=z=2 From Eqs. (i), (ii) and (iii), we have −β+α−2=−1 ⇒α−β=1...(iv) −3−β+2α=0 ⇒2α−β=3 ...(v) −α+β+2=1 ⇒−α+β=−1....(vi) Solving Eqs. (iv) and (v), we get α−β−2α+β=1−3 ⇒α=−2 ⇒α=2 On putting α=2 in Eq. (iv), 2−β=1 ⇒β=1 ∴(α,β)=(2,1)