For the circuit shown below, taking the op-amp as ideal, the output voltage V_{out} in terms of the input voltages V_1, V_2 \text { and } V_3 is
(a) 1.8 V_1+7.2 V_2-V_3 (b) 2 V_1+8 V_2-9 V_3
(c) 7.2 V_1+1.8 V_2-V_3 (d) 8 V_1+2 V_2-9 V_3
For the given circuit, at the two nodes we have
\begin{aligned}V_{ A } & =\left\lgroup \frac{4}{5} V_1+\frac{1}{5} V_2 \right\rgroup \\V_{ B } & =V_3 \\V_{\text {out }} & =\square 9 V_{ B }+10 V_{ A }=\square 9 V_3+8 V_1+2 V_2\end{aligned}