If $D =$ $\left| {\,\begin{array}{*{20}{c}}{\frac{1}{z}}&{\frac{1}{z}}&{ - \frac{{(x + y)}}{{{z^2}}}}\\{ - \frac{{(y + z)}}{{{x^2}}}}&{\frac{1}{x}}&{\frac{1}{x}}\\{ - \frac{{y(y + z)}}{{{x^2}z}}}&{\frac{{x + 2y + z}}{{xz}}}&{ - \frac{{y(x +y)}}{{x{z^2}}}}\end{array}\,} \right|$ then, the incorrect statement is

  • A

    $D$ is independent of $x$

  • B

    $D$ is independent of $y$

  • C

    $D$ is independent of $z$

  • D

    $D$ is dependent on $x, y, z$

Similar Questions

Prove that $\left|\begin{array}{ccc}a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$

If $a, b, c > 0 \, and \, x, y, z \in R$ , then the determinant $\left|{\begin{array}{*{20}{c}}{{{\left( {{a^x}\, + \,\,{a^{ - x}}} \right)}^2}}&{{{\left( {{a^x}\, - \,\,{a^{ - x}}} \right)}^2}}&1\\{{{\left( {{b^y}\, + \,\,{b^{ - y}}} \right)}^2}}&{{{\left( {{b^y}\, - \,\,{b^{ - y}}} \right)}^2}}&1\\{{{\left( {{c^z}\, + \,\,{c^{ - z}}} \right)}^2}}&{{{\left( {{c^z}\, - \,\,{c^{ - z}}} \right)}^2}}&1\end{array}} \right|$ $=$

Prove that $\left|\begin{array}{ccc}b+c & a & a \\ b & c+a & b \\ c & c & a+b\end{array}\right|=4 a b c$

$\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}} \right| = 0$ then $\sin \,4\theta $ equal to

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}4&{ - 6}&1\\{ - 1}&{ - 1}&1\\{ - 4}&{11}&{ - 1\,}\end{array}} \right|$is