If ${a^{ - 1}} + {b^{ - 1}} + {c^{ - 1}} = 0$ such that $\left| {\,\begin{array}{*{20}{c}}{1 + a}&1&1\\1&{1 + b}&1\\1&1&{1 + c}\end{array}\,} \right| = \lambda $, then the value of $\lambda $is

If ${a^2} + {b^2} + {c^2} = - 2$ and $f(x) = \left| {\begin{array}{*{20}{c}}{1 + {a^2}x}&{(1 + {b^2})x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{1 + {b^2}x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{(1 + {b^2})x}&{1 + {c^2}x}\end{array}} \right|$ then $f(x)$ is a polynomial of degree

If $\Delta=\left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \\ 2 x-3 & 3 x-4 & 4 x-5 \\ 3 x-5 & 5 x-8 & 10 x-17\end{array}\right|=$ $Ax ^{3}+ Bx ^{2}+ Cx + D ,$ then $B + C$ is equal to

$\left| {\,\begin{array}{*{20}{c}}{b + c}& a& a\\b& {c + a}& b\\c& c& {a + b}\end{array}\,} \right| = $

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&1&1\\{b + c}&{c + a}&{a + b}\\{b + c - a}&{c + a - b}&{a + b - c}\end{array}\,} \right|$ is

Prove that

$\Delta=\left|\begin{array}{ccc}
a+b x & c+d x & p+q x \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|=\left(1-x^{2}\right)\left|\begin{array}{lll}
a & c & p \\
b & d & q \\
u & v & m
\end{array}\right|$