For the system of linear equations $a x+y+z=1$, $x+a y+z=1, x+y+a z=\beta$, which one of the following statements is NOT correct ?
It has infinitely many solutions if $\alpha=2$ and $\beta=-1$
It has no solution if $\alpha=-2$ and $\beta=1$
$x+y+z=\frac{3}{4}$ if $\alpha=2$ and $\beta=1$
It has infinitely many solutions if $\alpha=1$ and $\beta=1$
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
How many values of $k $ , systeam of linear equations $\left( {k + 1} \right)x + 8y = 4k\;,\;kx + \left( {k + 3} \right)y$$ = 3k - 1$ has no solutions.
If $\left| {\,\begin{array}{*{20}{c}}{x + 1}&3&5\\2&{x + 2}&5\\2&3&{x + 4}\end{array}\,} \right| = 0$, then $ x =$
If $a_i^2 + b_i^2 + c_i^2 = 1,\,i = 1,2,3$ and $a_ia_j + b_ib_j +c_ic_j = 0$ $\left( {i \ne j,i,j = 1,2,3} \right)$ then the value of determinant $\left| {\begin{array}{*{20}{c}}
{{a_1}}&{{a_2}}&{{a_3}} \\
{{b_1}}&{{b_2}}&{{b_3}} \\
{{c_1}}&{{c_2}}&{{c_3}}
\end{array}} \right|$ is
The value of the determinant $\left| {\,\begin{array}{*{20}{c}}{10!}&{11!}&{12!}\\{11!}&{12!}&{13!}\\{12!}&{13!}&{14!}\end{array}\,} \right|$ is