If \( D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2+x & 1 \\ 1 & 1 & 2+y \end{vmatrix} \), for \( x \neq 0 \) and \( y \neq 0 \), then \( D \) is:
Step-by-step Solution:
The given determinant is: \[ \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2+x & 1 \\ 1 & 1 & 2+y \end{array} \right|. \] Using the row operations \( C_1 \rightarrow C_1 - C_2 \) and \( C_2 \rightarrow C_2 - C_3 \), we get: The value of the determinant is: \[ (1+x)(1+y), \] hence it is divisible by both \( (1+x) \) and \( (1+y) \).