If $A$ is a square matrix of order $3 \times 3$,then $|KA|$ is equal to

If $A$ is a square matrix of order $n$ and $A = kB$,where $k$ is a scalar,then $|A|=$

$\left| \begin{array}{ccc} a - b & b - c & c - a \\ x - y & y - z & z - x \\ p - q & q - r & r - p \end{array} \right| = $

The value of $\left|\begin{array}{lll}1990 & 1991 & 1992 \\ 1991 & 1992 & 1993 \\ 1992 & 1993 & 1994\end{array}\right|$ is

By using properties of determinants,show that:
$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$

Let $A$ be a square matrix of order $3 \times 3$,then $|5A| = $ (in $|A|$)