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The value of ∣xaaaxaaax∣\begin{vmatrix}x & a & a\\a & x & a\\a & a & x\end{vmatrix} is equal to:

Solution

✅ Correct Option: 2

Add all rows to R1R_1, giving a common factor (x+2a)(x+2a) in the first row.

Then R2→R2−aR1R_2\to R_2-aR_1 and R3→R3−aR1R_3\to R_3-aR_1 leave a triangular block with diagonal entries (x−a)(x-a) and (x−a)(x-a).

Hence the determinant is (x+2a)(x−a)2(x+2a)(x-a)^2.

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