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Question : 178 of 180
Marks:
+1,
-0
Solution:
Let
Δ=|| 1 | 1 | 1 |
| a | b | c |
| a2−bc | b2−ca | c2−ab |
| On applying
C3→C3−C2 and
C2→C2−C1, we get
Δ=||
| 1 | 0 | 0 |
| a | b−a | c−b |
| 0 | (b2−a2) | (c2−b2) |
| a2−bc | +c(b−a) | +a(c−b) |
| =1[(b−a)(c−b)(c+b+a)−(c−b)(b−a)(b+a+c)]=0
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