If $a,b,c$ are in $A.P$., then the value of $\left| {\,\begin{array}{*{20}{c}}{x + 2}&{x + 3}&{x + a}\\{x + 4}&{x + 5}&{x + b}\\{x + 6}&{x + 7}&{x + c}\end{array}\,} \right|$ is

Let $P=\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{array}\right]$ and $I$ be the identity matrix of order $3$ . If $\left.Q=q_{i j}\right]$ is a matrix such that $P^{50}-Q=I$, then $\frac{q_{31}+q_{32}}{q_{21}}$ equals

By using properties of determinants, show that:

$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$

If $a,b,c$ are positive integers, then the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}{{a^2} + x}&{ab}&{ac}\\{ab}&{{b^2} + x}&{bc}\\{ac}&{bc}&{{c^2} + x}\end{array}\,} \right|$ is divisible by

Using properties of determinants, prove that:

$\left| {\begin{array}{*{20}{l}}
  {\sin \alpha }&{\cos \alpha }&{\cos (\alpha  + \delta )} \\ 
  {\sin \beta }&{\cos \beta }&{\cos (\beta  + \delta )} \\ 
  {\sin \gamma }&{\cos \gamma }&{\cos (\gamma  + \delta )} 
\end{array}} \right| = 0$

$\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}} \right| = 0$ then $\sin \,4\theta $ equal to