If $A^{2} + B^{2} + C^{2} = 0$ , then what is the value of the following?

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

Question

Question

This question was previously asked in

NDA-I (Mathematics) Official Paper (Held On: 13 Apr, 2025)

Answer 1

Concept:

When A² + B² + C² = 0, it implies A = B = C = 0 (since the squares of real numbers are non-negative).

Substitute the values of A, B, and C for determinant calculation into the matrix

Calculation:

$\begin{vmatrix} 1& cos0& cos0\\ cos0&1&cos0 \\ cos0&cos0&1 \end{vmatrix} $

Since, Cos0 =1

Thus Matrix becomes

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

Now determinant = 1[(1×1 - 1×1)] - 1[(1×1 - 1×1)] + 1[(1×1 - 1×1)]

= = 1(0) - 1(0) + 1(0) = 0

∴ The value of the determinant is 0.

Hence, the correct answer is Option 2.

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