DeterminantHard
Question
If a ≠ b, then the system of equations ax + by + bz = 0, bx + ay + bz = 0, bx + by + az = 0 will have a non-trivial solution if
Options
A.a + b = 0
B.a + 2b = 0
C.2a + b = 0
D.a + 4b = 0
Solution
Here ᐃ = 
for non-trivial solution if ᐃ = 0.
C1 → C1 + C2 + C3
(a + 2b)
= 0
R2 → R2 - R1 & R3 → R3 - R1
⇒ (a + 2b)
= 0
⇒ (a + 2b) (a - b)2 = 0
Here a ≠ b ∴ (a + 2b) = 0
for non-trivial solution if ᐃ = 0.
C1 → C1 + C2 + C3
(a + 2b)
= 0R2 → R2 - R1 & R3 → R3 - R1
⇒ (a + 2b)
= 0⇒ (a + 2b) (a - b)2 = 0
Here a ≠ b ∴ (a + 2b) = 0
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