Determinant ∆=a(ei−fh)−b(di−fg)+c(dh−eg) Now, For our matrix, a=2,b=3+i,c=−1,d=3−i,e=0,f=i,g=−1,h=−i,i=1 calculate the subdeterminants ⇒ei−fh=(0)(1)−(i)(−i)=0−(−1)=1 ⇒di−fg=(3−i)(1)−(i)(−1)=3−i+i=3 ⇒dh−eg=(3−i)(−i)−(0)(−1)=−3i+i2=−3i−1=−1−3i ⇒∆=2(1)−(3+i)(3)+(−1)(−1−3i) ⇒∆=2−9−3i+1+3i ⇒∆=−6+0i Since we are given that ∆=A+iB comparing the real and imaginary parts, we find: A=−6 and B=0 Thus A+B=−6+0=−6