$\left|\begin{array}{ccc}1 & bc+ad & b^2c^2+a^2d^2 \\ 1 & ca+bd & c^2a^2+b^2d^2 \\ 1 & ab+cd & a^2b^2+c^2d^2\end{array}\right|=$

  • A
    $(a-b)(b-c)(c-d)(a-d)(a-c)(d-b)$
  • B
    $(a-b)(a-c)(b-c)(b-d)(a-d)(c-d)$
  • C
    $(a-b)(a-c)(a-d)(b-c)(b-d)(d-c)$
  • D
    $(a-b)(b-c)(c-d)(b-d)$

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Similar Questions

$\left| {\begin{array}{*{20}{c}} 1 & 5 & \pi \\ {{\log }_e}e & 5 & {\sqrt 5 } \\ {{\log }_{10}}10 & 5 & e \end{array}} \right| = $

यदि $\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ a & b & c \\ l & m & n\end{array}\right|=(-1)^K\left|\begin{array}{ccc}m & n & l \\ b & c & a \\ \beta & \gamma & \alpha\end{array}\right|$, तो $K$ का न्यूनतम मान है

सारणिकों के गुणधर्मों का उपयोग करके और बिना विस्तार किए सिद्ध कीजिए कि:
$\left|\begin{array}{lll}1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b)\end{array}\right|=0$

सारणिकों के गुणों का उपयोग करके सिद्ध कीजिए कि:
$\left|\begin{array}{ccc}\alpha & \alpha^{2} & \beta+\gamma \\ \beta & \beta^{2} & \gamma+\alpha \\ \gamma & \gamma^{2} & \alpha+\beta\end{array}\right|=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$

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View Solution

$\left| \begin{array}{ccc} 13 & 16 & 19 \\ 14 & 17 & 20 \\ 15 & 18 & 21 \end{array} \right| = $

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