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NDA 2 2024 Maths, question 9

Topic: Matrices and determinants

Q9

If

Z=13∣i2i12i3i2313∣=x+iy; i=−1Z=\frac{1}{3}\left| \begin{matrix} i & 2i & 1 \\ 2i & 3i & 2 \\ 3 & 1 & 3 \end{matrix} \right|=x+iy;\ i=\sqrt{-1}

then what is modulus of ZZ equal to?

  1. (a) 11
  2. (b) 2\sqrt{2}Correct answer
  3. (c) 22
  4. (d) 3\sqrt{3}

Solution.

Answer: (b) 2\sqrt{2}

  1. Expand the determinant along the first row: i(9i−2)−2i(6i−6)+(2i−9i)=3+3i.
  2. Dividing by 3 gives Z=1+i.
  3. Thus |Z|=√(1²+1²)=√2.