If $f(x) = \left| \begin{array}{ccc} x-3 & 2x^2-18 & 3x^3-81 \\ x-5 & 2x^2-50 & 4x^3-500 \\ 1 & 2 & 3 \end{array} \right|$,then $f(1)f(3) + f(3)f(5) + f(5)f(1)$ is equal to

The solutions of $\operatorname{det}(A-\lambda I_2)=0$ are $4$ and $8$,where $A=\begin{bmatrix} 2 & 3 \\ x & y \end{bmatrix}$. Then:

The least value of the product $xyz$ for which the determinant $\left| \begin{array}{ccc} x & 1 & 1 \\ 1 & y & 1 \\ 1 & 1 & z \end{array} \right|$ is non-negative,is

$\begin{aligned} & \text { If }\left|\begin{array}{ccc}n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2 \\ (n+2)^2 & (n+3)^2 & (n+4)^2\end{array}\right|=\Delta \text { and } \\ & \left|\begin{array}{ccc}1 & -4 & 7 \\ -2 & 3 & -5 \\ 3 & x & -3\end{array}\right|=2 \Delta+1, \text { then } x=\end{aligned}$

The determinant $\left| \begin{array}{ccc} a & b & a - b \\ b & c & b - c \\ 2 & 1 & 0 \end{array} \right|$ is equal to zero if $a, b, c$ are in

If $\left| \begin{matrix} -6 & 1 & \lambda \\ 0 & 3 & 7 \\ -1 & 0 & 5 \end{matrix} \right| = 5948$,then $\lambda$ is: