If matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,then $|A|^{-1}$ is equal to

Given that, $a \alpha^2+2 b \alpha+c \neq 0$ and that the system of equations
$\begin{aligned} & (a \alpha+b) x+a y+b z=0 \\ & (b \alpha+c) x+b y+c z=0 \\ & (a \alpha+b) y+(b \alpha+c) z=0\end{aligned}$
has a non-trivial solution, then $a, b$ and $c$ lie in

The value of $x$ such that the matrix $\begin{bmatrix} x & 2 & 3 \\ 4 & 5 & 6 \\ 2 & 3 & 5 \end{bmatrix}$ is not invertible is

If $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$ and $|A^{3}| = 125$,then $\alpha$ is equal to

Let $P$ be a matrix of order $3 \times 3$ such that all the entries in $P$ are from the set $\{-1, 0, 1\}$. Then,the maximum possible value of the determinant of $P$ is:

$\left|\begin{array}{ccc}\sqrt{3} & 2 \sqrt{5} & \sqrt{5} \\ \sqrt{15} & 5 & \sqrt{10} \\ 3 & \sqrt{15} & 5\end{array}\right|=$