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begin vmatrix 1 a 1 1 1 1 1 b 1 1 1 1 1 c 1 1 1 1
Question:
$ \begin{vmatrix} 1+a & 1 & 1 & 1 \\ 1 & 1+b & 1 & 1 \\ 1 & 1 & 1+c & 1 \\ 1 & 1 & 1 & 1=d \end{vmatrix} $ is equal to
CUET (PG) - 2023
CUET (PG)
Updated On:
Mar 12, 2025
$ \frac{1}{abcd}(1+\frac1{a}+\frac1{b} + \frac1{c} + \frac1{d}) $
abcd(1+a+b+c+d)
$ abcd(1+\frac1{a}+\frac1{b} + \frac1{c} + \frac1{d}) $
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The Correct Option is
D
Solution and Explanation
The Correct answer is option (D) : $ abcd(1+\frac1{a}+\frac1{b} + \frac1{c} + \frac1{d}) $
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