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Question : 11 of 100
Marks:
+1,
-0
Solution:
‌||=0‌C2→C2−C1,C3→C3−C1 || a | 1−a | 1−a |
| 1 | b−1 | 0 |
| 1 | 0 | c−1 |
|=0 ‌a(b−1)(c−1)−(1−a)(c−1)+(1−a)(1−b)=0‌a(1−b)(1−c)+(1−a)(1−c)+(1−a)(1−b)=0‌‌+‌+‌=0‌⇒−1+‌+‌+‌=0‌⇒‌+‌+‌=1
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