If $ \begin{vmatrix} 2 & 3+i & -1 \\ 3-i & 0 & i \\ -1 & -i & 1 \end{vmatrix} = A + iB $

where $i= $\sqrt{-1}$$ , then what is A+B equal to?

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This question was previously asked in

NDA-I (Mathematics) Official Paper (Held On: 13 Apr, 2025)

Answer 1

Calculation:

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$

⇒ $di−fg=(3−i)(1)−(i)(−1)=3−i+i=3$

⇒ $dh−eg=(3−i)(−i)−(0)(−1)=−3i+i 2=−3i−1=−1−3i$

⇒ Δ = $2(1)−(3+i)(3)+(−1)(−1−3i)$

⇒ Δ = $2−9−3i+1+3i$

⇒ $Δ=−6+0i$

Since we are given that $$Δ=A+iB$$ comparing the real and imaginary parts, we find:

A = -6 and B = 0

Thus A + B = -6 + 0 = - 6

Hence, the Correct answer is Option 2.

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