Suppose that A, B and C are matrices of order $m \times n$, $n \times 5$ and $5 \times q$ respectively. The restriction on $m$, $n$ and $q$ so that AB-BC is defined are

If the system of linear equations

$x + 2y - 3z = 1$

$(2p+1)y + z = 2$

$3x + 3z = 5$

has a unique solution, then p can not be equal to

If $A = \begin{bmatrix} 2 & 1 & -1 \\ 0 & 3 & -2 \\ k & 1 & 0 \end{bmatrix}$ is non-singular matrix, then the value(s) of $k$ belongs to:

[Where $R$ is the set of real numbers]

For what value of $k$, the following system have a unique solution? (where $\mathbb{R}$ is set of real numbers)

$x + y + z = 1$

$2x + 3y + 4z = 3$

$x - y + kz = 5$

If the matrix $A = \begin{bmatrix} \alpha & \beta & \gamma \\ 0 & 0 & 2 \\ 3 & -2 & 0 \end{bmatrix}$ is a skew symmetric matrix, then the value of $(\alpha + \beta + \gamma)^2$ is:

If $x = -4$ is a root of $\begin{vmatrix}x & 2 & 3 \\ 1 & x & 1 \\ 3 & 2 & x\end{vmatrix} = 0$, then the sum of the other 2 roots is

If A is a matrix of order $3 \times 3$ such that $|A| = 4$ and $|k \cdot adj\, A| = 32|A|, k \in \mathbb{R}$; then the value of k is: