If \(x, y\) and \(z\) are non-zero distinct numbers, then \(\begin{vmatrix}x+y&y+z&z+x\\z&x&y\\1&1&1\end{vmatrix}\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0
Let \(\Delta\) be the determinant. Applying the row operation \(R_1 \to R_1 + R_2\):
\(\displaystyle \Delta = \begin{vmatrix}x+y+z&x+y+z&x+y+z\\z&x&y\\1&1&1\end{vmatrix}\)
Taking \(x + y + z\) common from \(R_1\):
\(\displaystyle \Delta = (x + y + z)\begin{vmatrix}1&1&1\\z&x&y\\1&1&1\end{vmatrix}\)
Now \(R_1\) and \(R_3\) are the same, and a determinant with two identical rows is 0. So \(\Delta = 0\).