Electronic – Negative voltage on both inverting and non-ínverting inputs of a differential amplifier gives non-zero output

Like markrages says (and like you did in your second schematic) you need to bias the non-inverting input. The document says that

"the input impedance = R1||R2 for minimum error due to input bias current"

That's how it should be, but they don't do it! If you leave it floating it will assume \$GND\$ or \$V_{CC}\$, depending on the opamp's design. The 100k\$\Omega\$ value you use is not the right one, though. For optimal bias this should be equal to 10k\$\Omega\$||12k\$\Omega\$ = 5.6k\$\Omega\$. This will affect the frequency response, since your input is now a high-pass RC filter, with

\$ F_C = \dfrac{1}{2 \cdot \pi \cdot R \cdot C} \$

This will also be the response of your amplifier if you add a capacitor in the feedback path:

\$R\$ goes to \$V_{CC}/2\$ instead of \$GND\$.

From a DC point of view R1 and C1 aren't present, so the output bias will be the same as the input bias (voltage follower). This way of biasing is preferable over connecting R1 to \$V_{CC}/2\$, because you would have to take the voltage divider resistors into account to calculate R1. That's unless you use a "hard" \$V_{CC}/2\$ with a low impedance, like from a voltage regulator.

Same goes for R. If you do want to use a voltage divider you can use resistor values equal to 2 \$\cdot\$ R, then you can eliminate R altogether.

Equations:

\$ F_C = \dfrac{1}{2 \cdot \pi \cdot R1 \cdot C1} = \dfrac{1}{2 \cdot \pi \cdot R \cdot C} \$

You're correct; this will output 100 mV as drawn. The reason for this is the amplifier outputs whatever voltage is required to set its inputs equal. Since one input is fixed at 100 mV, the other one must be driven to 100 mV. This happens when the output is at 100 mV. Since every node in the circuit is at 100 mV, no current will flow through the resistors (well, ignoring any current flowing in to the op amp input pins).