Let $A$ be an $n \times n$ (square) matrix. $A$ is invertible if there exists $n \times n$ matrix $B$ such that
$$ AB = BA = I_n $$
If it exists, $B$ is unique and we say $B = A^{-1}$
Invertible matrices $A$ and their corresponding transformations $T_A$ must satisfy these conditions
$T_A$ is one-to-one (2 inputs do not map to the same output)
Column vectors of $A$ are linearly independent (from above)
$T_A$ is onto (range is all of the codomain)
Column vectors of $A$ span all of $\mathbb{R}^n$, meaning $\text{span}(\vec{a_1}, \cdots, \vec{a_n}) = \mathbb{R}^n$ (from above)
Reduced echelon form of $A$ has pivot positions in every column (linear independence) and row (range spans codomain)
$A$ is row-equivalent to the identity matrix $I_n$
The determinant of $A$ is non-zero $|A| \neq 0$
Determinant of a 2x2 matrix is
$$ \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc $$
and the inverse is
$$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} $$