The existance of the unique solution of the system of equations$2x + y + z = \beta $ , $10x - y + \alpha z = 10$ and $4x+ 3y-z =6$ depends on

Evaluate the determinants : $\left|\begin{array}{cc}2 & 4 \\ -5 & -1\end{array}\right|$

The system of equations $x + y + z = 6$, $x + 2y + 3z = 10,x + 2y + \lambda z = \mu $, has no solution for

If $a,b,c$ be positive and not all equal, then the value of the determinant $\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right|$ is

$2x + 3y + 4z = 9$,$4x + 9y + 3z = 10,$$5x + 10y + 5z = 11$ then the value of $ x$ is

Let $M$ and $N$ be two $3 \times 3$ matrices such that $M N=N M$. Further, if $M \neq N^2$ and $M^2=N^4$, then

$(A)$ determinant of $\left( M ^2+ MN ^2\right)$ is $0$

$(B)$ there is a $3 \times 3$ non-zero matrix $U$ such that $\left( M ^2+ MN ^2\right) U$ is the zero matrix

$(C)$ determinant of $\left( M ^2+ MN ^2\right) \geq 1$

$(D)$ for a $3 \times 3$ matrix $U$, if $\left( M ^2+ MN ^2\right) U$ equals the zero matrix then $U$ is the zero matrix