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GUJCET · Maths · Determinants
If \(\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1+y & 1+2 y & 1 \\ 1+z & 1+z & 1+3 z\end{array}\right|\) \(=10 k x y z\left(3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\) then,
k = ____________ \(\left(\right.\) where \(\left., x y z \neq 0,3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \neq 0\right)\)
- A \(\frac{1}{5}\)
- B 2
- C 5
- D 1
Answer & Solution
Correct Answer
(A) \(\frac{1}{5}\)
Step-by-step Solution
Detailed explanation
\(\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1+y & 1+2 y & 1 \\ 1+z & 1+z & 1+3 z\end{array}\right|\) \(\xrightarrow{C_1 \to C_1-C_3, C_2 \to C_2-C_3}\) \(\left|\begin{array}{ccc}x & 0 & 1 \\ y & 2y & 1 \\ -2z & -2z & 1+3z\end{array}\right|\) \(x(2y(1+3z) - (-2z)) - 0 + 1(-2yz - 2y(-2z))\)
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