Suppose $D = \left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right|$ and $D' = \left| {\,\begin{array}{*{20}{c}}{{a_1} + p{b_1}}&{{b_1} + q{c_1}}&{{c_1} + r{a_1}}\\{{a_2} + p{b_2}}&{{b_2} + q{c_2}}&{{c_2} + r{a_2}}\\{{a_3} + p{b_3}}&{{b_3} + q{c_3}}&{{c_3} + r{a_3}}\end{array}\,} \right|$, then
$D' = D$
$D' = D(1 - pqr)$
$D' = D(1 + p + q + r)$
$D' = D(1 + pqr)$
Let $A=\left(\begin{array}{ccc}{[x+1]} & {[x+2]} & {[x+3]} \\ {[x]} & {[x+3]} & {[x+3]} \\ {[x]} & {[x+2]} & {[x+4]}\end{array}\right),$ where $[t]$ denotes the greatest integer less than or equal to $\mathrm{t}$. If $\operatorname{det}(\mathrm{A})=192$, then the set of values of $\mathrm{x}$ is the interval
If the sum of squares of all real values of $\alpha$, for which the lines $2 x-y+3=0,6 x+3 y+1=0$ and $\alpha x+2 y-2=0$ do not form a triangle is $p$, then the greatest integer less than or equal to $\mathrm{p}$ is $.........$
Let $\mathrm{A}$ be a square matrix of order $3 \times 3$ , then $|\mathrm{k A}|$ is equal to
If the following system of linear equations
$2 x+y+z=5$
$x-y+z=3$
$x+y+a z=b$
has no solution, then :
If $\omega$ is one of the imaginary cube roots of unity, then the value of the determinant $\left| {\begin{array}{*{20}{c}}1&{{\omega ^3}}&{{\omega ^2}}\\ {{\omega ^3}}&1&\omega \\{{\omega ^2}}&\omega &1\end{array}} \right|$ $=$