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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Determinants

NDA 2024 · Mathematics · Determinants

If Z=(1)/(3)| array ccc i & 2 i & 1 2 i & 3 i & 2 3 & 1 & 3 array...

If \mathrm{Z}=\frac{1}{3}\left|i2i12i3i2313\begin{array}{ccc}i & 2 i & 1 \\ 2 i & 3 i & 2 \\ 3 & 1 & 3\end{array}\right|=x+i y ; i=\sqrt{-1} then what is modulus of Z equal to?

Options

  1. A11
  2. B2\sqrt{2}
  3. C22
  4. D3\sqrt{3}

Correct answer

B. 2\sqrt{2}

Step-by-step solution

We have given, \mathrm{Z}=\frac{1}{3}\left|i2i12i3i2313\begin{array}{ccc}\mathrm{i} & 2 \mathrm{i} & 1 \\ 2 \mathrm{i} & 3 \mathrm{i} & 2 \\ 3 & 1 & 3\end{array}\right|⇒Z=13[i(9i−2)−2i(6i−6)+1(2i−9i)]⇒Z=13(3+3i)=1+i⇒ Modulus of  z=12+12=2\begin{aligned} & \Rightarrow \quad Z=\frac{1}{3}[i(9 i-2)-2 i(6 i-6)+1(2 i-9 i)] \\ & \Rightarrow \quad Z=\frac{1}{3}(3+3 i)=1+i \\ & \Rightarrow \quad \text{ Modulus\text{ of } } z=\sqrt{1^2+1^2}=\sqrt{2}\end{aligned}$$
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