0
$\begingroup$

A math teacher taught a shortcut for calculating the inverse or the derivative of a linear rational function of the form.

$$ R(x) = \frac{ax+b}{cx+d} $$

By first writing it in a matrix form,

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

Then, the inverse of linear function is given by,

$$ adj (A) $$

and the derivative is given by

$$ \frac{det(A)}{(cx+d)^2}$$

Why is that? How did he come up with those formulas? I am having trouble trying to find the reference, or source of this. I know that they are true, but I'd like to know how he this was found, especially, the inverse formula.

$\endgroup$
  • $\begingroup$ just apply quotient rule $\endgroup$ – JJR May 29 '17 at 15:34
  • $\begingroup$ Well, yeah I know, but what about the inverse? $\endgroup$ – user3855929 May 29 '17 at 15:40
0
$\begingroup$

I don't know how the teacher came up with those formulas but I can prove them.

First, if $$y=\frac{ax+b}{cx+d}$$ then $$x=\frac{dy-b}{-cy+a}.$$ So, the corresponding matrix is

$$\begin{bmatrix}d&-b\\-c&a\end{bmatrix}.$$ At the same time

$$\operatorname{adj}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}d&-c\\-b&a\end{bmatrix}^T=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}.$$

Second, the derivative of $y$ is

$$y'=\frac{da-bc}{(cx+d)^2}$$

and the determinant of $A$ is exactly $da-bc$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.