Using properties of determinants prove that: |(x,a,a),(a,x,a),(a,a,x) = (x +2a)(x-a)^2

Question

Using properties of determinants prove that:

\(\begin{vmatrix} x&a & a \\[0.3em] a & x & a \\[0.3em] a & a & x \end{vmatrix}\) = (x+2a)(x -a)²

Answer 1

\(\begin{vmatrix} x&a & a \\[0.3em] a & x & a \\[0.3em] a & a & x \end{vmatrix}\)

= \(\begin{vmatrix} x+2a &x+2a &x+ 2a \\[0.3em] a & x & a \\[0.3em] a & a & x \end{vmatrix}\) [R₁’ = R₁ + R₂ + R₃]

= (x + 2a)(x - a)[x - ( - a) + ( - a - 0) + ( - a)] [expansion by first row] 

= (x + 2a)(x - a)(x + a - a - a) = (x + 2a)(x - a)²