A \\A\times A^{-1}=A^{-1}\times A=I $$
$$ \begin{aligned} \left[\begin{matrix}x\\y\end{matrix}\right]^T &=\left[\begin{matrix}x &y \end{matrix}\right] \\ \begin{bmatrix} a & b \\ c & d \end{bmatrix}^T&=\begin{bmatrix} a&c\\b&d \end{bmatrix} \end{aligned} $$
$$ \begin{bmatrix}a&b \\c&d\end{bmatrix}^{-1} =\frac{1}{a d-b c} \overbrace{\begin{bmatrix}d & -b\\-c & a\end{bmatrix}}^{\text{adjoint matrix}}=\frac{1}{\det} \begin{bmatrix}d & -b\\-c & a\end{bmatrix} \\ \det{\left(\begin{bmatrix}a&b \\c&d\end{bmatrix}\right)}=\left\lvert \begin{bmatrix}a&b \\c&d\end{bmatrix}\right\lvert=ad-bc $$
there is not always an inverse if the determinant = 0 there is no inverse function
give a matrix:
$$ \begin{bmatrix} a_{1,1}&a_{1,2}&a_{1,3} \\ a_{2,1}&a_{2,2}&a_{2,3} \\ a_{3,1}&a_{3,2}&a_{3,3} \end{bmatrix} $$
We can create another matrix, the cofactor matrix where each factor can be defined as:
$$ \begin{aligned} c_{1,1}&=\left\lvert\begin{bmatrix} a_{2,2}&a_{2,3} \\ a_{3,2}&a_{3,3} \end{bmatrix}\right\lvert =a_{2,2}a_{3,3}-a_{2,3}a_{3,2} \\ c_{2,1}&=-\left\lvert\begin{bmatrix} a_{1,2}&a_{1,3} \\ a_{3,2}&a_{3,3} \end{bmatrix}\right\lvert =-(a_{1,2}a_{3,3}-a_{1,3}a_{3,2}) \end{aligned} $$
The cofactors are positive and negative in the following fashion:
$$ \begin{bmatrix} +&-&+ \\ -&+&- \\ +&-&+ \end{bmatrix} $$