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3 and 4 .Determinants and Matrices
easy
सारणिक $\left| {\,\begin{array}{*{20}{c}}{10!}&{11!}&{12!}\\{11!}&{12!}&{13!}\\{12!}&{13!}&{14!}\end{array}\,} \right|$ का मान होगा
A
$2\,(10!\,\,11!)$
B
$2\,(10\,!\,\,13\,!)$
C
$2\,(10!\,\,11!\,\,12!)$
D
$2\,(11\,!\,\,12!\,\,13!)$
Solution
(c) $\left| {\,\begin{array}{*{20}{c}}{10!}&{11!}&{12!}\\{11!}&{12!}&{13!}\\{12!}&{13!}&{14!}\end{array}\,} \right|\, = 10!\,\,\,11!\,\,\,12!\,\,\,\,\left| {\,\begin{array}{*{20}{c}}1&{11}&{11 \times 12}\\1&{12}&{12 \times 13}\\1&{13}&{13 \times 14}\end{array}\,} \right|$
संक्रिया ${R_2} \to {R_2} – {R_1}$तथा$\,{R_3} \to {R_3} – {R_1}$ से,
$10!\,11!\,12!\,\left| {\,\begin{array}{*{20}{c}}1&{11}&{11 \times 12}\\0&1&{24}\\0&2&{50}\end{array}\,} \right| = 2(10!\,\,\,11!\,\,\,12!)$.
Standard 12
Mathematics
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