Inverse Matrix and Function

The inverse of $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ , so $a' = \frac{d}{ad - bc}$ , $b' = \frac{-b}{ad - bc}$ , $c' = \frac{-c}{ad - bc}$ , and $d' = \frac{a}{ad - bc}$ .

Since $f(x) = \frac{ax + b}{cx + d}$ , $x = \frac{af^{-1}(x) + b}{cf^{-1}(x) + d}$ , which solves to $f^{-1}(x) = \frac{dx - b}{-cx + a}$ = $\frac{\frac{d}{ad - bc}x + \frac{-b}{ad - bc}}{\frac{-c}{ad - bc}x + \frac{a}{ad - bc}}$ .

Substituting $a'$ , $b'$ , $c'$ , and $d'$ gives $f^{-1} = \frac{a'x + b'}{c'x + d'}$ , so the statement is true .

We have a group homomorphism $\Big( T \begin{bmatrix} a & b \\ c & d \end{bmatrix}\Big)(x)=\frac{ax+b}{cx+d}$ from $GL_2(\mathbb{R})$ to the group of linear fractional maps over $\mathbb{R}$ . Thus $(T(A))^{-1}=T(A^{-1})$ is $\boxed{True}$ .

Am I correct in suspecting that this holds for all invertible $2 \times 2$ matrices $A$ , not merely those with determinant 1 ?