$\left| {\,\begin{array}{*{20}{c}}4&{ - 6}&1\\{ - 1}&{ - 1}&1\\{ - 4}&{11}&{ - 1\,}\end{array}} \right|= . . . $
$\left| {\,\begin{array}{*{20}{c}}4&{ - 6}&1\\{ - 1}&{ - 1}&1\\{ - 4}&{11}&{ - 1\,}\end{array}} \right|= . . . $
- A
$-75$
- B
$25$
- C
$0$
- D
$-25$
Similar Questions
$\left| {\,\begin{array}{*{20}{c}}{a + b}&{b + c}&{c + a}\\{b + c}&{c + a}&{a + b}\\{c + a}&{a + b}&{b + c}\end{array}\,} \right| = K\,\,\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right|\,,$ તો $K = $
$\left| {\,\begin{array}{*{20}{c}}{a + b}&{b + c}&{c + a}\\{b + c}&{c + a}&{a + b}\\{c + a}&{a + b}&{b + c}\end{array}\,} \right| = K\,\,\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right|\,,$ તો $K = $
$\left| {\,\begin{array}{*{20}{c}}1&1&1\\{bc}&{ca}&{ab}\\{b + c}&{c + a}&{a + b}\end{array}\,} \right|$ =
$\left| {\,\begin{array}{*{20}{c}}1&1&1\\{bc}&{ca}&{ab}\\{b + c}&{c + a}&{a + b}\end{array}\,} \right|$ =
$\left| {\,\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}\\{ab}&{{b^2} + {x^2}}&{bc}\\{ca}&{bc}&{{c^2} + {x^2}}\end{array}\,} \right|$ એ . . . વડે વિભાજ્ય છે .
$\left| {\,\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}\\{ab}&{{b^2} + {x^2}}&{bc}\\{ca}&{bc}&{{c^2} + {x^2}}\end{array}\,} \right|$ એ . . . વડે વિભાજ્ય છે .
નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{ccc}x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y\end{array}\right|=2(x+y+z)^{3}$
નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{ccc}x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y\end{array}\right|=2(x+y+z)^{3}$
જો ${U_n} = \left| {\,\begin{array}{*{20}{c}}n&1&5\\{{n^2}}&{2N + 1}&{2N + 1}\\{{n^3}}&{3{N^2}}&{3N}\end{array}\,} \right|$ તો $\sum\limits_{n = 1}^N {{U_n},} $ મેળવો.
જો ${U_n} = \left| {\,\begin{array}{*{20}{c}}n&1&5\\{{n^2}}&{2N + 1}&{2N + 1}\\{{n^3}}&{3{N^2}}&{3N}\end{array}\,} \right|$ તો $\sum\limits_{n = 1}^N {{U_n},} $ મેળવો.