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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. - Mathematics

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Sum

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  R→ - R1

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

⇒ (3a - x )(4x2) = 0

⇒ x = 0 or 3a.

Concept: Properties of Determinants
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