Question

If \( D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2+x & 1 \\ 1 & 1 & 2+y \end{vmatrix} \) for \( x \neq 0, y \neq 0 \), then D is

• A. Divisible by x and y
• B. Divisible by x but not by y
• C. Divisible by (1+x) and (1+y)
• D. Divisible by (1+x) but not (1+y)

Hint:

When evaluating determinants, always use row or column operations to create as many zeros as possible. The goal is to get a triangular matrix, as its determinant is simply the product of the diagonal entries, which makes the calculation trivial and reveals the factors immediately.

Answer 1

The Correct Option is C

We are asked to find the factors of the determinant D. The determinant is given by: \[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2+x & 1 \\ 1 & 1 & 2+y \end{vmatrix} \] To simplify, we use column operations. Let's apply \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \). This will create zeros in the first row. \[ D = \begin{vmatrix} 1 & 1-1 & 1-1 \\ 1 & (2+x)-1 & 1-1 \\ 1 & 1-1 & (2+y)-1 \end{vmatrix} \] This simplifies to: \[ D = \begin{vmatrix} 1 & 0 & 0 \\ 1 & 1+x & 0 \\ 1 & 0 & 1+y \end{vmatrix} \] This is a lower triangular matrix. The determinant of a triangular matrix is the product of its main diagonal elements. \[ D = 1 \cdot (1+x) \cdot (1+y) = (1+x)(1+y) \] Since the value of the determinant D is \( (1+x)(1+y) \), it is divisible by both \( (1+x) \) and \( (1+y) \). This corresponds to option (C), which matches the provided answer key. My previous analysis of this problem was incorrect.