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UPSC NDA II 2024 Mathematics Paper

Question : 9 of 120

Marks: +1, -0

\rm z=\frac{1}{3}\begin{vmatrix}i&2i&1\\\ 2i&3i&2\\\ 3&1&3\end{vmatrix}=x+iy;i=\sqrt{-1}

then what is modulus of Z equal to?

Solution:

Concept:

Evaluating a determinant with complex entries and finding the modulus of the resulting complex number.

Explanation:

Compute the determinant:

\rm \begin{vmatrix} i & 2i & 1 \\ 2i & 3i & 2 \\ 3 & 1 & 3 \end{vmatrix}

Expand along the first row:

i(3i \cdot 3 - 2 \cdot 1) - 2i(2i \cdot 3 - 2 \cdot 3) + 1(2i \cdot 1 - 3i \cdot 3)

Simplify each term:

i(9i - 2) = 9i^2 - 2i = -9 - 2i

;

-2i(6i - 6) = -12i^2 + 12i = 12 + 12i

;

1(2i - 9i) = -7i

Sum:

(-9 - 2i) + (12 + 12i) + (-7i) = 3 + 3i

Thus

\rm z = \frac{1}{3}(3 + 3i) = 1 + i

Modulus:

|z| = \sqrt{1^2 + 1^2} = \sqrt{2}

Answer:

\sqrt{2}

(Option B).

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