Q. 334.5( 8 Votes )

Prove the followi

Answer :

Let


Multiplying a, b and c to R1, R2 and R3, we get




Dividing C1, C2 and C3 with a, b and c, we get



Recall that the value of a determinant remains same if we apply the operation Ri Ri + kRj or Ci Ci + kCj.


Applying R1 R1 – R2, we get




Applying R1 R1 – R3, we get




Applying C2 C2 – C1, we get




Applying C3 C3 – C1, we get




Expanding the determinant along R1, we have


Δ = 0 + (2c2)[(b2)(a2 + b2 – c2)] + (–2b2)[–(c2)(c2 + a2 – b2)]


Δ = 2b2c2(a2 + b2 – c2) + 2b2c2(c2 + a2 – b2)


Δ = 2b2c2 [(a2 + b2 – c2) + (c2 + a2 – b2)]


Δ = 2b2c2[2a2]


Δ = 4a2b2c2


Thus,


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