The system of equations $\begin{array}{l}\alpha x + y + z = \alpha - 1\\x + \alpha y + z = \alpha - 1\\x + y + \alpha z = \alpha - 1\end{array}$ has no solution, if $\alpha $ is
Not $ -2$
$1$
$-2$
Either $ -2$ or $ 1$
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 :
$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2} - bc}\\1&b&{{b^2} - ac}\\1&c&{{c^2} - ab}\end{array}\,} \right| = $
$\left| {\,\begin{array}{*{20}{c}}1&a&b\\{ - a}&1&c\\{ - b}&{ - c}&1\end{array}\,} \right| = $
Consider the system of linear equation $x+y+z=$ $4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda^2 z=\mu^2+15$, where $\lambda, \mu \in R$. Which one of the following statements is $NOT$ correct?
Let $S=\left\{A=\left(\begin{array}{lll}0 & 1 & c \\ 1 & a & d \\ 1 & b & e\end{array}\right): a, b, c, d, e \in\{0,1\}\right.$ and $\left.|A| \in\{-1,1\}\right\}$, where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is. . . . .