Calculate the input voltage for this circuit if Vo = -11V
I guess that Vo is 11 V instead of 10 V to mislead you. With the same purpose, Rc is included...
So, I suggest that as your teacher tries to mislead you, ask him/her such a &quot;counter question&quot;: Why is the gain of a non-inverting amplifier one unit greater than that of an inverting amplifier?
Your teacher will have to answer you because his/her job is not only to test you but, above all, to explain circuit phenomena... to make you love circuits... Let's try to guess what the answer would be.
In both circuits, there is a circuit (voltage divider) of two resistors (R1 and R2) in series, through which the same current I flows. It creates voltage drops (VR1 = I.R1 and VR2 = I.R2) across them. As though, these voltage drops are connected by an &quot;electrical transmission&quot; I so V2/R2 = V1/R1 and what is most useful, V2/V1 = R2/R1.
In both circuits, the op-amp adjusts the voltage VR1 equal to the input voltage Vin so the voltage VR2 = Vin.R2/R1 is a scaled copy of the input voltage. If it was used as an output voltage in both circuits, they would have the same gain of R2/R1.
However, there is a problem if we try to connect the load in parallel to R2 because it will divert some of the current. In addition, the load must be grounded. Let's see how these problems are solved in both circuit configurations.
In the inverting amplifier, the op-amp makes a mirror copy of VR2 at its output... and we use it as a (buffered and grounded) output voltage Vout = -VR2 = -Vin.R2/R1.
In the non-inverting amplifier, the op-amp produces a total voltage Vin + VR2 = Vin + Vin.R2/R1 and applies it across the resistor network... and we use this total voltage as a (buffered and grounded) output voltage. So, in this case, the output voltage contains another input voltage;  that is why this &quot;1&quot; appears in the expression.
A few years ago, I asked such a RG question and it provoked a heated discussion.