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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.
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Solution
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.
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