The roots of the equation $\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\\1&{1 + x}&1\\1&1&{1 + x}\end{array}\,} \right| = 0$ are
$0, -3$
$0, 0, -3$
$0, 0, 0, -3$
None of these
$A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right],$ then show that $|3 A|=27|A|$.
If $\left| {\begin{array}{*{20}{c}}{a\, + \,1}&{a\, + \,2}&{a\, + \,p}\\{a\, + \,2}&{a\, +\,3}&{a\, + \,q}\\{a\, + \,3}&{a\, + \,4}&{a\, + \,r}\end{array}} \right|$ $= 0$ , then $p, q, r$ are in :
The value of the determinant $\left| {\,\begin{array}{*{20}{c}}{10!}&{11!}&{12!}\\{11!}&{12!}&{13!}\\{12!}&{13!}&{14!}\end{array}\,} \right|$ is
If the system of equations, $a^2 x - ay = 1 - a$ & $bx + (3 - 2b) y = 3 + a$ possess a unique solution $x = 1, y = 1$ then :
If a system of the equation ${(\alpha + 1)^3}x + {(\alpha + 2)^3}y - {(\alpha + 3)^3} = 0$ and $(\alpha + 1)x + (\alpha + 2)y - (\alpha + 3) = 0,x + y - 1 = 0$ is constant. what is the value of $\alpha $