Assuming that the op-amp in the circuit shown is ideal,

The output voltage $V_{0}$ is

\frac{5}{2} V_{1} - 3 V_{2}

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2 V_{1} - \frac{5}{2} V_{2}

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- 3 V_{1} + \frac{11}{2} V_{2}

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\frac{- 3}{2} V_{1} + \frac{7}{2} V_{2}

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Solution

The correct option is C $- 3 V_{1} + \frac{11}{2} V_{2}$
By applying KCL at node 1 we get

$\frac{V_{2} - V_{1}}{R} + \frac{V_{2}}{2 R} + \frac{V_{2} - V_{0}}{3 R} = 0$

$\frac{6 V_{2} - 6 V_{1} + 3 V_{2} + 2 V_{2} - 2 V_{0}}{6 R} = 0$

$11 V_{2} - 6 V_{1} - 2 V_{0} = 0$

$V_{0} = - 3 V_{1} + \frac{11}{2} V_{2}$

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