The value of \( \begin{vmatrix}x & a & a\\a & x & a\\a & a & x\end{vmatrix} \) is equal to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( (x+2a)(x-a)^{2} \)
Each row adds up to x + 2a, so \( C_1\to C_1+C_2+C_3 \) gives a common factor in the first column:
\( \displaystyle \Delta=(x+2a)\begin{vmatrix}1 & a & a\\1 & x & a\\1 & a & x\end{vmatrix} \)
\( R_2\to R_2-R_1 \) and \( R_3\to R_3-R_1 \) create zeros in the first column:
\( \displaystyle \Delta=(x+2a)\begin{vmatrix}1 & a & a\\0 & x-a & 0\\0 & 0 & x-a\end{vmatrix} \)
The determinant is now upper triangular, so it equals the product of the diagonal:
\( \displaystyle \Delta=(x+2a)(x-a)^2 \)
Shortcut: Put x = a: all rows become equal, so \( \Delta=0 \), and put a = 0: \( \Delta=x^3 \). Options (3) and (4) are not zero at x = a, and Option (1) gives \( x^2 \) at a = 0; only Option (2) passes both checks.