Question

Solve for x : `|("a"+"x","a"-"x","a"-"x"),("a"-"x","a"+"x","a"-"x"),("a"-"x","a"-"x","a"+"x")| = 0`, using properties of determinants. 

Answer 1

We have `|("a"+"x","a"-"x","a"-"x"),("a"-"x","a"+"x","a"-"x"),("a"-"x","a"-"x","a"+"x")| = 0` 

By C₁ → C₁ + C₂ + C₃

⇒ `|(3"a"-"x", "a"-"x","a"-"x"),("3a"-"x","a"+"x","a"-"x"),("3a"-"x","a"-"x","a"+"x")| = 0` 

⇒ `(3"a" -"x") |(1,"a"-"x","a"-"x"),(1,"a"+"x","a"-"x"),(1,"a"-"x","a"+"x")| = 0`

By R₂ → R₂ - R₁ and  R₃ → - R₁

⇒ `(3"a" -"x") |(1,"a"-"x","a"-"x"),(0, 2x, 0),(0, 0,2x)| = 0`

⇒ (3a - x )(4x²) = 0

⇒ x = 0 or 3a.