Prove the given :-

Asked by  | 20th May, 2012, 10:21: PM

Expert Answer:

Apply C1 -> C1 - C3 and C2 -> C2 - C3
On simplifying the so obtained determinant, you will get,
              |(b+c-a)         0              a2|
(a+b+c)2 |0             (c+a-b)           b2|
              |(c-a-b)     (c-a-b)     (a+b)2|
Apply R3 -> R3 - (R1 + R2)
And then apply C1 -> aC1+ C3, C2 -> bC2 + C3
On simplification, you will get the required result.

Answered by  | 21st May, 2012, 09:54: PM

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