Using the distributive property, which says that for any $a,b,c$,
$$a(b+c)=ab+ac,$$
we can see that
$$3 - 1 \cdot (23 - 7 \cdot 3) = 3+(-1)\cdot(23+(-7)(3))=$$
$$3+(-1)(23)+(-1)(-7)(3)=(-1)(23)+3+(7)(3)=-1\cdot 23 + 8 \cdot 3$$
In our case, we had $a=-1$, $b=23$, and $c=-21=(-7)(3)$.
There is an error in back-substitution going from $3)$ to $4),\,$ namely $\,3 + 1(3) = 2(3),\,$ not $\,4(3).$
I recommend against using the back-substitution version of the extended Euclidean algorithm since it often leads to errors such as above. Instead, using the form described here yields
$$\begin{array}{rrr}
2436 & 1 & 0\\
13 & 0 & 1\\
5 & 1 & -187\\
3 & -2 & 375\\
-1 & \color{#c00}5& \color{#0a0}{-937}\\
\end{array}$$
where each above line $\,\ a\ \ b\ \ c\ \,$ means that $\ a = 2436\, b + 13\, c.\ $ Therefore
$$ -1 \,=\, 2436\cdot \color{#c00}{5}+ 13(\color{#0a0}{-937})\quad $$
Multiplying the above by $\,-1\,$ yields that $\ {\rm mod}\ 2436\!:\ 13\cdot 937\equiv 1,\,$ so $\,13^{-1}\equiv 937$.
The linked post describes the algorithm in great detail, in a way that is easy to remember.
Here is another example computing $\rm\ gcd(141,19),\,$ with the equations written explicitly
$$\rm\begin{eqnarray}(1)\quad \color{#C00}{141}\!\ &=&\,\ \ \ 1&\cdot& 141\, +\ 0&\cdot& 19 \\
(2)\quad\ \color{#C00}{19}\ &=&\,\ \ \ 0&\cdot& 141\, +\ 1&\cdot& 19 \\
\color{#940}{(1)-7\,(2)}\, \rightarrow\, (3)\quad\ \ \ \color{#C00}{ 8}\ &=&\,\ \ \ 1&\cdot& 141\, -\ 7&\cdot& 19 \\
\color{#940}{(2)-2\,(3)}\,\rightarrow\,(4)\quad\ \ \ \color{#C00}{3}\ &=&\, {-}2&\cdot& 141\, + 15&\cdot& 19 \\
\color{#940}{(3)-3\,(4)}\,\rightarrow\,(5)\quad \color{#C00}{{-}1}\ &=&\,\ \ \ 7&\cdot& 141\, -\color{#0A0}{ 52}&\cdot& \color{#0A0}{19} \end{eqnarray}\qquad$$
Best Answer
I think you're confused as to what an inverse is. First, we take your equation
$$-35(4620)+1601(101)=1$$
and now look at this mod $4620$
$$1601(101)\equiv1\pmod{4620}$$
By this equation, $1601$ and $101$ are inverses of one another mod $4620$. When we multiply one by the other, the result is equivalent to $1$.