If \( \begin{vmatrix}2+x&y&z\\x&2+y&z\\x&y&2+z\end{vmatrix}=0 \), then the value of \( x+y+z \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → −2
Adding all three columns to the first (\( C_{1}\to C_{1}+C_{2}+C_{3} \)) makes every entry of the first column \( 2+x+y+z \), which can be taken out as a common factor:
\( \displaystyle \Delta=(2+x+y+z)\begin{vmatrix}1&y&z\\1&2+y&z\\1&y&2+z\end{vmatrix} \)
With \( R_{2}\to R_{2}-R_{1} \) and \( R_{3}\to R_{3}-R_{1} \):
\( \displaystyle \Delta=(2+x+y+z)\begin{vmatrix}1&y&z\\0&2&0\\0&0&2\end{vmatrix} \)
\( =4(2+x+y+z) \)
Since \( \Delta=0 \),
\( x+y+z=-2 \)