Solve ad-bc=2*1-3*1=

$$a d - b c = 2 \cdot 1 - 3 \cdot 1 =$$

Solve for a

$$\left\{\begin{matrix}a=\frac{bc-1}{d}\text{, }&d\neq 0\\a\in \mathrm{R}\text{, }&b=\frac{1}{c}\text{ and }c\neq 0\text{ and }d=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply 2 and 1 to get 2.

$$ad-bc=2-3\times 1$$

Multiply 3 and 1 to get 3.

$$ad-bc=2-3$$

Subtract 3 from 2 to get -1.

$$ad-bc=-1$$

Add bc to both sides.

$$ad=-1+bc$$

The equation is in standard form.

$$da=bc-1$$

Divide both sides by d.

$$\frac{da}{d}=\frac{bc-1}{d}$$

Dividing by d undoes the multiplication by d.

$$a=\frac{bc-1}{d}$$

Solve for b

$$\left\{\begin{matrix}b=\frac{ad+1}{c}\text{, }&c\neq 0\\b\in \mathrm{R}\text{, }&a=-\frac{1}{d}\text{ and }d\neq 0\text{ and }c=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply 2 and 1 to get 2.

$$ad-bc=2-3\times 1$$

Multiply 3 and 1 to get 3.

$$ad-bc=2-3$$

Subtract 3 from 2 to get -1.

$$ad-bc=-1$$

Subtract ad from both sides.

$$-bc=-1-ad$$

The equation is in standard form.

$$\left(-c\right)b=-ad-1$$

Divide both sides by -c.

$$\frac{\left(-c\right)b}{-c}=\frac{-ad-1}{-c}$$

Dividing by -c undoes the multiplication by -c.

$$b=\frac{-ad-1}{-c}$$

Divide -1-ad by -c.

$$b=\frac{ad+1}{c}$$