If $\Delta = \left| {\,\begin{array}{*{20}{c}}a&b&c\\x&y&z\\p&q&r\end{array}\,} \right|$, then $\left| {\,\begin{array}{*{20}{c}}{ka}&{kb}&{kc}\\{kx}&{ky}&{kz}\\{kp}&{kq}&{kr}\end{array}\,} \right|$=
$\Delta $
$k\Delta $
$3k\Delta $
${k^3}\Delta $
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
If $a,b,c$ are positive integers, then the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}{{a^2} + x}&{ab}&{ac}\\{ab}&{{b^2} + x}&{bc}\\{ac}&{bc}&{{c^2} + x}\end{array}\,} \right|$ is divisible by
Using the property of determinants and without expanding, prove that:
$\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0$
If $\left| {\begin{array}{*{20}{c}} {a - b}&{b - c}&{c - a} \\ {b - c}&{c - a}&{a - b} \\ {c - a + 1}&{a - b}&{b - c} \end{array}} \right| = 0$ ,$\left( {a,b,c \in R - \left\{ 0 \right\}} \right),$ then
The value of $\left|\begin{array}{lll}(a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1\end{array}\right|$ is