If $\left[\begin{array}{ccc}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]$ has no inverse,then the real value of $x$ is

If ${a^2} + {b^2} + {c^2} + ab + bc + ca \leq 0$ for all $a, b, c \in R$,then find the value of the determinant $\left| {\begin{array}{*{20}{c}} {{(a + b + c)}^2} & {{a^2} + {b^2}} & 1 \\ 1 & {{(b + c + 2)}^2} & {{b^2} + {c^2}} \\ {{c^2} + {a^2}} & 1 & {{(c + a + 2)}^2} \end{array}} \right|$.

If $k > 1$ and the determinant of the matrix $A^2$,where $A = \begin{bmatrix} k & k\alpha & \alpha \\ 0 & \alpha & k\alpha \\ 0 & 0 & k \end{bmatrix}$,is $k^2$,then $|\alpha|$ is equal to

If $x^2+y^2+z^2 \neq 0, \quad x=cy+bz, \quad y=az+cx$ and $z=bx+ay$,then $a^2+b^2+c^2+2abc$ is equal to

If $\left| {\begin{array}{*{20}{c}}{y + z}&{x - z}&{x - y}\\{y - z}&{z + x}&{y - x}\\{z - y}&{z - x}&{x + y}\end{array}} \right| = kxyz$,then the value of $k$ is

If $a, b$ and $c$ are real numbers,and $\Delta=\begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}=0$,show that either $a+b+c=0$ or $a=b=c$.