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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