二阶方阵求逆公式

今天在某教材上看到一个二阶方阵的求逆公式 想推导一下
（包不规范的 练练$\LaTeX$

$$\text{If}\space \mathbf{M}= \begin{bmatrix} a & b\\ c & d \end{bmatrix} \text{, then}\space \mathbf{M}^{-1}=\frac{1}{\det \mathbf{M}} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix}$$

$\bold{Proof}\space\bold{1.14.514}$
$\text{Let}$

$$\mathbf{M}= \begin{bmatrix} a & b\\ c & d \end{bmatrix}$$

$\text{we want to find}\space\mathbf{M}^{-1}\space\text{so that}$

$$\mathbf{M}\mathbf{M}^{-1}=\mathbf{I}$$

$\text{Let}$

$$\mathbf{M} \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} x_0\\ y_0 \end{bmatrix} \tag{*}$$

$\text{which is equivalent as}$

$$\begin{cases} ax+by=x_0\\ cx+dy=y_0 \end{cases}$$

$\text{solved as}$

$$\begin{cases} x=\frac{d}{ad-bc}x_0+\frac{-b}{ad-bc}y_0\\ y=\frac{-c}{ad-bc}x_0+\frac{a}{ad-bc}y_0 \end{cases}$$

$\text{so we have}$

$$\begin{aligned} \begin{bmatrix} x\\ y \end{bmatrix} &= \begin{bmatrix} \frac{d}{ad-bc} & \frac{-b}{ad-bc}\\ \frac{-c}{ad-bc} & \frac{a}{ad-bc} \end{bmatrix} \begin{bmatrix} x_0\\ y_0 \end{bmatrix}\\ &= \frac{1}{ad-bc} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} \begin{bmatrix} x_0\\ y_0 \end{bmatrix}\\ &= \frac{1}{\det\mathbf{M}} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} \begin{bmatrix} x_0\\ y_0 \end{bmatrix} \end{aligned}$$

$\text{back to}\space(*)\text{, we have}$

$$\mathbf{M} \frac{1}{\det\mathbf{M}} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} \begin{bmatrix} x_0\\ y_0 \end{bmatrix} = \begin{bmatrix} x_0\\ y_0 \end{bmatrix}$$

$\text{so}$

$$\mathbf{M} \frac{1}{\det\mathbf{M}} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix} = \mathbf{I}$$

$\text{here comes}$

$$\mathbf{M}^{-1}= \frac{1}{\det\mathbf{M}} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix}$$

$\text{Q.E.D.}$