D = | array lll a & 1 & 1 1 & b & 1 1 & 1 & c array | = 0, then the value of 1 1-a + 1…
Mathematics · JEE Advanced · NTA Exams — Matrices and Determinants
D = \[\left|\begin{array}{lll}
a & 1 & 1 \\
1 & b & 1 \\
1 & 1 & c
\end{array}\right|\] = 0, then the value of \(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\) is :
- – 1
- 0
- 1
- none of these
Answer
(C) 1
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