The value of the determinant \(\begin{vmatrix}1 & x & y\\ 1 & x + y & y\\ 1 & x & x + 2y\end{vmatrix}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(y(x + y)\)
The first column is all ones, so subtracting the first row from the others creates zeros. Applying \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\):
\( \displaystyle \Delta = \begin{vmatrix}1 & x & y\\ 0 & y & 0\\ 0 & 0 & x + y\end{vmatrix} \)
This is a triangular determinant, so it equals the product of the diagonal entries:
\(\Delta = 1 \cdot y \cdot (x + y)\)
\(\Delta = y(x + y)\)