$\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\\1&{1 + y}&1\\1&1&{1 + z}\end{array}\,} \right| = $

  • A

    $xyz\left( {1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)$

  • B

    $xyz$

  • C

    $1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$

  • D

    $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$

Similar Questions

$\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$ માટે ગુણધર્મ $2$ ની ચકાસણી કરો. 

ધારો કે  $a-2 b+c=1$ છે . જો $f(x)=\left|\begin{array}{lll}{x+a} & {x+2} & {x+1} \\ {x+b} & {x+3} & {x+2} \\ {x+c} & {x+4} & {x+3}\end{array}\right|,$ હોય તો  . . . 

  • [JEE MAIN 2020]

$\left| {\,\begin{array}{*{20}{c}}0&{{b^3} - {a^3}}&{{c^3} - {a^3}}\\{{a^3} - {b^3}}&0&{{c^3} - {b^3}}\\{{a^3} - {c^3}}&{{b^3} - {c^3}}&0\end{array}\,} \right| = . . $

$\left| {\,\begin{array}{*{20}{c}}{b + c}& a& a\\b& {c + a}& b\\c& c& {a + b}\end{array}\,} \right| = $

જો $a,b,c$ એ અસમાન હોય તો $\Delta = \left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3} + 1}\\b&{{b^2}}&{{b^3} + 1}\\c&{{c^2}}&{{c^3} + 1}\end{array}\,} \right|= 0$ માટે . . . .શરતનું પાલન થવું જોઈએ.

  • [IIT 1985]