Determinant Δ=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⇒Δ=2(1)−(3+i)(3)+(−1)(−1−3i)⇒Δ=2−9−3i+1+3i⇒Δ=−6+0iSince we are given that Δ=A+iB comparing the real and imaginary parts, we find:A=−6 and B=0Thus A+B=−6+0=−6