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NDA 1 2025 Maths, question 10

Topic: Matrices and determinants

Q10

If ∣23+i−13−i0i−1−1−1−i1∣=A+iB\left| \begin{matrix} 2 & 3 + i & - 1 \\ 3 - i & 0 & i - 1 \\ - 1 & - 1 - i & 1 \end{matrix} \right| = A + iB, where i=−1i = \sqrt{- 1}, then what is A+BA + B equal to?

  1. (a) -10
  2. (b) -6Correct answer
  3. (c) 0
  4. (d) 6

Solution.

Answer: (b) -6

  1. Expanding along the first row, the determinant is 2[−(i−1)(−1−i)]−(3+i)[(3−i)+(i−1)]−[(3−i)(−1−i)]2\left\lbrack - (i - 1)( - 1 - i) \right\rbrack - (3 + i)\left\lbrack (3 - i) + (i - 1) \right\rbrack - \left\lbrack (3 - i)( - 1 - i) \right\rbrack.
  2. The three contributions are −4- 4, −6−2i- 6 - 2i and 4+2i4 + 2i, giving −6- 6.
  3. Hence A=−6A = - 6, B=0B = 0 and A+B=−6A + B = - 6.