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

Let $f (x) =$ $\left| {\begin{array}{*{20}{c}}{1\, + \,{{\sin }^2}x}&{{{\cos }^2}x}&{4\,\sin \,2x}\\{{{\sin }^2}x}&{1\, + \,{{\cos }^2}x}&{4\,\sin \,2x}\\{{{\sin }^2}x}&{{{\cos }^2}x}&{1\, + \,4\,\sin \,2x}\end{array}} \right|$, then the maximum value of $f (x) =$

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

If $\mathrm{a, b, c},$ are in $\mathrm{A.P}$, then the determinant

$\left|\begin{array}{lll}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+2 b \\ x+4 & x+5 & x+2 c\end{array}\right|$ is

Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|=2\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$

The value of $\sum\limits_{n = 1}^N {{U_n},} $ if ${U_n} = \left| {\,\begin{array}{*{20}{c}}n&1&5\\{{n^2}}&{2N + 1}&{2N + 1}\\{{n^3}}&{3{N^2}}&{3N}\end{array}\,} \right|$ is