Step 3: Multiply by the scalar (1 over the
determinant):

Let's call the number "n"

if we write in an equation what you said it's like this:

n*2+4-n-3=n+1.

first i will move all the "n"s to the left and all the numbers to the right

n*2-n-n=1+3-4

which is

0=0

this means that for EVERY number, if you double it, add four, subtract three, you will get this number plus one.