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Question : 100 of 160
Marks:
+1,
-0
Solution:
Let
∆=[| a−b−c | 2a | 2a |
| 2b | b−c−a | 2b |
| 2c | 2c | c−a−b |
]Applying
R1→R1+R2+R3 and taking common
(a+b+c) from
R1=(a+b+c)|| 1 | 1 | 1 |
| 2b | b−c−a | 2b |
| 2c | 2c | c−a−b |
|Applying
C2→C2−C1 and
C3→C3−C1,
‌=(a+b+c)|| 1 | 0 | 0 |
| 2b | −b−c−a | 0 |
| 2c | 0 | −a−b−c |
|‌=(a+b+c)[(−b−c−a)(−a−b−c)]‌=(a+b+c)3
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