If $|A|$  denotes the value of the determinant of the square matrix $A$ of order $3$ , then $ |-2A|=$

Let $a,b,c$ be positive real numbers. The following system of equations in $x, y$  and $ z $ $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} – \frac{{{z^2}}}{{{c^2}}} = 1$, $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1, – \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$ has

If $A = \left[ {\begin{array}{*{20}{c}}
1&1\\
1&1
\end{array}} \right]$ and $\det ({A^n} – I) = 1 – {\lambda ^n}\,,\,n \in N$ then $\lambda $ is

If the system of linear equations $2 x+3 y-z=-2$  ; $x+y+z=4$  ; $x-y+|\lambda| z=4 \lambda-4$  (where $\lambda \in R$), has no solution, then

The value of $\lambda $ for which the system of equations $2x – y – z = 12,$ $x – 2y + z = – 4,$ $x + y + \lambda z = 4$ has no solution is