For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with $ad - bc \neq 0$:

$A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.$

For example, $\begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}^{-1} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$ since $\det = 6 - 5 = 1$.

$\text{diag}(d_1, \ldots, d_n)^{-1} = \text{diag}(1/d_1, \ldots, 1/d_n)$, provided all $d_i \neq 0$. If any $d_i = 0$, the matrix is singular.

$I_n^{-1} = I_n$.

$A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$ is singular: $\det A = 4 - 4 = 0$. Equivalently, the columns $(1, 2)^T$ and $(2, 4)^T$ are linearly dependent, so $A$ has a nontrivial kernel: $A(-2, 1)^T = (0, 0)^T$.

$R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ has inverse $R_{-\theta} = R_\theta^T = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$.

Rotation matrices are orthogonal: $R^T R = I$, so $R^{-1} = R^T$.

An upper triangular matrix is invertible iff all diagonal entries are nonzero. Its inverse is also upper triangular:

$\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & -2/3 \\ 0 & 1/3 \end{pmatrix}.$

Elementary matrices (corresponding to elementary row operations) are invertible:

• Row swap: $(E_{ij})^{-1} = E_{ij}$ (swapping twice returns to original).
• Row scaling by $c \neq 0$: inverse scales by $1/c$.
• Adding $c$ times row $j$ to row $i$: inverse subtracts $c$ times row $j$ from row $i$.

$(AB)^{-1} = B^{-1}A^{-1}$ (order reverses, like taking off socks and shoes).

$(ABC)^{-1} = C^{-1}B^{-1}A^{-1}$.

$(A^T)^{-1} = (A^{-1})^T$. Proof: $A^T (A^{-1})^T = (A^{-1}A)^T = I^T = I$.

$(A^k)^{-1} = (A^{-1})^k$, often written $A^{-k}$.

To find $A^{-1}$, augment $[A \mid I]$ and row-reduce to $[I \mid A^{-1}]$.

Starting from $A = \begin{pmatrix} 1 & 2 \\ 3 & 7 \end{pmatrix}$: apply $R_2 \to R_2 - 3R_1$ to get pivots, then $R_1 \to R_1 - 2R_2$ to reach RREF. The right half of the augmented matrix becomes $A^{-1} = \begin{pmatrix} 7 & -2 \\ -3 & 1 \end{pmatrix}$.

Verify: $\begin{pmatrix} 1 & 2 \\ 3 & 7 \end{pmatrix}\begin{pmatrix} 7 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.

If $A$ is invertible, the unique solution to $Ax = b$ is $x = A^{-1}b$.

$\begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} \implies \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}.$

$\text{GL}_2(\mathbb{R}) = \{A \in M_{2 \times 2}(\mathbb{R}) \mid ad - bc \neq 0\}$. This is an open, dense subset of $M_{2 \times 2}(\mathbb{R}) \cong \mathbb{R}^4$, with the singular matrices forming a hypersurface of measure zero.