The value of the determinant \(\begin{vmatrix}x + 3 & x & x\\ x & x + 3 & x\\ x & x & x + 3\end{vmatrix}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(27(x + 1)\)
Every row has the same sum \(3x + 3\). Applying \(C_1 \to C_1 + C_2 + C_3\) and taking \(3x + 3\) out of \(C_1\):
\( \displaystyle \Delta = (3x + 3)\begin{vmatrix}1 & x & x\\ 1 & x + 3 & x\\ 1 & x & x + 3\end{vmatrix} \)
Applying \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\):
\( \displaystyle \Delta = (3x + 3)\begin{vmatrix}1 & x & x\\ 0 & 3 & 0\\ 0 & 0 & 3\end{vmatrix} \)
This is a triangular determinant, equal to the product of its diagonal entries:
\(\Delta = (3x + 3) \times 9\)
\(\Delta = 27(x + 1)\)
Check with \(x = 0\): the determinant is \(3 \times 3 \times 3 = 27\), and \(27(0 + 1) = 27\).