Inverting amplifier vs noninverting voltage amplifier
simulate this circuit &ndash; Schematic created using CircuitLab

Here, \$V_{sig}\$ is the signal of interest. It is a high-impedance source, represented by the series resistance \$R_{sig}\$. This resistance will experience a voltage drop proportional to the current through it, by Ohm's law. Thus, if we want to measure what \$V_{sig}\$ is, without introducing error due to \$R_{sig}\$, we need to measure it with a high-impedance load, that is, a load that does not require much current from the signal source.

It's easy to see how OA2, the non-inverting amplifier accomplishes that. One of the rules of the ideal op-amp is that no current flows through the inputs (input impedance is infinite). Current from \$V_{sig}\$ has no other place to go, so there can be no current. If current through \$R_{sig}\$ is zero, by Ohm's law, so too is its voltage.

On the other hand, there is a path for current in the inverting configuration, OA1, through \$R_{sig}\$, \$R_2\$, and \$R_3\$. Another rule of ideal op-amps with negative feedback is that their inputs always have the same voltage. Since the + input is connected to ground, the - is effectively the same. It isn't actually connected to ground, but since our ideal op-amp can sink or source however much current is necessary to keep it at ground potential, we can consider it a virtual ground for the purposes of our analysis.

So, \$V_{sig}\$ is effectively connected to ground through \$R_{sig}\$ and \$R_2\$, forming a voltage divider:







simulate this circuit

The voltage in the middle, the one that would correspond to the input terminal on a non-inverting op-amp that didn't have the source drawn in, isn't \$V_{sig}\$. Rather, it's given by the voltage divider equation:

$$V_{measured} = V_{sig} \frac{R_2}{R_2+R_{sig}}$$

This is the voltage the inverting op-amp amplifies by the textbook gain of \$\frac{-R_3}{R_2}\$. That is:

$$\begin{align} \require{cancel}
V_{out} &amp;= V_{measured} \frac{-R3}{R2}\\
&amp;= V_{sig} \frac{\cancel{R_2}}{R_2+R_{sig}} \frac{-R_3}{\cancel{R_2}}\\
&amp;= V_{sig} \frac{-R_3}{R_2+R_{sig}}
\end{align}$$

If \$R_{sig} \ll R_2\$, then it's not significant. In this case, it's very significant.

Another, perhaps simpler way to think of it is this: \$R_{sig}\$ and \$R_2\$ are in series, so their resistances add. OA1 has no way of knowing that the resistance it sees between its (-) input and \$V_{sig}\$ is two series resistors, and not just one. So effectively, the resistor there is \$20k\Omega\$. The gain of OA1 is effectively:

$$ \frac{-R_3}{R_2+R_{sig}} $$

Which, as you can see, is a term in the equation above.