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UPSC NDA I 2015 Mathematics Paper

Question : 24 of 120

Marks: +1, -0

The value of

\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{vmatrix}

Solution:

Concept:

Elementary row operations do not change the value of a determinant.

Explanation:

Start with the given determinant:

\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{vmatrix}

Apply row operations:

R_2 \to R_2 - R_1

and

R_3 \to R_3 - R_1

This gives:

\begin{vmatrix} 1 & 1 & 1 \\ 0 & x & 0 \\ 0 & 0 & y \end{vmatrix}

Now expand along the first column

C_1

1 \cdot (x \cdot y - 0 \cdot 0) - 0 + 0 = xy

Thus the determinant simplifies directly to

xy

Answer:

xy

(Option C).

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