The value of the determinant \(\begin{vmatrix}x + 2 & x + 3 & x + 5\\ x + 4 & x + 6 & x + 9\\ x + 8 & x + 11 & x + 15\end{vmatrix}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-2\)
Subtracting a multiple of one row from another row does not change the value of a determinant.
Applying \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\):
\( \displaystyle \Delta = \begin{vmatrix}x + 2 & x + 3 & x + 5\\ 2 & 3 & 4\\ 6 & 8 & 10\end{vmatrix} \)
Applying \(R_3 \to R_3 - 3R_2\):
\( \displaystyle \Delta = \begin{vmatrix}x + 2 & x + 3 & x + 5\\ 2 & 3 & 4\\ 0 & -1 & -2\end{vmatrix} \)
Expanding along the first row:
\(\Delta = (x + 2)(-6 + 4) - (x + 3)(-4 - 0) + (x + 5)(-2 - 0)\)
\(\Delta = -2x - 4 + 4x + 12 - 2x - 10\)
\(\Delta = -2\)
The terms in x cancel, so the value is \(-2\) for every x.