These questions are from Stillwells The four Pillars of Geometry and in the context of linear fractional functions in projective geometry.
Solving for $x$ is straightforward where I get $x= \frac{b-yd}{yc-a}$ by some algebraic manipulation. But I can't spot where my manipulations assumed $ad-bc\ne 0$. What am I missing?
I have a hunch that another way to approach this could connected to the determinant of a matrix being zero and the impact that can have on whether or not an equation can be solved, but i'm not sure about that. I'm also looking to understand a bit more intuitively why $ad-bc\ne 0$ in general for linear fractional functions.
EDIT: Excerpt from the text regarding division by zero
" The linear fractional functions $f(x)= \frac{ax+b}{cx+d}$ we have used to describe projective mappings of lines are actually defective is the variable $x$ only runs through the set $\mathbb{R}$ of real numbers. For example, the function $f(x)=\frac{1}{x}$ we used to map points of the line L1 onto points of the line L2 does not in fact map all points. It cannot set the point $x=0$ anywhere because $\frac{1}{0}$ is undefined; not can it sent any point to $y=0$ because $0 \ne \frac{1}{x}$ for any real x. This defect is neatly fixed by extending the function $f(x)=\frac{1}{x}$ to an new object $x=\infty$ and declaring that $\frac{1}{\infty}=0$ and $\frac{1}{0}=\infty$. The new object $\infty$ is none other than the point at infinity of the line L1, which is supposed to map to the point $0$ on L2. Likewise, id $\frac{1}{0}= \infty$, the point 0 on L1 is sent to the point $\infty$ on L2, as it should be."
You don't show your work, so it is impossible to spot where you used the assumption.
From the givens, $cx+d \neq 0$, so we can multply both sides of the equation by $cx+d$ without destroying all information in the equation. $$ y (cx+d) = ax+b $$ Move $x$-containing terms left and others right. $$ c x y - ax = b - yd $$ Un-distribute $x$s. $$ (c y - a)x = b - yd $$ Now, assuming $cy - a \neq 0$, $$ x = \frac{b - yd}{cy-a} \text{.} $$ However, \begin{align*} cy - a &\neq 0 \\ c \frac{ax+b}{cx+d} - a \frac{cx+d}{cx+d} &\neq 0 \\ \frac{acx+bc - acx - ad}{cx+d} &\neq 0 \\ \frac{bc - ad}{cx+d} &\neq 0 \text{.} \end{align*} We already know $cx+d \neq 0$, so this is defined and is satisfied as long as the numerator on the left-hand side is nonzero, that is, when $bc-ad \neq 0$. Multiplying through by $-1$, when $ad - bc \neq 0$.
So you discover $ad - bc \neq 0$ is equivalent to $cy - a \neq 0$, which is necessary for inverting the relation between $x$ and $y$. So, this map isn't a bijection unless $ad -bc \neq 0$.