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UPSC NDA I 2025 Mathematics Paper

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Question : 11 of 120

Marks: +1, -0

If

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

, then what is the value of the following?

\begin{vmatrix} 1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix}

-1
0
1
2

Solution:

When

A ^2+ B ^2+ C ^2=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 & \cos 0 & \cos 0 \\ \cos 0 & 1 & \cos 0 \\ \cos 0 & \cos 0 & 1 \end{vmatrix}

Since,

\cos 0 = 1

Thus Matrix becomes

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

Now determinant

=1[(1 \times 1-1 \times 1)]-1[(1 \times 1-1 \times 1)]+1[(1 \times 1-1 \times 1)]

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

\therefore

The value of the determinant is 0 .

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