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UPSC NDA I 2025 Mathematics Paper

Question : 11 of 120

Marks: +1, -0

A^{2}+B^{2}+C^{2}=0

\begin{vmatrix} 1 & cosC & cosB \\ cosC & 1 & cosA \\ cosB & cosA & 1 \end{vmatrix}

Solution:

Concept:

Given

A^2 + B^2 + C^2 = 0

, each square is non‑negative, so all must be zero:

A = B = C = 0

Explanation:

Substitute

A = B = C = 0

into the determinant.

The matrix becomes

\begin{vmatrix} 1 & \cos 0 & \cos 0 \\ \cos 0 & 1 & \cos 0 \\ \cos 0 & \cos 0 & 1 \end{vmatrix}

Since

\cos 0 = 1

, the matrix is all ones:

\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix}

Compute the determinant:

1(1\cdot1 - 1\cdot1) - 1(1\cdot1 - 1\cdot1) + 1(1\cdot1 - 1\cdot1) = 1(0) - 1(0) + 1(0) = 0

Answer:

The value of the determinant is

0

, so the correct option is B.

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